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Stream: theory: category theory

Topic: ✔ Is h_0 really an isofibration?

Daniel Teixeira (Oct 23 2023 at 22:00):

I'm thinking about the functor $h_0:\mathsf{Cat}\to\mathsf{Set}$ that takes a (small) category to its set of isomorphisms classes, and it seems that it is either trivially an isofibration, or trivially not.

To see what I mean, consider the lifting problem at hand:

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We are picking a category $\mathcal C$ and a bijection $f:h_0(\mathcal C)\xrightarrow{\sim} X$. There seems to be a trivial lift: the identity!

But not quite - we don't have $h_0(\mathcal C) \mathbf{=} X$ - just a bijection. So maybe it's not a fibration at all.

What is the solution to the conundrum?

Daniel Teixeira (Oct 23 2023 at 22:08):

To illustrate the issue, let the square pick the discrete category $\mathcal C$ on $\{0,1\}$ and the bijection $f:\{\{0\},\{1\}\}\to \{3, 4\}$. In the latter I considered the sets $\{0\}$ and $\{1\}$ which are the isomorphism classes of $0$ and $1$.

Then we can try to consider to form the discrete category $\mathcal D$ on $\{3,4\}$, but actually $h_0(\mathcal D) = \{\{3\},\{4\}\}$, not $\{3,4\}$.

Daniel Teixeira (Oct 23 2023 at 22:08):

Am I overthinking this, or is it a fundamental issue?

Brendan Murphy (Oct 23 2023 at 22:20):

I wouldn't say you're overthinking this. It seems like what you've noticed is that the functor h0 is essentially surjective but is not surjective on objects. It's essentially surjective because we have a "section" given by forming the discrete category on a set. But this is only a section up to natural isomorphism, so it only gives essential surjectivity and not surjectivity. To see that the functor h0 isn't strictly surjective we classify its image: a set is in the image of h0 iff it forms a partitions of the union of its elements. A set like {{3, 4}, {4, 5}} will not be in the image. And for isofibrations essential surjectivitity implies surjectivity on objects, so this proves h0 is not an isofibration

Kevin Arlin (Oct 23 2023 at 22:20):

The thing that's tricky about this question is that $h_0$ is more of an anafunctor than a functor: we don't naturally have a completely specific set in mind for $h_0(\mathcal C).$ That said, there is a standard choice which is that the elements of $h_0(\mathcal C)$ are subsets of $\mathcal C,$ namely the isomorphism classes themselves. With this specific choice in hand, we can make annoyingly specific set-theoretic observations about $h_0(\mathcal C)$ that break the isofibration property. For instance, we know that the intersection of any two elements of $h_0(\mathcal C)$ is empty, since isomorphism classes are disjoint! So if $X$ is any set containing two elements with a nonempty intersection, your lift won't exist.

It's true that this functor is an "isofibration up to isomorphism", but the same argument shows that this is true for every functor. So it doesn't do anything for us.

Isofibrations are hard to get one's head around since they depend so critically on notions like equality of sets that we're used to forgetting about in category theory. But generally, the vibe of an isofibration $p:E\to B$ is that an object of $E$ contains a chosen object of $B$ which $p$ projects it to, and that however you build up an object of $E$ from an object of $B,$ that process is invariant under isomorphism in $B.$ This explains the problem with $h_0,$ which is that we don't build a category by starting with the set of its isomorphism classes of objects. This also explains the standard partial solution, if you really wanted a closely related isofibration: replace $\mathsf{Cat}$ with the (equivalent) category of categories equipped with a set $X$ and an isomorphism between $X$ and the set of connected components.

Daniel Teixeira (Oct 24 2023 at 14:05):

Thanks Brendan and Kevin! Yours answers clear the fog and leave food for thought.

Notification Bot (Oct 24 2023 at 14:05):

Daniel Teixeira has marked this topic as resolved.