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Stream: learning: questions

Topic: Bimodules vs Spans: General solution

Eric Forgy (Jan 16 2021 at 18:32):

Eric Forgy said:

Hi :wave:

I'm coming back to this stream after putting a lot of elbow grease into:

Bimodules vs Spans

• Puzzle 1
• Puzzle 2
• Puzzle 3

There, we learned that in $(\mathsf{Set^{op}},\times,*)$:

• A monoid object $A$ is a set with multiplication given by the diagonal map.
• An $A$-module $X$ is a function $X\to A.$
• An $(A,B)$-module $X$ is a span $A\leftarrow X\rightarrow B.$
• Tensoring an $(A,B)$-bimodule $X$ with a $(B,C)$-bimodule $Y$ is an $(A,C)$-bimodule $X\times_B Y$ corresponding to the composition of the respective spans

Good morning :coffee:

I'm still trying to pull all this together.

I'll try to explain / generalize what I learned in my words to see If I understand.

Given a monoidal category $(C,\otimes,1)$:

• Every object in $C$ is a monoid object in $(C^\mathsf{op},\otimes, 1).$
• Every span $R\leftarrow M\rightarrow S$ in $C$ is an $(R,S)$-bimodule in $(C^\mathsf{op},\otimes, 1).$

Since every morphism $f:M\to S$ is equivalent to a span
${}$
$M\overset{1}{\leftarrow}M\overset{f}{\rightarrow} S,$
${}$
this also implies:

• Every morphism $f:M\to S$ in $C$ is a left $S$-module in $(C^\mathsf{op},\otimes, 1).$

Right modules are a little trickier.

• Every span $R\leftarrow M\overset{1}{\rightarrow} M$ is a right $R$-module in $(C^\mathsf{op},\otimes, 1).$

Eric Forgy (Jan 16 2021 at 18:37):

This reminds of a cool tidbit:

• A category internal to some category $C$ is the same as a monad in $\mathsf{Span}(C).$

from https://www.math.uni-hamburg.de/home/schreiber/SpanMod.pdf.

Eric Forgy (Jan 16 2021 at 18:42):

I don't know if this is great notation, but I think it means:
${}$
$\mathsf{Span}(C,\otimes,1) \cong \mathsf{Bim}(C^\mathsf{op},\otimes,1).$

Eric Forgy (Jan 16 2021 at 18:47):

I am probably spewing nonsense at this point, but that makes me think:
${}$
$\mathsf{Cospan}(C,\otimes,1) \cong \mathsf{Bim}(C,\otimes,1).$

Eric Forgy (Jan 16 2021 at 18:50):

That actually makes a LOT of sense to me, but if it was true, I'd think I would have seen it in the 10s of papers I've been skimming the last couple of weeks.

Eric Forgy (Jan 16 2021 at 18:53):

I think it should be correct though :thinking:

Eric Forgy (Jan 16 2021 at 19:09):

I think the puzzles show
${}$
$\mathsf{Span(Set},\times,*)\cong \mathsf{Bim(Set^{op}},\times,*).$

Eric Forgy (Jan 16 2021 at 19:14):

If
${}$
$\mathsf{Cospan}(C,\otimes,1) \cong \mathsf{Bim}(C,\otimes,1),$
${}$
it would be a good exercise to show
${}$
$\mathsf{Cospan(Set},\times,$) \cong \mathsf{Bim(Set},\times,).
${}$
That might be within my grasp :sweat_smile:

Eric Forgy (Jan 16 2021 at 19:31):

I'm starting to see how this all ties together (I think) :blush:

Eric Forgy (Jan 16 2021 at 19:45):

Btw, I can hear John's voice in my head, so let me explain what I mean by $\mathsf{Bim}(C,\otimes,1)$ :sweat_smile:

The $(C,\otimes,1)$ part just means a monoidal category.

$\mathsf{Bim}(C,\otimes,1)$ is a bicategory whose

• Objects are monoid objects in $(C,\otimes,1)$
• Morphisms are $(R,S)$-bimodules in $(C,\otimes,1)$
• 2-morphisms are bimodule homomorphisms

Eric Forgy (Jan 16 2021 at 19:49):

Similarly, $\mathsf{Cospan}(C,\otimes,1)$ is a bicategory whose

• Objects are monoid objects in $(C,\otimes,1)$
• Morphisms are cospans $R\rightarrow M\leftarrow S$ in $(C,\otimes,1)$
• 2-morphisms are the usual maps between cospans

Eric Forgy (Jan 16 2021 at 20:06):

It is easy to see that
${}$
$\sout{\mathsf{Cospan}(C^\mathsf{op},\otimes,1) \cong \mathsf{Bim}(C^\mathsf{op},\otimes,1)}$
${}$
because every object of $\sout{(C,\otimes,1)}$ is a monoid object in $\sout{(C^\mathsf{op},\otimes,1),}$ but that should also mean
${}$
$\sout{\mathsf{Cospan}(C,\otimes,1) \cong \mathsf{Bim}(C,\otimes,1)}$
${}$
more generally.

Eric Forgy (Jan 16 2021 at 20:16):

This ties in with my other stuff because a functor
${}$
$\mathbb{N}\to\mathsf{Span(Set)}$
${}$
picks out a monoid object in $\mathsf{Span(Set)}$ and
${}$
$[\mathbb{N},\mathsf{Span(Set)}]$
${}$
is a category of monoid objects in $\mathsf{Span(Set)}.$
${}$
More generally,
${}$
$[\mathbb{N},C]$
${}$
is a category of monoid objects in $C.$
${}$
(I think)

Eric Forgy (Jan 16 2021 at 20:22):

It would be pretty cool if
${}$
$[\mathbb{N},\mathsf{Span(Set)}] \cong \mathsf{Span}([\mathbb{N},\mathsf{Set}],\times,*).$

John Baez (Jan 16 2021 at 20:23):

Eric Forgy said:

Good morning :coffee:

I'm still trying to pull all this together.

I'll try to explain / generalize what I learned in my words to see If I understand.

