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Stream: learning: questions

Topic: Bimodules vs Spans: Puzzle 3

Eric Forgy (Jan 14 2021 at 11:39):

(Note: This stream was split from Bimodules vs Spans: Puzzle 1 and Bimodules vs Spans: Puzzle 2. Some comments here refer to those streams and contains links to the respective posts.)

Puzzle 3: What’s a bimodule of a pair of monoid objects in $\mathsf{Set^{op}}$?

[Edit: This solution is completely wrong. See Attempt #2 below.]

A bimodule would be a set $X$ with compatible left and right actions.

Let
${}$
$\sout{L = \mathsf{End}(X)\times X}$
${}$
and
${}$
$\sout{R = \mathsf{End^{op}}(X)\times X}$
${}$
so that the actions can be written
${}$
$\sout{\alpha_L: L\to X\quad\text{and}\quad\alpha_R: R\to X.}$
${}$

Together, the actions form a cospan
${}$
$\sout{L\rightarrow X\leftarrow R}$
${}$
and pulling this back enforces their compatibility resulting in a span
${}$
$\sout{L\leftarrow L\times_X R\rightarrow R.}$
${}$
We can go one step further by forming the spans
${}$
$\sout{\mathsf{End}(X)\leftarrow L\rightarrow X}$
${}$
$\sout{X\leftarrow R\rightarrow \mathsf{End^{op}}(X)}$
${}$
so that the composition of these two spans is
${}$
$\sout{\mathsf{End}(X)\leftarrow L\times_X R\rightarrow\mathsf{End^{op}}(X).}$

Morgan Rogers (he/him) (Jan 14 2021 at 11:49):

Ahh Eric, check what monoidal product is in play in $\mathbf{Set}^{\mathrm{op}}$!! A "monoid object in [monoidal category $\mathcal{C}$ with monoidal product $\otimes$ and monoidal unit $I$]" needs maps $M \otimes M \to M$ and $I \to M$

Morgan Rogers (he/him) (Jan 14 2021 at 11:50):

John Baez said:

A set is a monoid object in the monoidal category $(\mathsf{Set}^{\mathrm{op}}, +)$.

We can define bimodules for monoid objects in any monoidal category.

A span of sets is a bimodule in $(\mathsf{Set}^{\mathrm{op}},+)$.

John mentioned already what monoidal product you should be using here in the message quoted above :wink:

Eric Forgy (Jan 14 2021 at 11:53):

Hi Morgan :wave:

I'm afraid I don't follow :thinking:

I was going by what the original post said:

Let’s work in the category $\mathsf{Set^{op}}$, where a morphism $f:S→T$ is a function from $T$ to $S$. This is a monoidal category with $×$ as its tensor product. So, we can talk about monoid objects in $\mathsf{Set^{op}}$.

Oscar Cunningham (Jan 14 2021 at 11:58):

Hmmm, does $\times$ there mean the product in $\mathbf{Set}$ or in $\mathbf{Set}^\mathrm{op}$?

Eric Forgy (Jan 14 2021 at 12:15):

I certainly made some mistakes (and hope to learn from them), but I don't think I got that wrong :sweat_smile:

This comment makes me think we really want $\times$ for $\mathsf{Set^{op}}.$

Eric Forgy (Jan 14 2021 at 12:17):

So I was using $(\mathsf{Set^{op}}, \times, *)$ (where $* = \{\bullet\}$, i.e. the terminal set).

The dual problem with $(\mathsf{Set},\sqcup,\emptyset)$ is also interesting :blush:

John Baez (Jan 14 2021 at 15:52):

Eric Forgy said:

Puzzle 1: What’s a monoid object in $\mathsf{Set^{op}}$?

The "object" is a set $X$, but for this to be a monoid object, we need a multiplication

$\mu: X\times X\to X.$

and a unit

$\eta: *\to X$

satisfying the conditions. I don't think the answer is complete without specifying those :thinking:

---

Attempt #1: Before looking at Puzzle #2

I think a multiplication $X\times X\to X$ in $\mathsf{Set^{op}}$ is a function $X\to X\times X$ in $\mathsf{Set}$ and the only function that would make sense to me is
${}$
$x\mapsto (x,x)\in X\times X.$
${}$
A unit $*\to X$ in $\mathsf{Set^{op}}$ is a function $X\to *$ in $\mathsf{Set}$ and that is unique since $*$ is terminal.

A quick doodle confirms these satisfy associativity and unit laws.

