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Stream: learning: questions

Topic: Bimodules vs Spans: Puzzle 2

Eric Forgy (Jan 14 2021 at 11:38):

(Note: This stream was split from Bimodules vs Spans: Puzzle 1.)

Puzzle 2: What’s a module of a monoid object in $\mathsf{Set^{op}}$?

We're looking for an object, i.e. set, $X$ and a monoid object $A$ together with an action
${}$
$\alpha: A\times X\to X.$

Attempt #1

I think this action should be
${}$
$\sout{\mathsf{eval}: [X,X]\times X\to X,\quad \mathsf{eval}: (f,x) \mapsto f(x).}$
${}$
This is true for every set $X$ in $\mathsf{Set^{op}}$ so every set $X$ is an $\mathsf{End}(X)$-module.

Attempt #2: After looking at Puzzle #3

Referring to Attempt #3 in Puzzle #1, there are actually two kinds of modules in $\mathsf{Set^{op}}$ corresponding to the two types of monoid objects.
${}$
1. A left $\mathsf{End}(X)$-module is a set $X$ and an action $\sout{\alpha_L : \mathsf{End}(X)\times X\to X.}$
2. A right $\mathsf{End}(X)$-module is a set $X$ and an action $\sout{\alpha_R : \mathsf{End^{op}}(X)\times X\to X.}$

${}$
In both cases, the action is function evaluation.

Eric Forgy (Jan 14 2021 at 21:56):

I see my mistake. My intuition wasn't too bad, but as often the case, I missed the mark :bow_and_arrow: :bulls_eye:

My action is too big. It contains all possible actions. The set $A$ is important. Hookoodanode :shrug:

The way I should probably think of it is the action
${}$
$\sout{\alpha: A\times X\to X}$
${}$
is more like a map
${}$
$\sout{\alpha: A\to\mathsf{End}(X).}$
${}$
For each $a\in A$, we get an endofunction $\sout{\alpha(a,-): X\to X.}$

Eric Forgy (Jan 14 2021 at 22:15):

Attempt #3

Now that I see it, is is almost embarassing. The answer is pretty much embedded in the puzzle.

An $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is a set $X$ and a function
${}$
$\sout{\alpha: A\times X\to X.}$

:face_palm:

Eric Forgy (Jan 15 2021 at 00:13):

As mentioned in my solution to Puzzle 3, an even better solution to Puzzle 2 is the following:

A left $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is a span
Left $A$-Module

A right $B$-module $X$ of a monoid object $B$ in $\mathsf{Set^{op}}$ is a span
Right $B$-module

An $(A,B)$-module $X$ of monoid objects $A,B$ in $\mathsf{Set^{op}}$ is a span

which can be seen as the composition of left and right modules, where a left $A$-module is a $(A,*)$-bimodule and a right $A$-module is a $(*,B)$-bimodule.

John Baez (Jan 15 2021 at 00:18):

Eric Forgy said:

---

Attempt #3

Now that I see it, is is almost embarassing. The answer is pretty much embedded in the puzzle.

An $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is a set $X$ and a function
${}$
$\alpha: A\times X\to X.$

No, that's not true. An action of a monoid object $A$ in $\mathsf{Set}$ is a set $X$ and a function $\alpha: A\times X\to X$ such that 2 equations hold (which I'm too lazy to write now). But I'm asking about an action of a monoid object in $\mathsf{Set}^{\rm{op}}$!

John Baez (Jan 15 2021 at 00:19):

So you have to turn around an arrow in your answer, and you need to take those 2 equations into account.

But there are also other surprises, which show up when you think a bit harder about this using Puzzle 1.

Eric Forgy (Jan 15 2021 at 00:23):

Thank you :pray:

I'll keep at this. I'm hoping this will pay off :sweat_smile:

John Baez (Jan 15 2021 at 00:25):

The ultimate answer here is incredibly simple, but it takes some thought to get it.

Eric Forgy (Jan 15 2021 at 00:28):

Ok. An action
${}$
$\alpha: A\times X\to X$
${}$
in $\mathsf{Set^{op}}$ is a function $X\to A\times X$ :thinking:

Eric Forgy (Jan 15 2021 at 00:31):

Is that right at least? Or do I need to swap products too, i.e. $X\to A\sqcup X$?

John Baez (Jan 15 2021 at 00:33):

All these puzzles are taking place in the category $\mathsf{Set}^{\mathrm{op}}$ where the monoidal structure is $\times$, defined by

$X \times Y = \{(x,y): x \in X, y \in Y\}$

John Baez (Jan 15 2021 at 00:33):

I just answered your question. I hope you see how I answered it.

John Baez (Jan 15 2021 at 00:34):

I didn't want to just say "yes" or "no"... I wanted to remind you of what you need to know.

Eric Forgy (Jan 15 2021 at 00:35):

I obviously have zero intuition when it comes to $\mathsf{op}$ stuff.

John Baez (Jan 15 2021 at 00:37):

Well, so what's the answer to your question "Is that right at least?"

John Baez (Jan 15 2021 at 00:37):

Yes? Or no?