Given a monoidal category $(C,\otimes,1)$:

• Every object in $C$ is a monoid object in $(C^\mathsf{op},\otimes, 1).$

No! If you prove this for $(C,\otimes,1) = (\mathsf{Set}, \times, \ast)$ you'll see what special features of this example are being used. You should really do this, since it's not hard, it's enlightening, and it lets you prove a more general result. I sketched the key steps earlier. The proof does not work for all monoidal categories.

The last part actually obvious: since every monoidal category is the opposite of its opposite, if every object in the opposite of every monoidal category were a monoid object, then every object in every monoidal category would be a monoid object!

And by the way, the result for $C = \mathsf{Set}$ is actually much better than what you just stated:

Theorem. Every object of $(\mathsf{Set}^{\text{op}}, \times, \ast)$ is a monoid object in a unique way. This way makes it a commutative monoid. Making every object into a monoid object in this way, every morphism in $(\mathsf{Set}^{\text{op}}, \times, \ast)$ is a homomorphism of monoid objects.

Eric Forgy (Jan 16 2021 at 20:25):

And... I just had this open https://ncatlab.org/nlab/show/Eckmann-Hilton+argument :sweat_smile:

John Baez (Jan 16 2021 at 20:31):

Eric Forgy said:

This ties in with my other stuff because a functor
${}$
$\mathbb{N}\to\mathsf{Span(Set)}$
${}$
picks out a monoid object in $\mathsf{Span(Set)}$ and
${}$
$[\mathbb{N},\mathsf{Span(Set)}]$
${}$
is a category of monoid objects in $\mathsf{Span(Set)}.$

I don't think that's true. But maybe you're using words in a somewhat different way than I do. For example, you are thinking of $\mathbb{N}$ as a category somehow, but there are at least two really famous ways to do this, and you're not saying which one you're using. However, I don't think your claims are true in either interpretation.

Maybe I'm wrong.

If you try to prove your claim here, we'll see what's going on.

Eric Forgy (Jan 16 2021 at 20:33):

So is the correct way to generalize what we learned in your puzzles to generalize to commutative monoidal categories? Is this true:

Given a commutative monoidal category $(C,\otimes,1)$:

• Every object in $C$ is a monoid object in $(C^\mathsf{op},\otimes, 1).$
• Every span $R\leftarrow M\rightarrow S$ in $C$ is an $(R,S)$-bimodule in $(C^\mathsf{op},\otimes, 1).$

Eric Forgy (Jan 16 2021 at 20:37):

Hopefully something good can be salvaged from this mess I created :sweat_smile:

Eric Forgy (Jan 16 2021 at 20:42):

I need to refine $\mathbb{N}$ because I am taking your advice and going to treat $\mathsf{Span(Set)}$ as a bicategory rather than a category with isomorphism classes. I'm guessing it will involve $1$ as a monad somehow.

Eric Forgy (Jan 16 2021 at 21:29):

I don't know if this is the right $\mathbb{N}$, but it is kind of neat.

Let $\mathbb{N}$ denote the bicategory with

• One object $\bullet.$
• Morphisms are spans $\bullet\overset{m}{\leftarrow}\bullet\overset{m}{\rightarrow}\bullet,$ where $m$ is a natural number.
• 2-morphisms are maps between spans $(\bullet\overset{m_1}{\leftarrow}\bullet\overset{m_1}{\rightarrow}\bullet)\overset{m}{\Rightarrow}(\bullet\overset{m_2}{\leftarrow}\bullet\overset{m_2}{\rightarrow}\bullet)$ where $m$ is a natural number and $m_1 = m + m_2.$

Eric Forgy (Jan 16 2021 at 21:40):

Composition of spans

Eric Forgy (Jan 16 2021 at 21:43):

2-morphism

John Baez (Jan 16 2021 at 21:44):

Eric Forgy said:

So is the correct way to generalize what we learned in your puzzles to generalize to commutative monoidal categories? Is this true:

Given a commutative monoidal category $(C,\otimes,1)$:

• Every object in $C$ is a monoid object in $(C^\mathsf{op},\otimes, 1).$
• Every span $R\leftarrow M\rightarrow S$ in $C$ is an $(R,S)$-bimodule in $(C^\mathsf{op},\otimes, 1).$

No. I already said that the first one is false - see below for my explanation of why it's false. Did I write that all for nothing?

The second is false too.

For example, they're both false when $(C^\mathsf{op},\otimes, 1) = (\mathsf{Set}, \times, 1)$. That's easy to show, and you should show it.

John Baez said:

Eric Forgy said:

Good morning :coffee:

I'm still trying to pull all this together.

I'll try to explain / generalize what I learned in my words to see If I understand.

Given a monoidal category $(C,\otimes,1)$:

• Every object in $C$ is a monoid object in $(C^\mathsf{op},\otimes, 1).$

No! If you prove this for $(C,\otimes,1) = (\mathsf{Set}, \times, \ast)$ you'll see what special features of this example are being used. You should really do this, since it's not hard, it's enlightening, and it lets you prove a more general result. I sketched the key steps earlier.

The proof does not work for all monoidal categories. This is actually obvious: since every monoidal category is the opposite of its opposite, if every object in the opposite of every monoidal category were a monoid object, then every object in every monoidal category would be a monoid object!

And by the way, the result for $C = \mathsf{Set}$ is actually much better than what you just stated:

Theorem. Every object of $(\mathsf{Set}^{\text{op}}, \times, \ast)$ is a monoid object in a unique way. This way makes it a commutative monoid. Making every object into a monoid object in this way, every morphism in $(\mathsf{Set}^{\text{op}}, \times, \ast)$ is a homomorphism of monoid objects.

Eric Forgy (Jan 16 2021 at 21:46):

John Baez said:

Eric Forgy said:

So is the correct way to generalize what we learned in your puzzles to generalize to commutative monoidal categories? Is this true:

Given a commutative monoidal category $(C,\otimes,1)$:

• Every object in $C$ is a monoid object in $(C^\mathsf{op},\otimes, 1).$
• Every span $R\leftarrow M\rightarrow S$ in $C$ is an $(R,S)$-bimodule in $(C^\mathsf{op},\otimes, 1).$

No. I already said that the first one is false - see below for my explanation of why it's false. Did I write that all for nothing?

This is different than what I had before because I was restricting to "commutative" monoidal categories. I didn't ignore what you said. I thought I was incorporating it :thinking:

Eric Forgy (Jan 16 2021 at 21:49):

I was thinking a monoid in a commutative monoidal category should be a commutative monoid.