Right! And the really cool part is that this is the only possible choice of maps $\mu: X \to X \times X$ and $\eta: X \to 1$ obeying (co)associativity and the (co)unit laws. (We usually put the word "co" in here since the maps are going backwards.) So while in $\mathsf{Set}$ you need to specify $\mu$ and $\eta$ to say how we are treating our set $X$ as a monoid, in $\mathsf{Set}^{\mathrm{op}}$ you don't: there's only one choice!

John Baez (Jan 14 2021 at 15:57):

Morgan Rogers (he/him) said:

John Baez said:

A set is a monoid object in the monoidal category $(\mathsf{Set}^{\mathrm{op}}, +)$.

We can define bimodules for monoid objects in any monoidal category.

A span of sets is a bimodule in $(\mathsf{Set}^{\mathrm{op}},+)$.

John mentioned already what monoidal product you should be using here in the message quoted above :wink:

But my remark was probably confusing, because the coproduct + in $\mathsf{Set}^{\mathrm{op}}$ is exactly what we usually call the product $\times$ in $\mathsf{Set}$. Anyway, this is the monoidal product I want: the coproduct in $\mathsf{Set}^{\mathrm{op}}$, which we should probably write as $\times$ because it's the familiar set

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

which is the product in $\mathsf{Set}$.

Morgan Rogers (he/him) (Jan 14 2021 at 16:00):

It was my mistake, then, not Eric's! Thanks for clarifying.

John Baez (Jan 14 2021 at 16:06):

Sure!

Anyway, Eric was completely correct in his first attempt to solve Puzzle 1.

Here's a more general way to think about this puzzle. In any monoidal category where the monoidal structure is given by coproduct, every object $x$ becomes a monoid in a unique way.

The unit $\eta: 0 \to x$ is unique because $0$ is initial. The left and right unit laws force $\mu: x + x \to x$ to be the codiagonal or 'fold' map $\nabla: x + x \to x$.

John Baez (Jan 14 2021 at 16:11):

$\Delta$ is associative so $x$ becomes a monoid this way. In fact it's commutative.

John Baez (Jan 14 2021 at 16:12):

Even better, the maps $\eta$ are $\Delta$ are natural, every morphism in our category becomes a morphism of commutative monoids.

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

Btw, I've the split the original stream into 3 streams, one for each puzzle, so this is now the stream for Puzzle 3 :blush:

Eric Forgy (Jan 14 2021 at 23:18):

My answer above for Puzzle 3 is completely wrong, so starting over :face_palm:

Given a monoid object $A$, i.e. a set, a left $A$-module is a set $X$ together with a function
${}$
$\sout{\alpha_L: A\times X\to X.}$
${}$
Similarly, a right $B$-module is a set $X$ together with a function
${}$
$\sout{\alpha_R:X\times B\to X.}$

Attempt #2

Given monoid objects $A$ and $B$, an $(A,B)$-bimodule is a set $X$ together with compatible functions
${}$
$\sout{\alpha_L: A\times X\to X.}$
${}$
and
${}$
$\sout{\alpha_R:X\times B\to X.}$
${}$
These functions form a cospan
${}$
$\sout{A\times X\rightarrow X\leftarrow X\times B}$
${}$
that can be pulled back (enforcing compatibility) forming a span
${}$
$\sout{A\times X\leftarrow (A\times X)\times_X (X\times B)\rightarrow X\times B.}$
${}$
Hence, an $(A,B)$-bimodule $X$ in $\mathsf{Set^{op}}$ is a span
${}$
$\sout{A\times X\leftarrow (A\times X)\times_X (X\times B)\rightarrow X\times B.}$

Eric Forgy (Jan 14 2021 at 23:47):

Note this means the answer to Puzzle #2 is also a span of sets.

Let $B = *$, then
${}$
$\sout{*\times B\cong B}$
${}$
and
${}$
$\sout{S\times_X X\cong S}$
${}$
so that a $(A,*)$-bimodule is a span
${}$
$\sout{A\times X\overset{\mathsf{id}}{\leftarrow} A\times X\overset{\alpha_L}\rightarrow X,}$
${}$
but this is a left $A$-module so a left $A$-module is an $(A,*)$-bimodule.

Similarly, a right $B$-module is a $(*,B)$-bimodule.

Furthermore, an $(A,B)$-bimodule is the composition of a left $A$-module, i.e. $(A,*)$-bimodule, with a right $B$-module, i.e. $(*,B)$-bimodule.

Eric Forgy (Jan 14 2021 at 23:51):

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