Todd Trimble (Jan 15 2021 at 00:40):

John Baez said:

All these puzzles are taking place in the category $\mathsf{Set}^{\mathrm{op}}$ where the monoidal structure is $\times$, defined by

$X \times Y = \{(x,y): x \in X, y \in Y\}$

You have it now in the display line, but without that, it's potentially confusing because the cartesian product $\times$ in $\mathsf{Set^{op}}$ corresponds to the coproduct in $\mathsf{Set}$.

Eric Forgy (Jan 15 2021 at 00:44):

I don't know.

$X\times Y$ is a diagram and $\mathsf{op}$ 'ing it gives a different diagram so I am confused.

An action
${}$
$\alpha: A\times X\to X$
${}$
in $\mathsf{Set^{op}}$ is also a diagram. Since an arrow $S\to T$ in $\mathsf{Set^{op}}$ is an arrow $T\to S$ in $\mathsf{Set}$, there seems to be just enough ambiguity for me to get confused whether that should be $X\to A\times X$ or $X\to A\sqcup X$ in $\mathsf{Set}$ :pensive:

John Baez (Jan 15 2021 at 00:59):

Todd Trimble said:

John Baez said:

All these puzzles are taking place in the category $\mathsf{Set}^{\mathrm{op}}$ where the monoidal structure is $\times$, defined by

$X \times Y = \{(x,y): x \in X, y \in Y\}$

You have it now in the display line, but without that, it's potentially confusing because the cartesian product $\times$ in $\mathsf{Set^{op}}$ corresponds to the coproduct in $\mathsf{Set}$.

Yeah! So let's just all agree that we're using the product in Set to give $\mathsf{Set}^{\rm op}$ a monoidal structure which we are going to call $\times$.

John Baez (Jan 15 2021 at 01:00):

It's confusing no matter what you call it, so let's just call it $\times$.

John Baez (Jan 15 2021 at 01:02):

Eric Forgy said:

I don't know.

$X\times Y$ is a diagram and $\mathsf{op}$ 'ing it gives a different diagram so I am confused.

I said that that $X\times Y$ is our notation for the set $\{(x,y) : x \in X , y \in Y\}$. I wrote out that crud to emphasize that in these puzzles, that's what I always mean by $X \times Y$. Okay?

John Baez (Jan 15 2021 at 01:02):

I said it all before:

John Baez (Jan 15 2021 at 01:03):

John Baez said:

All these puzzles are taking place in the category $\mathsf{Set}^{\mathrm{op}}$ where the monoidal structure is $\times$, defined by

$X \times Y = \{(x,y): x \in X, y \in Y\}$

John Baez (Jan 15 2021 at 01:03):

So I told you what our monoidal category is:

A morphism from $X$ to $Y$ is a function $f: Y \to X$, and the tensor product is $\times$ as defined above!

John Baez (Jan 15 2021 at 01:03):

Once you know that, you know (in theory) what a monoid object is, in this monoidal category.

John Baez (Jan 15 2021 at 01:04):

Puzzle 1 was to figure out all possible monoid objects in this monoidal category.

John Baez (Jan 15 2021 at 01:05):

Puzzle 2, which relies heavily on puzzle 1, is to figure out all actions of monoid object in this monoidal category.

John Baez (Jan 15 2021 at 01:05):

So what's an action of a monoid object in this monoidal category?

Eric Forgy (Jan 15 2021 at 01:16):

Hold on...

I need to get my bearings before I can answer basic questions.

(You're talking to someone who has never really thought much about and has never worked with $\mathsf{op}$.)

Eric Forgy (Jan 15 2021 at 02:08):

In $\mathsf{Set}$ we have a product $\times$ and a coproduct $\sqcup.$ The product $\times$ is a universal diagram (a span actually) and the coproduct is a universal cospan. An object $X\times Y$ can be written in terms of elements of $X$ and $Y$ as
${}$
$X\times Y = \{(x,y) : x\in X, y\in Y\}.$
${}$
This object exists in both $\mathsf{Set}$ and $\mathsf{Set^{op}},$ but it is not the product of $X$ and $Y$ in $\mathsf{Set^{op}}$. It is actually their coproduct because a universal span in $\mathsf{Set}$ is a universal cospan in $\mathsf{Set^{op}}.$ More generally, a limit in $\mathsf{Set}$ is a colimit in $\mathsf{Set^{op}}$ and vise versa.

So, getting back to the puzzle, an action
${}$
$\alpha: A\times X\to X$
${}$
in $\mathsf{Set^{op}}$ is a morphism from the object $A\times X$ to the object $X,$ which corresponds to a function
${}$
$\alpha^\mathsf{op}: X\to A\times X$
${}$
from the object $X$ to the object $A\times X.$
${}$
The only function I can think of would pick an element $a\in A$ and we can write
${}$
$\sout{\alpha^\mathsf{op} = a: X\to A\times X,\quad x\mapsto (a,x)}.$

Eric Forgy (Jan 15 2021 at 02:28):

This feels a little weird, but if we have a function
${}$
$\alpha: A\times X\to X$
${}$
and choose an element $a\in A$, we can define a function
${}$
$\alpha^\mathsf{op}: X\to A\times X,\quad x\mapsto (a,\alpha(a,x)).$

John Baez (Jan 15 2021 at 03:16):

I don't know what you're doing. I suggest you start by writing down the definition of a module in the monoidal category I so painstakingly specified.