John Baez (Jan 16 2021 at 21:51):

Oh, okay, I see that's a different guess. It's still false though. You didn't do what I told you to: prove that result is true for $(C, \otimes, 1) = (\mathsf{Set}, \times , 1)$ and see what you are using about the category $\mathsf{Set}$ that makes the argument work.

Instead, I guess you added one extra assumption and hoped that would be enough. That's a slow way to make progress: make guesses and let me knock them down. It's sort of like throwing darts blindfolded and hoping one hits the target.

I could of course just tell you what's true, but I think it's better if you go through the argument for $(C, \otimes, 1) = (\mathsf{Set}, \times , 1)$ and see what's really going on. You'll see some very special features of this case are being used.

Eric Forgy (Jan 16 2021 at 22:01):

OK. I will work through it. Sorry. I thought I understood it. In the case of $\mathsf{Set/Set^{op}},$ we didn't need to think about left / right modules because they are the same in $\mathsf{Set^{op}}$ which must be related to commutativity somehow. So when you mentioned commutative monoids, I thought that was the answer and moved on to what I considered more pressing issues, but I'll try this proof. Thank you.

John Baez (Jan 16 2021 at 22:02):

Yes, doing that proof was Puzzle 1, and I think it's important!

Eric Forgy (Jan 16 2021 at 22:02):

I did Puzzle 1. I can do it again :sweat_smile:

John Baez (Jan 16 2021 at 22:03):

You may have done it - I'm not really sure - but if so, you didn't analyze what about $\mathsf{Set}$ you were using. That "analyzing what made the argument work" step is crucial for generalizing.

Eric Forgy (Jan 16 2021 at 22:03):

In $\mathsf{Set^{op}}$, a $\mu:X\times X\to X$ corresponds to $\mu^{op}: X\to X\times X$ in $\mathsf{Set}$ which is the diagonal map.

John Baez (Jan 16 2021 at 22:05):

Yes, now you're starting to do the analysis of the argument. If you do it carefully enough, you'll know exactly which sort of categories you can generalize the argument to. And of course you complete this process by proving that the argument generalizes.

Eric Forgy (Jan 16 2021 at 22:06):

The unit $\eta: *\to X$ in $\mathsf{Set^{op}}$ corresponds to the unique map $\eta^{op}: X\to *$ sending every element of $X$ to the one element $\bullet\in *$ (i.e., $*$ is the terminal set).

John Baez (Jan 16 2021 at 22:11):

Right. So figure out what properties of $\mu^{\text{op}}$ and $\eta^{\text{op}}$ are required to get a monoid object by this trick.

Eric Forgy (Jan 16 2021 at 22:17):

Associativity:
${}$
$x\overset{\mu^\mathsf{op}}{\mapsto} (x,x)\overset{\mu^\mathsf{op}\times\mathsf{id}}{\mapsto} ((x,x),x)$
${}$
$x\overset{\mu^\mathsf{op}}{\mapsto} (x,x)\overset{\mathsf{id}\times\mu^\mathsf{op}}{\mapsto} (x,(x,x))\overset{\alpha^\mathsf{op}}{\mapsto}((x,x),x)$
${}$
Check :check_mark:

Eric Forgy (Jan 16 2021 at 22:23):

Left unit:
${}$
$x\overset{\mu^\mathsf{op}}{\mapsto} (x,x)\overset{\eta^\mathsf{op}\times\mathsf{id}}{\mapsto} (\bullet,x)$
${}$
$x\overset{\lambda^\mathsf{op}}{\mapsto} (\bullet,x)$
${}$
Check :check_mark:

Eric Forgy (Jan 16 2021 at 22:25):

Right unit:
${}$
$x\overset{\mu^\mathsf{op}}{\mapsto} (x,x)\overset{\mathsf{id}\times\eta^\mathsf{op}}{\mapsto} (x,\bullet)$
${}$
$x\overset{\rho^\mathsf{op}}{\mapsto} (x,\bullet)$
${}$
Check :check_mark:

Eric Forgy (Jan 16 2021 at 22:27):

Confession: I didn't work out left and right units when I did Puzzle 1 and that would have helped with Puzzle 2 :face_palm:

John Baez (Jan 16 2021 at 22:32):

Good! Next, to solve Puzzle 1, you need to show that there's only one way to make any set into a monoid in $\mathsf{Set}^{\text{op}}$. You've found one way: why is this the only way?

Only after proving that can you confidently claim "a monoid in $(\mathsf{Set}^{\text{op}}, \times, 1)$ is just the same as a set" - no more, no less.

Eric Forgy (Jan 16 2021 at 22:41):

I think I should have started with the most general possibility and widdle it down.

So instead of
${}$
$\mu^{op}: x\mapsto (x,x)$
${}$
I'll look at a more general
${}$
$\mu^{op}: x\mapsto (f(x),g(x))$
${}$

Associativity:
${}$
$x\overset{\mu^\mathsf{op}}{\mapsto} (f(x),g(x))\overset{\mu^\mathsf{op}\times\mathsf{id}}{\mapsto} ((f(x),g(x)),g(x))$
${}$
$x\overset{\mu^\mathsf{op}}{\mapsto} (f(x),g(x))\overset{\mathsf{id}\times\mu^\mathsf{op}}{\mapsto} (f(x),(f(x),g(x)))\overset{\alpha^\mathsf{op}}{\mapsto}((f(x),f(x)),g(x))$

This can only be satisfied if $f = g$ so we've narrowed it down a little.

John Baez (Jan 16 2021 at 22:43):

Right, now you're starting to see the point of this whole puzzle.

John Baez (Jan 16 2021 at 22:44):

By the way, James Dolan gave me this puzzle once and I started just like you did now: with associativity.

Eric Forgy (Jan 16 2021 at 22:44):

Right unit:
${}$
$x\overset{\mu^\mathsf{op}}{\mapsto} (f(x),f(x))\overset{\mathsf{id}\times\eta^\mathsf{op}}{\mapsto} (f(x),\bullet)$
${}$
$x\overset{\rho^\mathsf{op}}{\mapsto} (x,\bullet).$

This can only be satisfied if $f(x) = x$, i.e. $f = \mathsf{id}_X.$

John Baez (Jan 16 2021 at 22:45):

Now you're really getting the point.

