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Stream: learning: questions

Topic: inverse of inverse

Leopold Schlicht (Dec 03 2021 at 17:05):

First question: Let $f$ and $g$ be two parallel 1-morphisms in a quasicategory. Then $f$ is homotopic to $g$ if and only if $g$ is homotopic to $f$. (Recall: $f$ is homotopic to $g$ if there is a 2-morphism/homotopy from $f$ to $g$.) Consider your favorite proof of that fact. Extract from that proof a map
$(-)^{-1}\colon \{\text{homotopies from }f\text{ to }g\}\to\{\text{homotopies from }g\text{ to }f\}, \sigma\mapsto \sigma^{-1}.$
Is that map a bijection "on the nose" in the sense that $(\sigma^{-1})^{-1}$ is equal to $\sigma$ or only up to some form of equivalence of homotopies? (How to define that notion of "equivalence" of homotopies?)

On the one hand, since this is "higher category theory" I would expect the answer to be "up to equivalence". On the other hand, if $f$ and $g$ are paths in a space with the same endpoints, then there is an on-the-nose bijection between homotopies from $f$ to $g$ and homotopies from $g$ to $f$: given a homotopy $H\colon [0,1]^2\to X$ from $f$ to $g$, the construction $(x,t)\mapsto H(x, 1-t)$ defines a homotopy $H^{-1}$ from $g$ to $f$ (and vice versa). And then we have $(H^{-1})^{-1}=H$ on the nose.

Second question: One can show that $f$ is homotopic to $g$ in a quasicategory $\mathcal C$ if and only if $f$ is homotopic to $g$ in $\mathcal C^{\mathrm{op}}$. Extract from the proof of that fact a map
$\{\text{homotopies from }f\text{ to }g\text{ in }\mathcal C\}\to\{\text{homotopies from }f\text{ to }g\text{ in }\mathcal C^{\mathrm{op}}\}.$
Is that map a bijection on the nose, only up to equivalence, or none of that?

Reid Barton (Dec 03 2021 at 17:23):

So there are a couple issues with the premises of the question. The first one is that there is a map to be extracted from the proof of the fact. There isn't really one because the proof invokes the hypothesis that the input is a quasicategory to choose a solution to some filling problem, and there's no specified way to make this choice.

Reid Barton (Dec 03 2021 at 17:26):

The second one is that there isn't even an "inverse-ness" relation between $\sigma$ and $\sigma^{-1}$, because in order for $\sigma^{-1}$ to be an inverse to $\sigma$ you also have to say how it is an inverse. (This already happens at the level of 1-morphisms as well: an inverse of $f : x \to y$ is $g : y \to x$ together with some additional data; at a minimum, a 2-morphism $\alpha : gf \simeq 1$.)

Reid Barton (Dec 03 2021 at 17:32):

The better way to think about this kind of question is to consider a space (i.e. simplicial set) of homotopies, say $H(f,g)$, whose 0-simplices we can presumably arrange to be your 2-simplices $\sigma : f \to g$, and also a space $I(f,g)$ consisting of triples $\sigma : f \to g$, $\tau : g \to f$, and coherence data that says that $\sigma$ and $\tau$ are inverse. Then construct a zigzag

$$H(f,g) \xleftarrow{p_1} I(f,g) \xrightarrow{p_2} H(g,f)$$

where $p_1$ and $p_2$ remember just $\sigma$, respectively $\tau$.

One then shows that $p_1$ and $p_2$ are acyclic Kan fibrations so that, if we want, we could choose a section $s_1$ of $p_1$ and consider the composite

$$H(f,g) \xrightarrow{s_1} I(f,g) \xrightarrow{p_2} H(g,f)$$

as a choice of your function "$(-)^{-1}$". However, there is nothing canonical about this choice. The well-determined notion is the span.

Leopold Schlicht (Dec 04 2021 at 18:11):

Thanks!
But I guess one can define maps on equivalence classes:

• $(-)^{-1}\colon \{\text{homotopies from }f\text{ to }g\text{ in }\mathcal C\}/\mathord{\simeq}\to\{\text{homotopies from }g\text{ to }f\text{ in }\mathcal C\}/\mathord{\simeq}$ and
• $(-)^{\mathrm{op}}\colon\{\text{homotopies from }f\text{ to }g\text{ in }\mathcal C\}/\mathord{\simeq}\to\{\text{homotopies from }f\text{ to }g\text{ in }\mathcal C^{\mathrm{op}}\}/\mathord{\simeq}$.

Are both of these maps $(-)^{-1}$ and $(-)^{\mathrm{op}}$ bijections?

Leopold Schlicht (Dec 04 2021 at 18:16):

Also, can one construct an $\infty$-category $\mathcal C$ in which there isn't a bijection
$\{\text{homotopies from }f\text{ to }g\text{ in }\mathcal C\}\cong\{\text{homotopies from }g\text{ to }f\text{ in }\mathcal C\}$
on the nose? As I said, for singular simplicial sets there should be such a bijection on the nose. That question is a bit artificial, but I'm curious nevertheless. (Same question in the other situation.)

Leopold Schlicht (Dec 17 2021 at 16:43):

Or are these questions again ill-defined? Then let me put it this way: Is the datum of a homotopy from $f$ to $g$ in $\mathcal C$ in some sense equivalent to the datum of a homotopy from $g$ to $f$ in $\mathcal C$? Also, is the datum of a homotopy from $f$ to $g$ in $\mathcal C^\mathrm{op}$ in some sense equivalent to the datum of a homotopy from $f$ to $g$ in $\mathcal C$? If yes, in which sense? If no, why not?