John Baez (Jan 15 2021 at 03:17):

Then, after you've written down the definition, including the two equations that need to hold, you could seek to analyze what a module in that monoidal category actually amounts to.

Jason Erbele (Jan 15 2021 at 04:00):

Eric Forgy said:

This feels a little weird, but if we have a function
${}$
$\alpha: A\times X\to X$

Since $\alpha$ is a morphism in $\mathrm{Set}^\mathrm{op}$ , this is technically the opposite of a function. Not sure if this will help you, but it might be easier to think about the honest-to-goodness functions $X \rightarrow A \times X$ of which there are many. Generally, crazy stuff could happen in both the $A$ part and in the $X$ part of $A \times X$, but you want to find the one(s) that play(s) well with the definition of a module when you op it/them.

Eric Forgy (Jan 15 2021 at 04:11):

John Baez said:

I don't know what you're doing. I suggest you start by writing down the definition of a module in the monoidal category I so painstakingly specified.

This is what I'm trying to do :pensive:

From the nLab:

Definition

Let $(\mathcal{V}, \otimes, I)$ be a [[monoidal category]] and $A$ a [[monoid object]] in $\mathcal{V}$, hence an object $A \in \mathcal{V}$ equipped with a multiplication morphism

$\cdot : A \otimes A \to A$

and a unit element

$e : I \to A$

satisfying the [[associativity law]] and the [[unit law]].

Definition

A (left) module over $A$ in $(\mathcal{V}, \otimes, I)$ is

• an [[object]] $N \in \mathcal{V}$

• equipped with a morphism

$\lambda : A \otimes N \to N$

in $\mathcal{V}$

such that this satisfies the axioms of an [[action]], in that the following are [[commuting diagrams]] in $\mathcal{V}$:

Puzzle 1 was about sorting out the monoid objects and I think I got that.

Puzzle 2 is defining the module. Looking at the definition above, the thing I need is an action
${}$
$\alpha: A\times X\to X.$
${}$

I was trying to define the action, but obviously failing :pensive:

Eric Forgy (Jan 15 2021 at 04:13):

Jason Erbele said:

Eric Forgy said:

This feels a little weird, but if we have a function
${}$
$\alpha: A\times X\to X$

Since $\alpha$ is a morphism in $\mathrm{Set}^\mathrm{op}$ , this is technically the opposite of a function. Not sure if this will help you, but it might be easier to think about the honest-to-goodness functions $X \rightarrow A \times X$ of which there are many. Generally, crazy stuff could happen in both the $A$ part and in the $X$ part of $A \times X$, but you want to find the one(s) that play(s) well with the definition of a module when you op it/them.

Thanks Jason. I was missing the "usual equations" :face_palm:

Eric Forgy (Jan 15 2021 at 04:16):

I understood the monoid object by looking at the opposite functions in $\mathsf{Set}.$ To understand an action in $\mathsf{Set^{op}}$, I was trying to look at the opposite function $X\to A\times X$, but I failed to consider the "usual equations". I'll keep trying...

John Baez (Jan 15 2021 at 04:25):

Remember, in an ordinary action of an ordinary monoid A on a set X we demand two equations, the left unit law

1x = x

and associativity

a(bx) = (ab)x

You should be familiar with these in from the definition of a module, which is an action of a monoid object in $(\mathsf{Vect}, \otimes)$ on an object in $(\mathsf{Vect}, \otimes)$.

So we abstract these and write these as commutative diagrams to get the definition of a monoid object in any monoidal category on any object of that category.

The point of Puzzle 2 is to see what these imply in our example $(\mathsf{Set}^{\rm op}, \times)$.

John Baez (Jan 15 2021 at 04:28):

I don't think you ever really solved Puzzle 1, which is to take the axioms for a monoid object and see what they imply in $(\mathsf{Set}^{\rm op}, \times)$. In my own answer I sketched the miracle that happens there: it turns out that for any set there's exactly one way to make it into a monoid object in $(\mathsf{Set}^{\rm op}, \times)$. To prove this you need to use the equations in the definition of monoid object.

A similar miracle happens in Puzzle 2, and the argument is similar.

Eric Forgy (Jan 15 2021 at 04:37):

Quick recap of Puzzle 1...

A monoid object in $\mathsf{Set^{op}}$ is, first of all, a set (what else could it be?) $A$. But we also need a multiplication
${}$
$\mu: A\times A\to A$
${}$
and a unit
${}$
$\eta:*\to A$
${}$
satisfying equations.

To understand $\mu$, you can look at the opposite function $A\to A\times A$. This is the diagonal map $a\mapsto (a,a)\in A\times A.$

To understand $\eta$, you can look at the opposite function $A\to *$. This just sends every element of $A$ to the one element in $*.$

A quick doodle confirms the equations are satisfied.