John Baez (Jan 16 2021 at 22:46):

The humble unit laws are the key here.

Eric Forgy (Jan 16 2021 at 22:46):

Yeah. I did this kind of exercise with James when trying to solve Puzzle 2 and definitely felt my brain muscles growing :muscle:

John Baez (Jan 16 2021 at 22:47):

In fact you didn't even need to use associativity!

John Baez (Jan 16 2021 at 22:47):

If you use both unit laws you see first $f = \mathrm{id}$ and then $g = \mathrm{id}$.

Eric Forgy (Jan 16 2021 at 22:47):

Cool. Yeah :+1:

Eric Forgy (Jan 16 2021 at 22:48):

Then associativity is free :+1:

John Baez (Jan 16 2021 at 22:48):

I don't know if there's a name for thing that has a multiplication that obeys the unit laws but may not be associative.

John Baez (Jan 16 2021 at 22:49):

But anyway, there's exactly one way to make any set into one of those things in $\mathsf{Set}^{\text{op}}$.

John Baez (Jan 16 2021 at 22:49):

And then it turns out to be a commutative monoid.

Eric Forgy (Jan 16 2021 at 22:55):

I feel dense, but I'm still missing the punchline :thinking:

Eric Forgy (Jan 16 2021 at 22:57):

It feels a little vacuous, but every set is a comonoid object in $\mathsf{Set}$ so every set is a monoid in $\mathsf{Set^{op}}.$

John Baez (Jan 16 2021 at 22:58):

You should say in a unique way to get the most for all your hard work.

John Baez (Jan 16 2021 at 22:59):

So now you've solved Puzzle 1.

John Baez (Jan 16 2021 at 22:59):

The next step is to analyze what about the category $\mathsf{Set}$ made this proof work?

John Baez (Jan 16 2021 at 22:59):

Then you can generalize to certain other categories.

John Baez (Jan 16 2021 at 22:59):

But not all monoidal categories, or even all symmetric monoidal categories.

Eric Forgy (Jan 16 2021 at 22:59):

Eric Forgy said:

It feels a little vacuous, but every set is a comonoid object in $\mathsf{Set}$ so every set is a monoid in $\mathsf{Set^{op}}.$

:point_of_information: ?

Eric Forgy (Jan 16 2021 at 23:00):

We need monoidal categories where every object is a comonoid object.

John Baez (Jan 16 2021 at 23:01):

That's like saying your result generalizes to categories where the result is true.

John Baez (Jan 16 2021 at 23:01):

Which is true, but scarcely helpful.

Eric Forgy (Jan 16 2021 at 23:01):

John Baez said:

That's like saying your result generalizes to categories where the result is true.

Can't argue with that :joy:

John Baez (Jan 16 2021 at 23:01):

Theorem. This theorem is true for all categories for which this theorem is true.

John Baez (Jan 16 2021 at 23:02):

That's a true theorem.

Eric Forgy (Jan 16 2021 at 23:02):

Right, so I need to get a "little" more specific :sweat_smile:

Eric Forgy (Jan 16 2021 at 23:04):

But all we needed for every set in $\mathsf{Set}$ to be a comonoid object was a $\mu^{op}$, $\eta^{op}$, $\lambda^{op}$ and $\rho^{op}$ satisfying left and right unit laws (and associativity, but that probably comes free for any case I'll ever see).

John Baez (Jan 16 2021 at 23:05):

Yes, that's like saying the result is true when it's true.

Eric Forgy (Jan 16 2021 at 23:05):

$\eta^{op}$, $\lambda^{op}$ and $\rho^{op}$ all come basically for free.

John Baez (Jan 16 2021 at 23:05):

You have to actually analyze your argument to see what facts about the category of sets you were using. And this may be hard, because you don't have a strong command of the various different properties of categories, so you may not recognize which ones are coming into play.

Eric Forgy (Jan 16 2021 at 23:05):

Oh! One restriction. $C$ must have a terminal object?

John Baez (Jan 16 2021 at 23:06):

Well, don't say it must have a terminal object. We're looking for sufficient conditions for the argument to work, not necessary ones. But yeah: you seemed to use something about it having a terminal object.

John Baez (Jan 16 2021 at 23:07):

But more than that, you used the fact that the unit object of your monoidal category is the terminal object!

Eric Forgy (Jan 16 2021 at 23:08):

I guess when you write $(C,\otimes,1)$, the $1$ is not always terminal?

John Baez (Jan 16 2021 at 23:08):

You should write $I$ if it's not terminal.

John Baez (Jan 16 2021 at 23:08):

Consider $(\mathsf{Vect}, \otimes, k)$. The field $k$ is not terminal.

Eric Forgy (Jan 16 2021 at 23:09):

Oops. I thought $k$ was terminal in $\mathsf{Vect}$ :thinking:

Eric Forgy (Jan 16 2021 at 23:12):

For every $V\in\mathsf{Vect},$ I thought we have a dual $V^*: V\to k.$

John Baez (Jan 16 2021 at 23:13):

I don't know what that has to do with it or even what it means. Since when is the dual $V^\ast$ a linear map? I thought it was a vector space!

John Baez (Jan 16 2021 at 23:13):

What does it mean for an object to be terminal?

Eric Forgy (Jan 16 2021 at 23:14):

For every object $c\in C$, there is a unique morphism $c\to 1.$

John Baez (Jan 16 2021 at 23:15):

Okay, good. So let's look at the category of real vector spaces, where $k = \mathbb{R}$. Is $\mathbb{R}$ terminal?

Eric Forgy (Jan 16 2021 at 23:18):

Eric Forgy said:

For every $V\in\mathsf{Vect},$ I thought we have a dual $V^*: V\to k.$

This is a typo. What I meant (which is also wrong) was that any $\alpha\in V^*$ is a morphism $\alpha: V\to k$, but the key is the "unique" part. That often gets me :pensive:

Eric Forgy (Jan 16 2021 at 23:19):

I can think of lots of morphisms $V\to\mathbb{R}$ (which means there is not a unique one).

