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Stream: learning: questions

Topic: ✔ A confusion with the source-target of structured cospans

Adittya Chaudhuri (Aug 30 2025 at 12:58):

Given a functor $F \colon \mathsf{A} \to \mathsf{X}$, @John Baez and Kenny Courser defined a structured cospan as a cospan in $\mathsf{X}$ of the form $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$ in the Section 2 of this paper. Now, when the category $\mathsf{A}$ and $\mathsf{X}$ have finite colimits and $F$ preserves them, then one can define a symmetric monoidal double category whose objects are objects of $\mathsf{A}$ and horizontal 1-morphisms from $a \in \mathsf{A}$ to $b \in \mathsf{A}$ consists of the structured cospans of the form $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$.

My question is the following:

Let $F(a)=F(a')$ and $F(b)=F(b')$ such that $a \neq a'$ and $b \neq b'$. Then, in a way, a horizontal 1-morphism $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$ has atleast two sources viz. $a$ and $a'$ and atleast two targets viz. $b$ and $b'$. I must be misunderstanding something fundamentally, as the above sounds a bit awkward.

However, if $F$ admits a right adjoint $G \colon \mathsf{X} \to \mathsf{A}$, then as shown by Baez-Courser here in page 6 that $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$ is same as the cospan $a \xrightarrow{i} G(x) \xleftarrow{o}b$ in $\mathsf{A}$. In that case, my confusion does not arise. However, in the same paper, Baez-Courser discussed that it is convenient to consider the structured cospan in the form of $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$.

Now, in this paper, @Evan Patterson and others considered the right-adjoint version of structured cospans to define open signed graphs (as horizontal 1-morphisms) in Proposition 2.4 and to define open signed categories (as horizontal 1-morphisms) in Proposition 2.10. So, in this case, my confusion about "multiple source-target of a horizontal 1-morphism" does not arise. However, in this paper @John Baez and @Jade Master defined open petri nets in the form of $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$. So, again, I am getting a bit confused with the said "source-target issue".

I have a feeling that I am misunderstanding something very fundamentally, which I am not able to see at the moment , where exactly I am misunderstanding. I apologise priorly, if my question sounds very naive.

John Baez (Aug 30 2025 at 13:42):

Adittya Chaudhuri said:

Given a functor $F \colon \mathsf{A} \to \mathsf{X}$, John Baez and Kenny Courser defined a structured cospan as a cospan in $\mathsf{X}$ of the form $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$ in the Section 2 of this paper. Now, when the category $\mathsf{A}$ and $\mathsf{X}$ have finite colimits and $F$ preserves them, then one can define a symmetric monoidal double category whose objects are objects of $\mathsf{A}$ and horizontal 1-morphisms from $a \in \mathsf{A}$ to $b \in \mathsf{A}$ consists of the structured cospans of the form $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$.

My question is the following:

Let $F(a)=F(a')$ and $F(b)=F(b')$ such that $a \neq a'$ and $b \neq b'$. Then, in a way, a horizontal 1-morphism $F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$ has at least two sources viz. $a$ and $a'$ and at least two targets viz. $b$ and $b'$. I must be misunderstanding something.

Right, you're misunderstanding something. It's impossible for a morphism to have two different sources or two different targets.

In category theory (and double category theory) we're not allowed do ask whether morphism from $a$ to $b$ is equal to a morphism from $a'$ to $b'$ unless $a = a'$ and $b = b'$. It's a 'type error' to ask that sort of question, as you're now doing.

One way to ensure that this is impossible to ask this sort of question is to decree a morphism in a category (or double category) to be a triple consisting of its source $a$, its target $b$ and then $f \in \text{hom}(a,b)$.

The best way may be to use some type theory, but I don't want to get into that.

So, let's take the first approach. Then, in the double category of structured cospans, a horizontal morphism is a triple consisting of an object $a \in \mathsf{A}$, an object $b \in \mathsf{A}$ and a diagram

$F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$

in $\mathsf{X}$.

So, it's impossible for this morphism to also have some source $a'$ unless $a' = a$, or target $b'$ unless $b' = b$.

In our paper, Kenny and I weren't so pedantic. We assumed the readers knew that in category theory it's against the rules to ask if a morphism in one homset is equal to a morphism in some other homset. So, we just said a horizontal morphism from $a$ to $b$ is a structured cospan

$F(a) \xrightarrow{i} x \xleftarrow{o} F(b)$

This defines the homset $\text{hom}(a,b)$. It's up to the reader to make sure the set of all horizontal morphisms is the disjoint union of these homsets.

Adittya Chaudhuri (Aug 30 2025 at 14:01):

Thanks very much. I got now where exactly I was misunderstanding.

Notification Bot (Aug 30 2025 at 14:01):

Adittya Chaudhuri has marked this topic as resolved.

Mike Shulman (Aug 30 2025 at 15:28):

FWIW, the definition of a category with a family of homsets doesn't require type theory.

John Baez (Aug 31 2025 at 10:09):

Yes. However, viewing that definition through the eyes of material set theory, one can still get distracted by questions like "what if the same morphism is an element of two distinct homsets?" - as @Adittya Chaudhuri was in this thread.

I believe nothing bad can happen with that definition even if a morphism is an element of two distinct homsets. You just have to keep in mind things like: in this framework, composition is not a single operation, but a bunch of different operations, so $f \circ g$ does not make sense unless you say which $\circ$ operation you mean. (The nLab article just writes $f \circ g$.)

However, all this is easily confusing and distressing to beginners.