Eric Forgy (Jan 15 2021 at 04:43):

For Puzzle 2...

A (left) $A$-module is a set $X$, a monoid object $A$ and an action
${}$
$\alpha: A\times X\to X$
${}$
satisfying equations.

To understand $\alpha$, you can look at the opposite function $X\to A\times X.$ This is a function that for a given $a\in A$ maps $x\mapsto (a,x).$

A quick doodle confirms the equations are satisfied.

Eric Forgy (Jan 15 2021 at 04:47):

Here is the doodle:

We need to check
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (a,x)\overset{\mu^\mathsf{op}\times\mathsf{id}}{\mapsto} (a,a,x)$
${}$
is the same as
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (a,x)\overset{\mathsf{id}\times \alpha^\mathsf{op}}{\mapsto} (a,a,x).$
${}$
Check :check_mark:

Eric Forgy (Jan 15 2021 at 04:50):

We also need to check
${}$
$x\mapsto (\bullet,x)$
${}$
is the same as
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (a,x) \overset{\eta^\mathsf{op}\times\mathsf{id}}{\mapsto} (\bullet,x).$
${}$
Check :check_mark:

Jason Erbele (Jan 15 2021 at 04:53):

Will any other way of choosing $\alpha$ satisfy these equations?

Eric Forgy (Jan 15 2021 at 04:53):

Maybe? :sweat_smile:

Eric Forgy (Jan 15 2021 at 04:53):

Actually, no. I don't think so.

Eric Forgy (Jan 15 2021 at 04:55):

This was my first thought, but I striked it out above and generalized to
${}$
$\alpha^\mathsf{op}: x\mapsto (a,\alpha^\mathsf{op}(a,x))$
${}$
but then this didn't satisfy the unit equations.

Jason Erbele (Jan 15 2021 at 04:55):

If there is another way, can you find it? If there isn't, can you prove that there is no other way of choosing $\alpha$ to satisfy these equations?

Eric Forgy (Jan 15 2021 at 04:56):

Also, because the diagonal map, there can only be one chosen $a\in A$. We have freedom which $a$ we choose, but once its chosen, that is your $\alpha.$

Jason Erbele (Jan 15 2021 at 04:58):

Sure. You have a collection of $\alpha$ s indexed by your choice of $a \in A$, but that's all one way of choosing $\alpha$. :upside_down:

Eric Forgy (Jan 15 2021 at 05:00):

I started with a more general
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (a,x')\overset{\mu^\mathsf{op}\times\mathsf{id}}{\mapsto} (a,a,x')$
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (a,x')\overset{\mathsf{id}\times \alpha^\mathsf{op}}{\mapsto} (a,a,x''),$
${}$
but that only commutes if $\alpha^\mathsf{op}(x) = (a,x).$

Jason Erbele (Jan 15 2021 at 05:06):

I'm just going to re-state this:
Jason Erbele said:

Generally, crazy stuff could happen in both the $A$ part and in the $X$ part of $A \times X$, but you want to find the one(s) that play(s) well with the definition of a module when you op it/them.

Eric Forgy (Jan 15 2021 at 05:10):

Sorry, I'm feeling dense. Did I miss something or is what I said correct? :sweat_smile:

To understand an action $\alpha: A\times X\to X$ in $\mathsf{Set^{op}}$, we can look at the opposite function $\alpha^\mathsf{op}: X\to A\times X$ in $\mathsf{Set}$ and this function is $x\mapsto (a,x)$, so $\alpha$ in $\mathsf{Set^{op}}$ is the opposite of this function. Right?

Jason Erbele (Jan 15 2021 at 05:23):

You're looking for an action $\alpha: A \times X \rightarrow X$. This is essentially the same as a function from $X$ to $A \times X$. I think you've got that point. But the point I'm making in the bit I quoted is that you can have a function $f: \{x,y,z\} \rightarrow \{a,b,c\} \times \{x,y,z\}$ that, let's say, maps $x$ to $(a,x)$, maps $y$ to $(c,z)$, and maps $z$ to $(a,z)$. So a function in general can have crazy stuff going on in both the $A$ and $X$ parts.

Eric Forgy (Jan 15 2021 at 05:27):

Right. That is why it is important to check "the equations" :blush:

Jason Erbele (Jan 15 2021 at 05:27):

Now that particular function, when opped, will not be (an instance of) the action you are looking for, and it looks like see why. But the point is that it seems you are ignoring a huge swath of potential actions.

Eric Forgy (Jan 15 2021 at 05:29):

Oh. Ok. Thanks. I was too dense to realize that is what you meant :sweat_smile:

I'll think some more :pray:

Jason Erbele (Jan 15 2021 at 05:31):

If you take the special case of $A = X$, you have already identified an important morphism (in puzzle 1) that looks like $A \times X \rightarrow X$. Have you checked the action equations on that one?

Eric Forgy (Jan 15 2021 at 05:37):

I think I see what you mean. Instead of choosing an element $a\in A$, we want a function $f:X\to A.$

Jason Erbele (Jan 15 2021 at 05:39):

Will something like that always work? Is there a way to express what you were doing with $a \in A$ as a special case of this new attempt?