Eric Forgy (Jan 16 2021 at 23:24):

So $k$ is not terminal because for any $V\in\mathsf{Vect_k}$, any two elements $\alpha,\beta\in V^*$ are both maps $\alpha,\beta: V\to k.$

David Egolf (Jan 16 2021 at 23:25):

Or I guess it's not terminal because in general there is more than one linear map from a vector space to the underlying field viewed as a vector space.

John Baez (Jan 16 2021 at 23:26):

Good. So is there a terminal object in $\mathsf{Vect}_k$ or not?

Eric Forgy (Jan 16 2021 at 23:26):

John Baez said:

Good. So is there a terminal object in $\mathsf{Vect}_k$ or not?

Yes! 0

Eric Forgy (Jan 16 2021 at 23:27):

0 is both initial and terminal.

John Baez (Jan 16 2021 at 23:27):

Okay, so $(\mathsf{Vect}_k, \otimes , k)$ is a nice example of a monoidal category that has a terminal object, but where the terminal object is not the unit object for the tensor product.

Eric Forgy (Jan 16 2021 at 23:28):

Got it! Thank you :blush:

John Baez (Jan 16 2021 at 23:28):

Such monoidal categories have a very different flavor than, say, $(\mathsf{Set}, \times, 1)$

Eric Forgy (Jan 16 2021 at 23:30):

So one condition is that for a monoidal category $(C,\otimes,I)$, we want $I$ to be terminal, but in that case we would write $(C,\otimes,1).$

Eric Forgy (Jan 16 2021 at 23:31):

I think that is good enough because there is always a diagonal map in a monoidal category.

John Baez (Jan 16 2021 at 23:31):

What is the diagonal map for $(\mathsf{Vect}, \otimes, k)$?

Eric Forgy (Jan 16 2021 at 23:32):

Oops! I thought the diagonal map was always defined as $x\mapsto x\otimes x.$

John Baez (Jan 16 2021 at 23:33):

Is that a linear map?

John Baez (Jan 16 2021 at 23:33):

Say, in $\mathsf{Vect}$?

Eric Forgy (Jan 16 2021 at 23:33):

No. It's not :face_palm:

John Baez (Jan 16 2021 at 23:33):

Okay. Also, you're writing down a map in terms of an "element" $x$, but for lots of categories the objects don't have "elements".

John Baez (Jan 16 2021 at 23:34):

So no: whoever told you every monoidal category has diagonal maps was lying.

Eric Forgy (Jan 16 2021 at 23:35):

So we have another condition. $C$ needs to have diagonal maps.

So far, we have:

• The unit of $C$ must be terminal.
• $C$ must have diagonal maps.

Eric Forgy (Jan 16 2021 at 23:38):

So I've just learned and unlearned some things :clap: Thank you :blush:

John Baez (Jan 16 2021 at 23:39):

Great! Now I think it'd be good to read this:

https://math.ucr.edu/home/baez/quantum/node4.html

starting around where I say "However, these two examples are very different, because the product in $\mathsf{Set}$ is 'cartesian' in a certain technical sense", and going on to the definition of diagonal.

John Baez (Jan 16 2021 at 23:41):

This has just the clues you need next.

Eric Forgy (Jan 16 2021 at 23:45):

Reading :+1: :book:

Eric Forgy (Jan 16 2021 at 23:59):

Ok. I've read and reread it a couple times. Unfortunately, I don't see how it gets us any new conditions other than

• The unit of $C$ must be terminal.
• $C$ must have diagonal maps.

Eric Forgy (Jan 17 2021 at 00:00):

The diagonal map relates to whether the product is Cartesian.

John Baez (Jan 17 2021 at 00:00):

Well, you said "C must have diagonal maps" without saying anything about what a "diagonal map" is.

John Baez (Jan 17 2021 at 00:01):

I.e., what properties do those "diagonal maps" need to have, for your argument to work?

John Baez (Jan 17 2021 at 00:01):

They can't just be any old morphisms $x \to x \otimes x$.

John Baez (Jan 17 2021 at 00:02):

So when you say "$C$ must have diagonal maps", that sounds good and it's on the right track, but what does it actually mean?

John Baez (Jan 17 2021 at 00:02):

The reading material provides the answer to that.

John Baez (Jan 17 2021 at 00:04):

When you're done, you'll be able to prove "if the monoidal category $(\mathsf{C}, \otimes, I)$ has such and such properties, every object in $(\mathsf{C}^{\text{op}}, \otimes, I)$ becomes a commutative monoid object in exactly one way".

David Egolf (Jan 17 2021 at 00:07):

I don't want to interrupt, but I think this is also helpful for understanding diagonal maps: https://ncatlab.org/nlab/show/diagonal+morphism

It seems like we have a diagonal map when we are in a category where we can take the (cartesian) product of pairs of objects.

John Baez (Jan 17 2021 at 00:09):

Yes, this is useful information about "diagonal morphisms".

John Baez (Jan 17 2021 at 00:10):

John Baez said:

When you're done, you'll be able to prove "if the monoidal category $(\mathsf{C}, \otimes, I)$ has such and such properties, every object in $(\mathsf{C}^{\text{op}}, \otimes, I)$ becomes a commutative monoid object in exactly one way".

The reading material I provided Eric says exactly what "such and such properties" are, though of course one has to dig them out and check they actually work.

David Egolf (Jan 17 2021 at 00:10):

I'm also noticing a suspicious similarity between the diagram describing the diagonal map and the diagram corresponding to the unit law conditions for a monoid in a monoidal category.

John Baez (Jan 17 2021 at 00:11):

Suspicious, suspicious.... very suspicious. :upside_down:

Eric Forgy (Jan 17 2021 at 00:13):

It looks like the diagonal is a 2-morphism in the bicategory $\mathsf{Span}(C)$ from the identity span
${}$
$X\overset{1}{\leftarrow} X\overset{1}{\rightarrow} X$
${}$
to the span
${}$
$X\overset{\pi_1}{\leftarrow} X\times X\overset{\pi_2}{\rightarrow} X$
${}$
:blush:

Eric Forgy (Jan 17 2021 at 00:17):

This is also interesting:

https://ncatlab.org/nlab/show/monoidal+category+with+diagonals

John Baez (Jan 17 2021 at 00:18):

Anyway, I'm hoping Eric will guess how to fill in the "such and such properties" here, to make this theorem true:

Theorem. If the monoidal category $(\mathsf{C}, \otimes, I)$ has such and such properties, every object in $(\mathsf{C}^{\text{op}}, \otimes, I)$ becomes a commutative monoid object in exactly one way.

