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Stream: learning: questions

Topic: an invertibility problem

Matteo Capucci (he/him) (Mar 28 2023 at 12:27):

Suppose I have a bounded poset $P$ and a functor $F:P \to \cal C$, where $\cal C$ is a category. Suppose I know $a \leq b : P$, and I also know $F(a \leq \top)$ is invertible. Then $F(b \leq\top)$ has a right inverse, given by $F(a \leq\top)^{-1}F(a \leq b)$.
Can I conclude $F(b \leq\top)$ is left invertible too?
I'm bothered because this fact seems germane to me but it doesn't seem to be true, which means my intuition is broken. I can't concoct an example where this could go wrong.

(My actual example is a bit more structured, so maybe that's what's blinding me. In particular $P$ is a fibered poset over $\cal C$ and $F$ is a comprehension, meaning in particular it is a right adjoint)

Spencer Breiner (Mar 28 2023 at 12:43):

Any time an isomorphism factors as $A\rightarrowtail B \twoheadrightarrow T$, the first factor is monic and the second is epic.

Matteo Capucci (he/him) (Mar 28 2023 at 13:01):

Uhm I don't think I understand what you mean

Matteo Capucci (he/him) (Mar 28 2023 at 13:01):

It seems tautological?

Matteo Capucci (he/him) (Mar 28 2023 at 13:02):

(During lunch I realized my intuition is broken because I think of P as 'monotonically' mapped to $\cal C$, in some sense. But it doesn't have to.)

Ralph Sarkis (Mar 28 2023 at 13:06):

If $a\leq b\leq \top$ is all the poset, then send it to $\mathbb N \hookrightarrow \mathbb Q \xrightarrow{\lfloor - \rfloor} \mathbb N$ inside $\mathbf{Set}$, the composite $F(a\leq \top)$ is an isomorphism (the identity) and it factors into mono-epi, but $F(b\leq \top)$ is not an isomorphism.

Matteo Capucci (he/him) (Mar 28 2023 at 13:08):

Ah, yes! That's the weird way to 'non-monotonically' map stuff.

Matteo Capucci (he/him) (Mar 28 2023 at 13:08):

I wonder, though, if I assume $F$ preserves monos (and in my case it does since it's a right adjoint), can I still have the same 'pathology'?

Matteo Capucci (he/him) (Mar 28 2023 at 13:09):

Ralph Sarkis said:

If $a\leq b\leq \top$ is all the poset, then send it to $\mathbb N \hookrightarrow \mathbb Q \xrightarrow{\lfloor - \rfloor} \mathbb N$ inside $\mathbf{Set}$, the composite $F(a\leq \top)$ is an isomorphism (the identity) and it factors into mono-epi, but $F(b\leq \top)$ is not an isomorphism.

Like, this works because $\lfloor - \rfloor$ is a quotient

Ralph Sarkis (Mar 28 2023 at 13:10):

In case $F$ preserves monos, $F(b\leq \top)$ will be epi and mono (all morphisms in a poset are mono), so you have to find a similar example in a category that is not balanced (in $\mathbf{Set}$ all epic monics are isos).

Morgan Rogers (he/him) (Mar 28 2023 at 15:54):

@Ralph Sarkis actually you can't, because split epi + mono always implies iso :wink:

Matteo Capucci (he/him) (Mar 28 2023 at 16:27):

Aaaah that's the trick

Reid Barton (Mar 28 2023 at 18:32):

Relevant: [[retract]]