Eric Forgy (Jan 15 2021 at 05:39):

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

Check :check_mark:

Eric Forgy (Jan 15 2021 at 05:40):

$x\mapsto (\bullet,x)$
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (f(x),x) \overset{\eta^\mathsf{op}\times\mathsf{id}}{\mapsto} (\bullet,x).$

Check :check_mark:

Eric Forgy (Jan 15 2021 at 05:43):

Jason Erbele said:

Will something like that always work? Is there a way to express what you were doing with $a \in A$ as a special case of this new attempt?

What I had before was equivalent to having just one $f:X\to A$ that sends every element of $X$ to the element $a\in A.$

Jason Erbele (Jan 15 2021 at 05:43):

Now for the big question: Is there any other way that you could pick an action?

Eric Forgy (Jan 15 2021 at 05:44):

Of course not!

(I wish I was that confident :sweat_smile: But I don't think so :blush: )

Eric Forgy (Jan 15 2021 at 05:44):

That seems to be the most general $\alpha^\mathsf{op}$ that will satisfy the conditions.

Jason Erbele (Jan 15 2021 at 05:48):

You've played with what I referred to as "craziness" that can happen in the $A$ part of $X \rightarrow A \times X$ with that $f : X \rightarrow A$. Where else did I mention "craziness" could happen in a function $X \rightarrow A \times X$ ?

Eric Forgy (Jan 15 2021 at 05:49):

Ok ok :blush:

Back to the drawing board. Thank you :raised_hands:

Eric Forgy (Jan 15 2021 at 05:54):

There is one step in the equations I'm not sure about. There is an arrow $I\otimes N\to N$, which gets reversed to $X\to *\times X$. I thought this should be $x\mapsto (\bullet, x)$ with no craziness, but if that is the case, then there can not be any craziness in the $X$ part of $X\to A\times X$ :thinking:

Eric Forgy (Jan 15 2021 at 05:56):

So if $\alpha^\mathsf{op}$ involved both a function $f:X\to A$ and a function $g:X\to X$ with
${}$
$\alpha: x\mapsto (f(x),g(x))$
${}$
then I need that identity equation to be $x\mapsto (\bullet,g(x))$ :thinking:

Eric Forgy (Jan 15 2021 at 05:58):

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

Check :check_mark:

Eric Forgy (Jan 15 2021 at 05:59):

$x\mapsto (\bullet,x)$
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (f(x),g(x)) \overset{\eta^\mathsf{op}\times\mathsf{id}}{\mapsto} (\bullet,g(x)).$

Doesn't check :x:

Eric Forgy (Jan 15 2021 at 06:02):

The only way to make it check is if the map $X\to *\times X$ knows about $g$ somehow :thinking:

Jason Erbele (Jan 15 2021 at 06:06):

So what can you conclude about what happens in the $X$ part?

Eric Forgy (Jan 15 2021 at 06:08):

I don't think $X\to *\times X$ can know about $g$, so the $X$ part cannot vary and we have
${}$
$\alpha^\mathsf{op}: x\mapsto (f(x),x).$

Jason Erbele (Jan 15 2021 at 06:09):

Is there anything else that can happen, then?

Eric Forgy (Jan 15 2021 at 06:09):

Not that I can think of :sweat_smile:

Jason Erbele (Jan 15 2021 at 06:10):

How confident are you in that answer?

Eric Forgy (Jan 15 2021 at 06:11):

I'm not confident about anything when it comes to this stuff :sweat_smile:

Eric Forgy (Jan 15 2021 at 06:11):

But fairly confident I guess.

Eric Forgy (Jan 15 2021 at 06:12):

The thing I'm not sure about is if that $X\to *\times X$ is allows to know about $g$, but I don't think it is. If that is the case, my confidence increases.

Jason Erbele (Jan 15 2021 at 06:13):

Okay. Under what circumstances could that map know about $g$? Are there any specific problems that could occur otherwise?

Jason Erbele (Jan 15 2021 at 06:14):

There is definitely one circumstance that you have identified where things do work. What does that map know about $g$ in that obviously good case?

Jason Erbele (Jan 15 2021 at 06:15):

Or more simply, what is $g$ in that case?

Eric Forgy (Jan 15 2021 at 06:16):

In the case that appears to work, $g = \mathsf{id}_X.$

Eric Forgy (Jan 15 2021 at 06:23):

In that case, we can write $\alpha^\mathsf{op}$ explicitly as
${}$
$\alpha^\mathsf{op} = (f\times\mathsf{id}_X)\circ \mu^\mathsf{op}.$

Jason Erbele (Jan 15 2021 at 06:27):

Right. In the simplest case where a function from $X$ to $X$ doesn't have to be id, $X$ has two elements. There are four functions in that case. What bad things happen with some of those?

Jason Erbele (Jan 15 2021 at 06:27):

(I like to think about simple concrete examples if I'm completely drawing a blank)

Eric Forgy (Jan 15 2021 at 06:28):

Looking up the definition on unitor on nLab, I see
${}$
$I\otimes N\to N$
${}$
is a natural isomorphism, so that in any case, I think any such $g$ should have an inverse.