John Baez (Jan 17 2021 at 00:19):

The reading material gives it away.

Eric Forgy (Jan 17 2021 at 00:19):

Looking at

https://ncatlab.org/nlab/show/cartesian+monoidal+category

it looks like a sufficient condition is that $C$ needs to be a cartesian monoidal category.

John Baez (Jan 17 2021 at 00:20):

Yes. In the article of mine that I pointed you to, I just called it a "cartesian category".

Eric Forgy (Jan 17 2021 at 00:20):

That is awesome. Thank you :raised_hands:

John Baez (Jan 17 2021 at 00:21):

The article explains how the cartesianness is connected to our ability to "duplicate" information using a diagonal, and "discard" information using a map to the terminal object.

Eric Forgy (Jan 17 2021 at 00:22):

Yes. That is cool. I've seen that before, but didn't appreciate it as much until now :blush:

John Baez (Jan 17 2021 at 00:22):

But the slick way to say all this is that every object in a cartesian category becomes a cocommutative comonoid in a unique way.

John Baez (Jan 17 2021 at 00:22):

You should really prove it, to see what's going on!

John Baez (Jan 17 2021 at 00:23):

Or if you prefer, prove this, which is just the same thing said more slowly:

Theorem. If the monoidal category $(\mathsf{C}, \otimes, I)$ is cartesian, every object in $(\mathsf{C}^{\text{op}}, \otimes, I)$ becomes a commutative monoid object in exactly one way.

Eric Forgy (Jan 17 2021 at 00:25):

In a cartesian monoidal category $(C,\times,1)$, we have our diagonal map and the unit is terminal. Those are the conditions we needed to show that every object in $C^\mathsf{op}$ is a commutative monoid.

John Baez (Jan 17 2021 at 00:27):

So you claim, but you never showed that - at least, not to me.

John Baez (Jan 17 2021 at 00:27):

Especially not the "in a unique way" part, which is the really cool part.

John Baez (Jan 17 2021 at 00:27):

You showed the "in a unique way" part when $C = \mathsf{Set}$, but not in general.

John Baez (Jan 17 2021 at 00:28):

I'm not trying to torture you, just pointing out that you haven't really dug to the bottom of this particular rabbit hole yet.

Eric Forgy (Jan 17 2021 at 00:29):

I showed it in terms of elements. Can I cheat and say we have elements $1\to C$ for any cartesian category? :smiley:

John Baez (Jan 17 2021 at 00:29):

The proof for a general cartesian $C$ is just as easy as it is for $\mathsf{Set}$, but it uses commutative diagrams, not "elements".

John Baez (Jan 17 2021 at 00:30):

Consider the example of $(\mathsf{Vect}, \times, 1)$. This is a perfectly nice cartesian category. How many elements does a vector space $V$ have, if an "element" is a morphism $1 \to V$? (And what's "1"?)

Eric Forgy (Jan 17 2021 at 00:33):

You have 1 "element" $1\overset{v}{\to} V$ for every $v\in V$ and those are infinite.

Eric Forgy (Jan 17 2021 at 00:34):

I was only kidding. I should try with commuting diagrams :blush:

John Baez (Jan 17 2021 at 00:35):

I hope you're kidding. $1 = 0$ in this case, and every vector space has just one "element".

John Baez (Jan 17 2021 at 00:36):

So "elements" are useless here.

Eric Forgy (Jan 17 2021 at 00:36):

I kind of feel like that was far enough down this rabbit hole for now. My clock is ticking ("feed the family" and all) so I think I will turn back to my favorite spans and bimodules for a while armed with this new knowledge :muscle:

John Baez (Jan 17 2021 at 00:36):

Okay. What you really need is not knowledge so much as power. They say knowledge is power, but in math that's not quite true.

John Baez (Jan 17 2021 at 00:37):

Skill is power.

Eric Forgy (Jan 17 2021 at 00:40):

True. I hear you. I need more skill. No doubt. I think I can also improve my skills while carefully working on the specific problems I need to solve though :pray:

Eric Forgy (Jan 17 2021 at 00:41):

I've got a nice bimodule I know how to work with and I have a nice span category I know how to deal with and I'm still hoping to bridge the two. This has been super helpful along those lines I think. Thank you :pray:

Eric Forgy (Jan 17 2021 at 00:42):

Eric Forgy said:

I don't know if this is the right $\mathbb{N}$, but it is kind of neat.

Let $\mathbb{N}$ denote the bicategory with

• One object $\bullet.$
• Morphisms are spans $\bullet\overset{m}{\leftarrow}\bullet\overset{m}{\rightarrow}\bullet,$ where $m$ is a natural number.
• 2-morphisms are maps between spans $(\bullet\overset{m_1}{\leftarrow}\bullet\overset{m_1}{\rightarrow}\bullet)\overset{m}{\Rightarrow}(\bullet\overset{m_2}{\leftarrow}\bullet\overset{m_2}{\rightarrow}\bullet)$ where $m$ is a natural number and $m_1 = m + m_2.$

I want to think about this guy a little bit.

Eric Forgy (Jan 17 2021 at 00:43):

Eric Forgy said:
Composition of spans

Eric Forgy said:
2-morphism

David Egolf (Jan 17 2021 at 00:47):

unit laws

diagonal morphism

When you feel like thinking about this again, I think comparing these two might be helpful, Eric. The first diagram is required to commute for us to have a monoid in a monoidal category (we're wondering what $\mu$ can make this happen), and the second diagram is guaranteed to have a unique $\Delta$ making it commute when we have binary products.

David Egolf (Jan 17 2021 at 00:48):

I haven't figured it out yet, but I suspect there's some way of taking the guaranteed $\Delta$ and "turning it around" so that it becomes the $\mu$ we need.

