3
$\begingroup$

Let $\mathcal{C}$ be an $\infty$-category, and let $u:x\rightarrow y$ be an edge. It seems reasonable to say that: The map $Map_{\mathbb{C}[\mathcal{C}]}(a,x)\rightarrow Map_{\mathbb{C}[\mathcal{C}]}(a,y)$ given by composition with $u\in Map_{\mathbb{C}[\mathcal{C}]}(x,y)_0$, where $\mathbb{C}[\mathcal{C}]$ is the simplical category associated to $\mathcal{C}$, is equivalent in the homotopy category of simplicial sets to the composite $\mathcal{C}_{\backslash x}\times_{\mathcal{C}}\{a\}\rightarrow \mathcal{C}_{\backslash u}\times_{\mathcal{C}}\{a\} \rightarrow \mathcal{C}_{\backslash y}\times_{\mathcal{C}}\{a\}$, where the first map is a section of the trivial fibration $\mathcal{C}_{\backslash x}\times_{\mathcal{C}}\{a\}\leftarrow \mathcal{C}_{\backslash u}\times_{\mathcal{C}}\{a\}$.

However, I am unable to come up with a proof, and have not found one in HTT. I would guess that one needs to use the straightening functor, as is done to identify the targets, but how to deal with the composition map of the simplicial category associated to $\mathcal{C}$?

EDIT: I am relatively convinced that it boils down to proving that the maps: $$(St_{\mathcal{C}}\mathcal{C}_{/f})(a)\rightarrow (St_{\mathcal{C}}\mathcal{C}_{/x})(a)\rightarrow Map_{\mathbb{C}[\mathcal{C}]}(a,x)\rightarrow Map_{\mathbb{C}[\mathcal{C}]}(a,y)$$ and $$(St_{\mathcal{C}}\mathcal{C}_{/f})(a)\rightarrow (St_{\mathcal{C}}\mathcal{C}_{/y})(a)\rightarrow Map_{\mathbb{C}[\mathcal{C}]}(a,y)$$ are homotopic, where $(St_{\mathcal{C}}\mathcal{C}_{/x})(a)\rightarrow Map_{\mathbb{C}[\mathcal{C}]}(a,x)$ corresponds to the map $f''$ in HTT 2.2.4 but I do not see any reasonnable way to proceed.

$\endgroup$
  • 1
    $\begingroup$ I’m having trouble parsing this- “C_/x” isn’t a Kan complex and doesn’t involve “a” so how could C_/x be the same ‘homotopy type’ as Map(a,x)? (Let alone the whole composite). $\endgroup$ – Dylan Wilson Jun 25 at 15:37
  • $\begingroup$ You are absolutely right, I was thinking about $Hom^{R}_{\mathcal{C}}(a,x)$ to use the notation of HTT. $\endgroup$ – user09127 Jun 25 at 15:50
3
$\begingroup$

Here's one way to do this. Consider the left fibration $p\colon C_{a/}\to C$, and pick a lift of $u\colon x\to y$ (let's assume that there is a map $a\to x $ otherwise it's trivial). When you have a cocartesian fibration $p\colon E \to B$, a $p$-cocartesian lift of a morphism $u\colon x \to y$ in the base category $B$ induce a map between the fibers $E_x \to E_y$ which is equivalent to the action of the functor $St(p)\colon B \to Set^+_{\Delta}$ applied to $u$ (in our simpler case, the map is a left fibration so this follows more or less from 2.1.1.4 in HTT). The map you mentioned is equivalent to the one you obtain in this way, so we are left with analyzing what the action of $St(p)(u)$ is. This is defined by $u\circ -\colon \mathfrak{C}(\bar{C})(*,x)\to \mathfrak{C}(\bar{C})(*,y)$, where $\bar{C}$ is the pushout of $\mathfrak{C}(C_{a/}) \to \mathfrak{C}(C^{\lhd}_{a/}) $ and $ \mathfrak{C}(C_{a/}) \to \mathfrak{C}(C)$. Since $\mathfrak{C}(C_{a/}) \to \mathfrak{C}(C^{\lhd}_{a/}) $ is a trivial cofibration, we can identify $\mathfrak{C}(\bar{C})(*,x)$ with $\mathfrak{C}(C)(a,x)$, and similarly for $y$, which concludes the proof.

$\endgroup$
  • $\begingroup$ Thank you you for your answer, it has helped me understand quite a few concepts in HTT more precisely. However I am pretty sure that it does not answer my question, as your are treating a "dual" case. $\endgroup$ – user09127 Jul 1 at 7:09
  • $\begingroup$ How is this "dual"? $\endgroup$ – Edoardo Lanari Jul 2 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.