Eric Forgy (Jan 15 2021 at 06:30):

Wikipedia calls it the right identity.

Jason Erbele (Jan 15 2021 at 06:32):

So it's not just any isomorphism, it's a natural isomorphism. That means more that just having an inverse. And the wikipedia entry gives you a further clue.

Eric Forgy (Jan 15 2021 at 06:33):

From this diagram,

Eric Forgy said:

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

Check :check_mark:

we see there is no restriction on $g$ so it is fine.

Eric Forgy (Jan 15 2021 at 06:34):

The diagram we need to worry about is this one:

Eric Forgy said:

$x\mapsto (\bullet,x)$
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (f(x),g(x)) \overset{\eta^\mathsf{op}\times\mathsf{id}}{\mapsto} (\bullet,g(x)).$

Doesn't check :x:

Eric Forgy (Jan 15 2021 at 06:37):

For that diagram to commute, we need
${}$
$x\mapsto (\bullet,g(x))$
${}$
$x\overset{\alpha^\mathsf{op}}{\mapsto} (f(x),g(x)) \overset{\eta^\mathsf{op}\times\mathsf{id}}{\mapsto} (\bullet,g(x))$

Jason Erbele (Jan 15 2021 at 06:38):

So what happens when you choose a different $g$ for your proposed action? And what $g$ did we already see definitely does work?

Eric Forgy (Jan 15 2021 at 06:38):

But the first line needs to be a natural isomorphism so if $X$ has two elements, then 2 of the possible 4 $g$ s will be invertible and two will not.

Jason Erbele (Jan 15 2021 at 06:39):

Right. In the two-element $X$ case there are two very bad situations where $g$ isn't an isomorphism.

Eric Forgy (Jan 15 2021 at 06:40):

One of the invertible ones is $\mathsf{id}_X.$

The other invertible one maps the one element to the other and vice versa.

The two bad ones map both elements to the same one element (two choices).

Jason Erbele (Jan 15 2021 at 06:42):

Does the swapping case give a natural isomorphism? (and what does that "natural" bit really mean, anyway?)

Eric Forgy (Jan 15 2021 at 06:44):

Two-sided inverse? :sweat_smile:

Jason Erbele (Jan 15 2021 at 06:45):

That's what the "isomorphism" part gives you. :wink:

Eric Forgy (Jan 15 2021 at 06:48):

I don't have a good enough intuition to answer your question about natural isomorphism. I think when there are just two elements, it might be natural (since there is no choice involved), but if there are more than 2 elements, maybe it isn't "natural"?

Jason Erbele (Jan 15 2021 at 06:51):

A simple intuition about whether a morphism is natural would be: is the morphism as boring as possible for that context?

Eric Forgy (Jan 15 2021 at 06:53):

I'm not confident, but I think in this situation, the only natural isomorphism $X\to X$ is the identity morphism.

Jason Erbele (Jan 15 2021 at 06:57):

If I had done a better job with this last part, you would be a lot more confident that this really is the only way to go:

Eric Forgy said:

$\alpha^\mathsf{op}: x\mapsto (f(x),x).$

Jason Erbele (Jan 15 2021 at 06:58):

Even if you could do something else for $g$, what you would get would be isomorphic to this.

Eric Forgy (Jan 15 2021 at 06:58):

Don't say that. How can you do better under the circumstances? I am the worst student ever :joy:

Eric Forgy (Jan 15 2021 at 07:00):

That sounds good enough for me! Move on! :smiley: :runner: :blush:

Eric Forgy (Jan 15 2021 at 07:01):

So an action $\alpha: A\times X\to X$ on $\mathsf{Set^{op}}$ is the same as a function $f:X\to A$ in $\mathsf{Set}.$

Jason Erbele (Jan 15 2021 at 07:01):

I don't think you qualify as the worst student. You are actually working at figuring stuff out. That's already lightyears beyond some of the students I have in the classes I teach.

Eric Forgy (Jan 15 2021 at 07:02):

Out of curisoity, how is that so obvious to everyone except me? :sweat_smile:

Eric Forgy (Jan 15 2021 at 07:03):

What is the trick I am missing?

Jason Erbele (Jan 15 2021 at 07:05):

It wasn't immediately obvious to me that that was the answer to the puzzle. It only seemed obvious to me in retrospect. You try everything (in this case literally), and only a few things end up actually working.

Eric Forgy (Jan 15 2021 at 07:08):

Yes. A lot of the way I learn things is a random walk. Try. Fail. Try again. This goes for everything in life. Not just mathematics. By making every possible mistake, you end up learning more than getting everything right the first time. I now know about 10,000 ways how NOT to get a module on $\mathsf{Set^{op}}$ :joy:

Jason Erbele (Jan 15 2021 at 07:09):

Though I might be tempted to phrase it as "an action on $\mathrm{Set}^\mathrm{op}$ is the same as the graph of a function." But that might be more opaque for someone less familiar with the algebraic definition of the graph of a function.

