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Stream: learning: questions

Topic: Intuition for cospans

Eric Forgy (Jan 21 2021 at 19:07):

Hello,

Over the past several weeks, I've developed some rudimentary intuition for spans. I think of them as generalizations of functions (i.e. multi-valued partial functions). An endomorphism in $\mathbf{Span(Set)}$ is a directed graph.

Now, I am hoping to develop a similar intuition for cospans. I can see that it is related to inputs and outputs, but my intuition kind of stops there at the moment. How should we think about cospans in terms of familiar things? "Familiar" being vaguely what someone with a degree in physics/engineering, but little to no exposure to CT, might know.

David Egolf (Jan 21 2021 at 19:17):

Can't you make a cospan into a span by working in an opposite category? Then maybe your intuition would apply already to that?

Eric Forgy (Jan 21 2021 at 19:23):

To some extent :blush:

For example,
${}$
$\mathbf{Span(Set)} \cong \mathbf{Cospan(Set^{op})},$
${}$
but how about $\mathbf{Cospan(Set)}$?

Maybe that helps more than I original thought if I try to think about $\mathbf{Span(Set}^\mathsf{op}),$ but my intuition is not so great with ${}^\mathsf{op}$ either :sweat_smile:

However, I supect their should be some ways to develop intuition directly with cospans and that is what I'm after :blush:

Jules Hedges (Jan 21 2021 at 19:29):

If you think of spans as "proof-relevant relations", then you can think of a cospan as a "proof-relevant corelation". (Like a relation is a subset of a cartesian product, a corelation is a quotient of a disjoint union. If you didn't meet them before, they're super intuitive once you've drawn a bunch of dots with some circles around them)

Eric Forgy (Jan 21 2021 at 19:38):

This has a nice intro :+1:

• Decorated Cospans

but the example with directed graphs seems special since a directed graph is already a span, so the example is a cospan of spans. Still helpful :+1:

This also has a nice discussion:

• Hypergraph Categories of Cospans

John Baez (Jan 21 2021 at 21:11):

Eric Forgy said:

To some extent :blush:

For example,
${}$
$\mathbf{Span}(\mathsf{Set}) \cong \mathbf{Cospan}(\mathrm{Set}^{\rm op}),$
${}$
but how about $\mathbf{Cospan(Set)}$?

Well,

$\mathbf{Cospan(\mathsf{Set})} \cong \mathbf{Span(\mathsf{Set}^{\rm op})}.$

When developing the theory of span or cospan (bi)categories, it's good to know that

$\mathbf{Span}(\mathsf{C}) \cong \mathbf{Cospan}(\mathsf{C}^{\rm op})$

so any theorem about span categories is also a theorem about cospan categories, and vice versa!

However, spans in $\mathsf{Set}$ are very different than cospans in $\mathsf{Set}$, so it's good to know a battery of theorems about each one.

Jade Master (Jan 22 2021 at 02:12):

Spans and cospans between two fixed sets X and Y are related as well. Given a cospan $X \to A \leftarrow Y$ you can get a span $X \leftarrow B \to Y$ where B contains an element (x,y) mapping to x and y whenever x and y map to the same element in the cospan. This is a 0-dimensional version of the comma category construction. Dually given a span $X \leftarrow B \to Y$, you can get a cospan by taking preimages...i .e. a cospan mapping x and y to e whenever there is an element in your span mapping to x and y. This is the 0-dimensional cocomma construction.

Jade Master (Jan 22 2021 at 02:17):

iirc. These two constructions form an idempotent adjunction and factor through Bool-profunctors between X and Y.

Jade Master (Jan 22 2021 at 02:18):

Check out the end of this article for more info: https://ncatlab.org/nlab/show/profunctor

Cole Comfort (Jan 22 2021 at 02:33):

Jade Master said:

Spans and cospans between two fixed sets X and Y are related as well. Given a cospan $X \to A \leftarrow Y$ you can get a span $X \leftarrow B \to Y$ where B contains an element (x,y) mapping to x and y whenever x and y map to the same element in the cospan. This is a 0-dimensional version of the comma category construction. Dually given a span $X \leftarrow B \to Y$, you can get a cospan by taking preimages...i .e. a cospan mapping x and y to e whenever there is an element in your span mapping to x and y. This is the 0-dimensional cocomma construction.

Is this not just the pullback and pushout?

Jade Master (Jan 22 2021 at 02:44):

That might be a simpler way to think about it...

Jade Master (Jan 22 2021 at 02:47):

Yeah I'm pretty sure that's right because comma and cocomma categories are lax pullbacks and pushouts.

John Baez (Jan 22 2021 at 05:17):

Jade Master said:

Spans and cospans between two fixed sets X and Y are related as well. Given a cospan $X \to A \leftarrow Y$ you can get a span $X \leftarrow B \to Y$ where B contains an element (x,y) mapping to x and y whenever x and y map to the same element in the cospan.

If I understand you correctly, this is called "taking the pullback".

A cospan $X \to A \leftarrow Y$ is precisely the sort of diagram you can take the pullback of, getting the span

$X \leftarrow X \times_A Y \to Y$

As you said, $X \times_A Y$ contains an element $(x,y)$ mapping to $x$ and $y$ whenever $x$ and $y$ map to the same element in $A$.

John Baez (Jan 22 2021 at 05:18):

Okay, so Cole already said this! I often respond to a post and then scroll down and see people already said the same thing. But I hope it's not too unpleasant to hear different people say the same thing in different ways.

Eric Forgy (Jan 22 2021 at 06:26):

:+1:

If a category $C$ has pullbacks, there is a bicategory $\mathbf{Span}(C)$ whose

• Objects are objects of $C$
• Morphisms are spans in $C$ with composition defined via pullback of the cospan connecting adjacent spans
• 2-morphisms are maps between spans (forming commutative diagrams)

${}$
If $(C,\times)$ is furthermore a cartesian monoidal category with pullbacks (hence finitely complete), then
${}$
$\mathbf{Span}(C,\times)\cong \mathbf{Bim}(C^\mathsf{op},\times)$
${}$
which relates spans to bimodules. This was a fun journey spelled out in another stream here.

Similarly, if a category $C$ has pushouts, you can form the obvious bicategory
${}$
$\mathbf{Cospan}(C)$
${}$
with composition of cospans given by the pushout of the span connecting adjacent cospans.

If $(C,+)$ is cocartesian monoidal with pushouts, then
${}$
$\mathbf{Cospan}(C,+)\cong\mathbf{Bim}(C,+).$
${}$
I find all this super fascinating which is why I'm trying to gain better intuition for cospans :blush:

Cole Comfort (Jan 22 2021 at 13:25):

Jade Master said:

That might be a simpler way to think about it...

I think it funny when I am studying something from a really abstract perspective without realizing that I already know it much better in more concrete terms.