Eric Forgy (Jan 17 2021 at 00:50):

Hi David. Yeah :+1:

In a cartesian cartegory, the tensor product is cartesian product so those two are pretty much exact opposites :blush:

Eric Forgy (Jan 17 2021 at 00:51):

$I\otimes M\cong M\cong M\otimes I.$

David Egolf (Jan 17 2021 at 00:51):

The thing worrying me is that in the top diagram we have $I \otimes M$ and $M \otimes I$ mapping into $M \otimes M$. This is a little different than having $M$ and $M$ mapping into $M \otimes M$, which is like what we have on the bottom (with the arrows turned around).

Eric Forgy (Jan 17 2021 at 00:52):

Eric Forgy said:

$I\otimes M\cong M\cong M\otimes I.$

:point_of_information: :blush:

David Egolf (Jan 17 2021 at 00:53):

Yeah those are isomorphic. But is the product of $M$ and $M$ isomorphic to the product of $I \otimes M$ and $M \otimes I$?

Eric Forgy (Jan 17 2021 at 00:56):

In a cartesian category, $I = 1$ (the terminal object) and in $\mathsf{Set}$, $1 = * = \{\bullet\}$, i.e. the one element set. AND... $\otimes = \times.$

David Egolf (Jan 17 2021 at 00:58):

I guess what I'm wondering is if $M \otimes M$ is a product of $I \otimes M$ and $M \otimes I$.

Eric Forgy (Jan 17 2021 at 00:58):

It depends on what your definition of "is" is :smile:

David Egolf (Jan 17 2021 at 00:59):

Does it satisfy the universal property of products with respect to those two objects?

Eric Forgy (Jan 17 2021 at 01:00):

It should. Yes. Try checking the diagrams :blush:

David Egolf (Jan 17 2021 at 01:02):

If it does, then we are in business! There should then be a unique $\mu^{op}:M \to M \otimes M$ making the flipped version of that unital law diagram commute, which is our "diagonal map".

Eric Forgy (Jan 17 2021 at 01:02):

Universal properties are unique "up to isomorphism".

Eric Forgy (Jan 17 2021 at 01:03):

So if you chase those diagrams with $M$ and then chase them again with $1\times M$, you should get the same thing up to isomorphism.

Eric Forgy (Jan 17 2021 at 01:06):

You probably know this, but elements in $1\times M$ are all of the form $(\bullet,m).$ For every $m\in M$ there is a $(\bullet,m)\in 1\times M$ and those two are basically the same thing since $m\mapsto (\bullet,m)$ and $(\bullet,m)\mapsto m$ is an isomorphism.

David Egolf (Jan 17 2021 at 01:07):

Yeah, that's true for when we can work in ordered pairs of "elements".

Eric Forgy (Jan 17 2021 at 01:08):

So an element of $(M\times 1)\times (1\times M)$ is of the form $((m_1, \bullet),(\bullet,m_2))$ which is the same as $(m_1,m_2)$ since there is an isomorphism between them.

David Egolf (Jan 17 2021 at 01:08):

Yeah, that seems right. But we'd need to generalize that argument to not rely on elements for it to be more general, I think.

Eric Forgy (Jan 17 2021 at 01:10):

David Egolf said:

Yeah, that's true for when we can work in ordered pairs of "elements".

True. More generally, we need to chase diagrams, but I have enough faith in
${}$
$(1\times M)\times (M\times 1)\cong M\times (M\times 1) \cong (1\times M)\times M \cong M\times M$
${}$
to not feel compelled to do that :sweat_smile:

David Egolf (Jan 17 2021 at 01:12):

Also, we'd need to check that the $\eta \otimes 1$ and $1 \otimes \eta$ when turned around act like projection functions. (So that $M \otimes M$ really is acting like a product).

Eric Forgy (Jan 17 2021 at 01:14):

Yeah. One of the whole purposes of monoidal categories, as I understand it, is that we can treat this stuff like familiar equations so that $1\times$ is really like multiplying by the identiy in normal algebra.

David Egolf (Jan 17 2021 at 01:15):

But if we accept all these things: $M \otimes M$ together with functions $\eta \otimes 1$ and $1 \otimes \eta$ is a coproduct of $I \otimes M$ and $M \otimes I$, and when we turn around the arrows it becomes a product in the opposite category of $I \otimes M$ and $M \otimes I$.

Then there is a unique "diagonal map" $\mu^{op}: M \to M \otimes M$ so that the corresponding morphism $\mu$ makes the identity law diagram commute.

Eric Forgy (Jan 17 2021 at 01:16):

Btw, I keep writing $\times$ instead of $\otimes$ because what we're trying to prove is only true for cartesian categories where $\otimes = \times.$

Eric Forgy (Jan 17 2021 at 01:21):

David Egolf said:

I haven't figured it out yet, but I suspect there's some way of taking the guaranteed $\Delta$ and "turning it around" so that it becomes the $\mu$ we need.

This is called "${}^\mathsf{op}$" :blush:

On the one hand, we have $\mu$ in $C^\mathsf{op}.$ On the other we have $\Delta$ in $C$. In what we're talking about $\mu^\mathsf{op} = \Delta.$ Put differently, $\mu$ in $C^\mathsf{op}$ is the opposite of the diagonal map $\Delta$ in $C$. That is the definition of $\mu.$

David Egolf (Jan 17 2021 at 01:22):

Summarizing a bit for myself: The key thing is this connection between the monoidal product and the cartesian product. When the monoidal product also enjoys the universal properties of the cartesian product (in some category) then we start getting unique morphisms to it. Unique morphisms to the monoidal product end up inducing a unique way to satisfy the unit laws.

Eric Forgy (Jan 17 2021 at 01:25):

So if a monoidal category $(C,\otimes,I)$ has a diagonal map and a unit that is the terminal object, i.e. $I = 1$, then every object of $C$ is a monoid object in $(C^\mathsf{op},\otimes,I).$

BUT... those conditions are precisely the definition of a cartesian monoidal category :blush:

Hence, in a cartesian category $(C,\times,1)$, every object of $C$ is a monoid object in $(C^\mathsf{op},\times,1).$

David Egolf (Jan 17 2021 at 01:26):

To have a diagonal map, I think you have to have at least binary products, right?

Eric Forgy (Jan 17 2021 at 01:27):

David Egolf said:

Summarizing a bit for myself: The key thing is this connection between the monoidal product and the cartesian product. When the monoidal product also enjoys the universal properties of the cartesian product (in some category) then we start getting unique morphisms to it. Unique morphisms to the monoidal product end up inducing a unique way to satisfy the unit laws.

