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Stream: theory: category theory

Topic: lemma in structured cospans

Daniel Plácido (Apr 11 2021 at 04:40):

I'm struggling with Lemma 2.2 in Structured cospans.

It seems that the category of morphisms of ${}_L\mathbb X$ should be a restriction of $\mathbb X_1$ to morphisms with source/target in the image of $L$. What happens, however, when $L$ isn't faithful?

e.g. when $f,f':a\to a'$ are such that $L(f) = L(f')$, a 2-cell such as

image.png

could have as source either $f$ or $f'$. To be very explicit, we could take $A = \mathbb X_1\sqcup \mathbb X_1$ and $L = S\sqcup S$. Do we make a choice?

Daniel Plácido (Apr 11 2021 at 14:52):

like if a 2-morphism between vertical morphisms $f:a\to a'$ and $g:b\to b'$ is a 2-cell like this, then if $L(f) = L(f')$ this same $\alpha$ is a 2-cell from $f'$ to $g$. this couldn't be the case because each 2-morphism should have only one (vertical) source

Jade Master (Apr 11 2021 at 15:53):

My understanding is that the 2-cell including f is a different 2-cell from the one containing f' even if L(f)=L(f'). I agree that this is kind of weird because these two 2-cells have essentially the same data however for what it's worth I can't think of any examples where L is not faithful.

John Baez (Apr 11 2021 at 15:53):

Daniel Plácido said:

It seems that the category of morphisms of ${}_L\mathbb X$ should be a restriction of $\mathbb X_1$ to morphisms with source/target in the image of $L$.

No. A vertical morphism ${}_L\mathbb X$ is a morphism in $\mathsf{A}$, like $f: a \to a'$ in your notation here. It's not a morphism in $\mathsf{X}$ that just happen to be equal $L(f) : L(a) \to L(a')$ for some $f$.

Similarly, a 2-morphism in ${}_L\mathbb X$

is a thing like this

John Baez (Apr 11 2021 at 15:55):

where $f: a \to b$ and $g: b \to b'$ are part of the structure of the 2-morphism. We should have said this more clearly: "of the form" is a vague, and that's what confused you! We're talking about a structure here, not a mere property.

John Baez (Apr 11 2021 at 15:57):

It has to be this way, because a 2-morphism knows its source, which in this case is $f: a \to a'$, and its target, which in this case is $g: b \to b'$.

Jade Master (Apr 11 2021 at 16:00):

Yeah to be honest I don't have any intuition where A is not Set so that you have a set of boundary terminals for your open systems. When you're working in Set it makes sense to have functions as the vertical 2-morphisms, regardless of what they map to in X. I think that even if A is not Set what you really care about for the vertical morphisms is the different ways you can change the boundary terminals. From that perspective if L isn't faithful it doesn't matter so much.

John Baez (Apr 11 2021 at 16:07):

Right. It's perfectly fine if L is not faithful. Kenny and I carefully thought about this issue. I'm annoyed that our explanation wasn't clear enough, so that now @Daniel Plácido is wondering what we meant. It's all about the ambiguity in these three words: "of the form".

Daniel Plácido (Apr 11 2021 at 23:12):

Jade Master said:

I agree that this is kind of weird because these two 2-cells have essentially the same data

that's what I had thought!

Daniel Plácido (Apr 11 2021 at 23:14):

John Baez said:

where $f: a \to b$ and $g: b \to b'$ are part of the structure of the 2-morphism

this really clears things up. so ${}_L\mathbb X$ is $\mathsf A$-huge! thanks for the explanation, John

John Baez (Apr 12 2021 at 02:31):

Sure! If I ever feel like fixing problems with this paper on the arXiv, I'll fix this one.