Jason Erbele (Jan 15 2021 at 07:11):

By "try everything," I don't necessarily mean making every possible mistake. We systematically looked at every possible function $X \rightarrow A \times X$ and found the ones that could work with the definition.

Eric Forgy (Jan 15 2021 at 07:12):

(A quick nLab search later) Fair enough, but if you know a function, you know its graph, right? So the important thing is to know the function and then you know the action.

Eric Forgy (Jan 15 2021 at 07:12):

Jason Erbele said:

By "try everything," I don't necessarily mean making every possible mistake. We systematically looked at every possible function $X \rightarrow A \times X$ and found the ones that could work with the definition.

That is only because you were here to guide me. Left to my own devices, I probably would have tried almost everything :sweat_smile:

Jason Erbele (Jan 15 2021 at 07:12):

Right. The graph of a function is "the same as" a function in the same way that an action on $\mathrm{Set}^\mathrm{op}$ is "the same as" a function.

Jason Erbele (Jan 15 2021 at 07:17):

A systematic approach to trying everything certainly makes it easier to know when you're done. I guess I should probably qualify that – everything that parses. If you try stuff that doesn't even parse, you will never finish.

Eric Forgy (Jan 15 2021 at 07:26):

So here is

Atempt #4
${}$

Puzzle 2: What’s a module of a monoid object in $\mathsf{Set^{op}}$?

${}$
A left $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is a function $f:X\to A.$

Eric Forgy (Jan 15 2021 at 07:35):

We see this by considering the definition of a module. A module in $\mathsf{Set^{op}}$ is a set $X$ and a monoid object $A$ together with an action
${}$
$\alpha: A\times X\to X$
${}$
satisfying some conditions.

This action can be understood by looking at the opposite function
${}$
$\alpha^\mathsf{op}: X\to A\times X$
${}$
in $\mathsf{Set}.$ Such a function is determined by a function $f:X\to A$ so that $\alpha^\mathsf{op}$ can be given explicitly in terms of $f$ (and $\mu^\mathsf{op}$) via
${}$
$\alpha^\mathsf{op} = (f\times\mathsf{id}_X)\circ\mu^\mathsf{op},$
${}$
where $\mu^\mathsf{op}$ is the diagonal map $X\to X\times X.$

Eric Forgy (Jan 15 2021 at 07:36):

The opposite of $\alpha^\mathsf{op}$ in $\mathsf{Set^{op}}$ is an action $A\times X\to X.$

John Baez (Jan 15 2021 at 16:04):

Eric Forgy said:

For Puzzle 2...

A (left) $A$-module is a set $X$, a monoid object $A$ and an action
${}$
$\alpha: A\times X\to X$
${}$
satisfying equations.

Here the arrow is not a function but a morphism in $\mathsf{Set}^{\rm op}$, right? (I never ask rhetorical questions.) I would avoid using the same kind of arrow for both functions and morphisms in $\mathsf{Set}^{\rm op}$; it's confusing. Personally I would only talk about functions.

To understand $\alpha$, you can look at the opposite function $X\to A\times X.$ This is a function that for a given $a\in A$ maps $x\mapsto (a,x).$

Is this your guess as to what $\alpha$ does? You're not making it clear that this is a guess. You seem to be making one guess $\alpha_a$ for each element $a\in A$. You seem to be guessing that for any choice of $a \in A$,

$\alpha_a(x) = (a,x)$

will give you an action of $A$.

Is that what you're claiming?

(You don't talk like a mathematician, so I often have trouble understanding you.)

Jason Erbele (Jan 15 2021 at 16:38):

John, I tried to guide Eric last night through this puzzle, and that does seem to be what he was claiming about what $\alpha$ does, before I jumped in. His final claim in this thread is at around 11:30 p.m. (in our time zone, at least).

Eric Forgy (Jan 15 2021 at 17:01):

Good morning :coffee:

Thanks again Jason. With Jason's help, my most recent attempt at a solution is three comments up, i.e. my last comment :blush:

I'm fairly confident this last one is correct.

Eric Forgy (Jan 15 2021 at 17:15):

Eric Forgy said:

So here is

---

Atempt #4
${}$

Puzzle 2: What’s a module of a monoid object in $\mathsf{Set^{op}}$?

${}$
A left $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is a function $f:X\to A.$

---

Eric Forgy (Jan 15 2021 at 18:30):

But then every function corresponds to a span, so another answer is:

A left $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is a span $X\leftarrow X\rightarrow A.$

Eric Forgy (Jan 15 2021 at 18:46):

I need to check, but I think a safe guess is that a right $B$-module is a span
${}$
$B\leftarrow X\rightarrow X$
${}$
and an $(A,B)$-bimodule is the composition of the two, i.e.
${}$
$B\leftarrow X\rightarrow A$
${}$
and the composition of an $(A,B)$-bimodule $X$ with a $(B,C)$-bimodule $Y$ is the composition of the two spans
${}$
$C\leftarrow Y\times_B X\rightarrow A.$
${}$
If this guess isn't correct, then something very similar looking to this should be correct (but I think this is correct).