No. There are two conditions and they are separate. You need:

1. Existence of diagonals
2. The unit element must be terminal

David Egolf (Jan 17 2021 at 01:27):

Yeah, I was summarizing the "existence of diagonals" part.

Eric Forgy (Jan 17 2021 at 01:28):

If you have both of those, you have a cartesian category.

David Egolf (Jan 17 2021 at 01:29):

The nlab puts it like this: the "monoidal structure is given by the category-theoretic product" in a cartesian category.

Eric Forgy (Jan 17 2021 at 01:30):

As John mentioned, those two conditions amount to the ability to duplicate and delete objects :blush:

David Egolf (Jan 17 2021 at 01:30):

I think that condition then implies both conditions 1 and 2

Eric Forgy (Jan 17 2021 at 01:31):

David Egolf said:

The nlab puts it like this: the "monoidal structure is given by the category-theoretic product" in a cartesian category.

This isn't enough. You need to also have $I=1.$ John gave a nice counter example. $(\mathsf{Vect},\times,1)$ is a perfectly fine monoidal category with cartesian product, but 1 is not terminal.

[Edit: Sorry.]

Eric Forgy (Jan 17 2021 at 01:33):

So $(\mathsf{Vect},\times,1)$ is not cartesian and vector spaces are not monoid objects in $\mathsf{Vect^{op}}.$

[Edit: Sorry.]

David Egolf (Jan 17 2021 at 01:34):

I think the monoidal unit can be thought of as the empty monoidal product. If the monoidal product coincides with the cartesian product, then the monoidal unit coincides with the empty cartesian product. We then have an object $I$ so that there is a unique morphism to it from every other object: the monoidal unit must be terminal.

David Egolf (Jan 17 2021 at 01:35):

Also note that the nlab suggests that the monoidal unit being terminal is a consequence:
"A cartesian monoidal category (usually just called a cartesian category), is a monoidal category whose monoidal structure is given by the category-theoretic product (and so whose unit is a terminal object)."

Eric Forgy (Jan 17 2021 at 01:35):

It is true that the "empty vector space" is terminal. But it is also initial. When an object is both initial and terminal, it is called a "zero object" 0.

David Egolf (Jan 17 2021 at 01:39):

At any rate, this has been fun. I now have a vague intuition for what a monoid in a monoidal category is, and I didn't before.

Eric Forgy (Jan 17 2021 at 01:39):

What would 1 be in $(\mathsf{Vect},\times,1)$? :blush:

Eric Forgy (Jan 17 2021 at 01:45):

What vector space could 1 possibly correspond to?

Todd Trimble (Jan 17 2021 at 01:45):

Eric Forgy said:

David Egolf said:

The nlab puts it like this: the "monoidal structure is given by the category-theoretic product" in a cartesian category.

This isn't enough. You need to also have $I=1.$ John gave a nice counter example. $(\mathsf{Vect},\times,1)$ is a perfectly fine monoidal category with cartesian product, but 1 is not terminal.

Vect carries more than one monoidal product, and one of them is the cartesian monoidal product where the monoidal unit is the terminal object (which here is the zero object).

FWIW, I think I can prove that if a monoidal product coincides with the cartesian product on objects, then the unit is forced to be terminal.

Eric Forgy (Jan 17 2021 at 01:46):

Oh. Hi Todd :wave:

Todd Trimble (Jan 17 2021 at 01:47):

Anyway, I don't know what you mean by John gave a counterexample. The $1$ here would be terminal.

Eric Forgy (Jan 17 2021 at 01:47):

Todd Trimble said:

FWIW, I think I can prove that if a monoidal product coincides with the cartesian product on objects, then the unit is forced to be terminal.

I believe it. David pointed out that the nLab seems to indicate that.

Eric Forgy (Jan 17 2021 at 01:48):

Todd Trimble said:

Anyway, I don't know what you mean by John gave a counterexample. The $1$ here would be terminal.

Yeah. I mispoke. In this case, 1=0 so it is both initial and terminal.

Eric Forgy (Jan 17 2021 at 01:51):

Thank you for fixing my mistake :pray:

Todd Trimble (Jan 17 2021 at 01:51):

As for the thing I said I thought I could prove: that takes a bit of effort. (I showed John a dual form of it on a different site maybe 45 days ago.) It's probably not an appropriate thing to get into at this stage of the discussion, but we could get into it later.

John Baez (Jan 17 2021 at 01:51):

David Egolf said:

I haven't figured it out yet, but I suspect there's some way of taking the guaranteed $\Delta$ and "turning it around" so that it becomes the $\mu$ we need.

Yes: If your category $\mathsf{C}$ is cartesian, every object $x$ in it has a diagonal $\Delta: x \to x \times x$, and then in $(\mathsf{C}^{\rm op}, \times, 1)$ this becomes a morphism from $x \times x$ to $x$ which, you can show, is an associative multiplication - the $\mu$ you want.

John Baez (Jan 17 2021 at 01:54):

It's fun to show that this multiplication really is associative.

Eric Forgy (Jan 17 2021 at 02:02):

So I guess in a general monoidal category, $(C,\otimes,I)$, the unit is not necessarily terminal (good to know!), but in a cartesian category, the unit is always terminal, but you could have 0=1 (also good to know) :thinking:

Eric Forgy (Jan 17 2021 at 02:03):

(my brain hurts, but in a good way :sweat_smile:)

John Baez (Jan 17 2021 at 02:07):

Yes, and don't forget that in the category with just one object the initial object is also terminal, products are also coproducts, etc. It's a good counterexample to keep in mind in case you think things need to be different.

Eric Forgy (Jan 17 2021 at 02:10):

You definitely upskilled me today (quite a feat!). Thank you John. I appreciate it :pray:

John Baez (Jan 17 2021 at 02:17):

Todd Trimble said:

FWIW, I think I can prove that if a monoidal product coincides with the cartesian product on objects, then the unit is forced to be terminal.

I should prove that sometime! The projection $p: I \times x \to I$ gives each object $x$ a morphism to the unit object $I$. So I should prove it's the only one.