Eric Forgy (Jan 15 2021 at 18:48):

I'm fairly confident about this, because:
${}$

• A left $A$-module should be equivalent to an $(A,*)$-bimodule
• A right $B$-module should be equivalent to a $(*,B)$-bimodule

Eric Forgy (Jan 15 2021 at 18:51):

A $($,)-bimodule is a monoid object :blush:

John Baez (Jan 15 2021 at 18:52):

Eric Forgy said:

But then every function corresponds to a span, so another answer is:

A left $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is a span $X\leftarrow X\rightarrow A.$

I don't believe that. I believe a left $A$-module $X$ of a monoid object $A$ in $\mathsf{Set^{op}}$ is the same as a span of the form

$X \stackrel{1}{\leftarrow} X\stackrel{f}{\rightarrow} A$

John Baez (Jan 15 2021 at 18:52):

In other words, the morphism from $X$ to itself in this span must be the identity.

John Baez (Jan 15 2021 at 18:53):

Such a span is a long-winded way of talking about a function $f: X \to A$, so I'd say the answer is "a function from $X$ to $A$".

Eric Forgy (Jan 15 2021 at 18:53):

Yes. That is what I meant. That is how you turn a function into a span (I've seen it denoted $f_*$).

John Baez (Jan 15 2021 at 18:54):

Okay, good. Please try to say what you mean... if you say "a span $X \leftarrow X \to A$", that means an arbitrary span of this form.

Eric Forgy (Jan 15 2021 at 18:55):

Ok. I still need to learn when I need to be explicit and when not. It seems you guys are always leaving off stuff like that :sweat_smile:

John Baez (Jan 15 2021 at 18:58):

We leave out unimportant things. You're leaving out crucial things. It's like I asked "what's 2+2" and you said "an even number", and then it turns out you meant 4. Part of my goal in this discussion is to try to teach you to communicate with mathematicians. This involves a higher level of precision.

Maybe part of the problem is this: when reviewing material one can leave out details while making it clear details are being left out, so the person reading it can ask questions. But when answering puzzles, or proving theorems (which is basically the same thing), one can't leave out details that make the difference between truth and falsity.

Eric Forgy (Jan 15 2021 at 18:59):

Ok. Thank you :pray:

Eric Forgy (Jan 15 2021 at 19:16):

I think it does help to think of modules in $\mathsf{Set^{op}}$ as spans though since:

• An $(A,B)$-bimodule is a span
• A left $A$-module is quivalent to an $(A,*)$-bimodule, i.e. span
• A right $B$-module is equivalent to a $(*,B)$-bimodule, i.e. a span

John Baez (Jan 15 2021 at 19:18):

Yeah, it's useful to think of modules as a special case of bimodules, and morphisms as a special case of spans.

John Baez (Jan 15 2021 at 19:19):

But if you want to give the royal answer to

"what's a module of a monoid object in $(\mathsf{Set}^{\rm op}, \times)$?"

that is, the simplest, clearest answer, it's:

"a function".

John Baez (Jan 15 2021 at 19:20):

There are lots of puzzles like this in category theory, where a fancy-sounding concept turns out to be something you already know. For such puzzles one always seeks the simplest answer.

John Baez (Jan 15 2021 at 19:20):

Similarly, the royal answer to

"what's a bimodule between monoid objects in $(\mathsf{Set}^{\rm op}, \times)$?"

is

"a span of sets"

or perhaps even more clearly:

"a pair of functions with the same domain".

Eric Forgy (Jan 15 2021 at 19:21):

John Baez said:

But if you want to give the royal answer to

"what's a module of a monoid object in $(\mathsf{Set}^{\rm op}, \times)$?"

that is, the simplest, clearest answer, it's:

"a function".

That is a left module.

What is a right module of a monoid in $(\mathsf{Set^{op}},\times)$?

John Baez (Jan 15 2021 at 19:21):

Also a function.

John Baez (Jan 15 2021 at 19:24):

I didn't specify left or right modules because for a commutative monoid object there's a natural bijection between left and right modules.

John Baez (Jan 15 2021 at 19:24):

So it doesn't make any difference in this particular question.

Eric Forgy (Jan 15 2021 at 19:25):

Ok. I wondered about that.

John Baez (Jan 15 2021 at 19:27):

For noncommutative monoid objects it's necessary to say "left" or "right" module.... except in some other situations where you have a recipe to turn left modules into right modules and vice versa, which I don't want to get into.

John Baez (Jan 15 2021 at 19:28):

But for commutative monoid objects it's rare to explicitly say whether the module is a left or right one. For example, nobody ever talks about "left vector spaces" and "right vector spaces".

Eric Forgy (Jan 15 2021 at 19:29):

Sure. That makes sense. Thank you.

John Baez (Jan 15 2021 at 19:31):

Sometimes I work with "quaternionic vector spaces" - and then you gotta distinguish between left and right vector spaces! (They're just left or right $\mathbb{H}$-modules, but they act so much like vector spaces in some ways that it's tempting to call them quaternionic vector spaces.)