5
$\begingroup$

Let $\mathit{Mfd}$ denote the category of smooth manifolds. Let $W$ denote all projections of the form $$M \times \mathbb{R} \to M.$$ Let $\mathit{Mfd}_W$ denote the Hammock localization of $\mathit{Mfd}$ at the class of maps $W$. Is the mapping complex $Map_W\left(M,N\right)$ in $\mathit{Mfd}_W$ between two manifolds $M$ and $N$ weakly equivalent to the space of maps from $M$ to $N$, i.e. do we have

$$Map_W\left(M,N\right) \simeq \underline{Hom}\left(Sing\left(M\right),Sing\left(N\right)\right)?$$

$\endgroup$

2 Answers 2

3
$\begingroup$

Ok, so I have an argument:

First note that the homotopy coherent nerve of $\mathit{Mfd}_W$ is equivalent to the quasicategory obtained by formally inverting $W$ in $N\left(\mathit{Mfd}\right)$- this holds in generality, as shown in a recent paper of Hinich (Proposition 2.2.1 of http://arxiv.org/abs/1311.4128). Let me just be lazy and write $\mathit{Mfd}_W$ also for this quasicategory.

Let $$y:\mathit{Mfd} \hookrightarrow Psh_\infty\left(\mathit{Mfd}\right)$$ be the Yoneda embedding into infinity presheaves. First, note that the localization $y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$ at the class of morphisms of the form $y\left(M \times \mathbb{R}\right) \to y(M)$ is canonically equivalent to $Psh_\infty\left(\mathit{Mfd}_W\right).$ To see this observe that the canonical functor $$\mathit{Mfd} \to Psh_\infty\left(\mathit{Mfd}\right) \to y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$$ sends $W$ to equivalences, so induces a functor $$\mathit{Mfd}_W \to y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right),$$ which one can show immediately by universal properties exhibits $y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$ as the free colimit completion of $\mathit{Mfd}_W$ in the world of quasicategories.

Let $$y_w:\mathit{Mfd}_W \hookrightarrow Psh_\infty\left(\mathit{Mfd}_W\right)\simeq y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$$ denote the Yoneda embedding. Uwinding what we've done, we have that $$y_w\left(M\right) \simeq h \circ y\left(M\right)$$ for all manifolds $M.$ Let $h$ denote the localization functor $$h:Psh_\infty\left(\mathit{Mfd}\right) \to y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right).$$ By Lemma 7.5 of http://arxiv.org/abs/1311.3188, we have that $h$ can be computed as:

$$h\left(F\right)\left(M\right) = \operatorname{hocolim} F\left(M \times \Delta^n_{ext}\right)$$

where $\Delta_{ext}:\Delta \to \mathit{Mfd}$ is the "standard cosimplicial manifold," with each $$\Delta^n_{ext} \cong \mathbb{R}^n.$$

Now, $$h\left(y\left(M\right)\right)\left(N\right) = \operatorname{hocolim} Hom\left(N \times \Delta^n_{ext},M\right).$$

This homotopy colimit is a simplicial diagram of sets (regarded as discrete simplicial sets), so the hocolim is simply the starting simplicial set $$Hom\left(N \times \Delta^\star_{ext},M\right).$$ Finally, (and here I guess I am being partly sloppy), this simplicial set is homotopy equivalent to $Sing(M^N),$ where $M^N$ is the compactly generated mapping space. (Incidentally, if anyone has a clean proof of this last claim, let me know).

Notice that $$h\left(y\left(M\right)\right)\left(N\right)\simeq Hom\left(y(N),ih\left(y\left(M\right)\right)\right),$$ where $i$ is the inclusion of $y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$ into $Psh_\infty\left(\mathit{Mfd}\right),$ and $$Hom\left(y(N),ih\left(y\left(M\right)\right)\right)\simeq Hom\left(h y(N), h y(M)\right)\simeq Hom\left(y^w(N),y^w(M)\right),$$ and finally $$Hom\left(y^w(N),y^w(M)\right)\simeq Hom_{\mathit{Mfd}_w}\left(N,M\right),$$ since the Yoneda embedding $y^w$ is full and faithful.

$\endgroup$
0
$\begingroup$

I believe the answer is yes.

Edit: I still believe the answer is yes! But my first paragraph is nonsense... I'll leave it here since it rephrases the problem.

Consider the Yoneda embedding $Mfld \hookrightarrow Sh(Mfld)$ where the right hand side is the homotopy theory (= $\infty$-category) of hypercomplete sheaves on the usual site of all manifolds. Localize both sides at the projections $X \times \mathbb{R} \rightarrow X$, which gives a map of $\infty$-categories $Mfld[W^{-1}] \rightarrow Sh(Mfld)[W^{-1}]$. This is fully faithful (put it in a square with the Yoneda embedding, which is fully faithful, and use 2-out-of-3).

So to compute this mapping space we are free to use the mapping spaces in $Sh(Mfld)$. But now we can take hypercovers of $M$ and $N$ by copies of $\mathbb{R}^n$ and so we see that these mapping spaces are weakly equivalent to a mapping space between two simplicial sets (simplicial "sets" since we can replace the $\mathbb{R}^n$'s by points because of the definition of W) whose realizations are homotopy equivalent to $M$ and $N$. If you like you can then replace these with the singular simplicial sets.

$\endgroup$
4
  • $\begingroup$ Thanks for the answer Dylan. I don't follow your "2-out-of-3" argument for full and faithfulness. I know how to prove the result if I can show that, but in some sense, this was the main obstacle, but maybe I am being silly. Can you explain this part in more detail? $\endgroup$ Commented Jul 4, 2014 at 12:44
  • 1
    $\begingroup$ (This boils down to showing that a manifold is the homotopy colimit of the Cech nerve of any cover of it, when computed in $Mfd[W^-1],$ but I don't see how to prove that without proving the result I'm after to begin with.) $\endgroup$ Commented Jul 4, 2014 at 12:46
  • 1
    $\begingroup$ (See mathoverflow.net/questions/172710/… btw, because if you have a way of showing this by abstract nonsense, it might help in that question as well :) ) $\endgroup$ Commented Jul 4, 2014 at 12:50
  • $\begingroup$ Oh dear you're right! My argument didn't make any sense- I think I was assuming something absolutely absurd in my head... like that $Mfld[W^{-1}]$ is a full subcategory of $Mfld$. hehe. doy... $\endgroup$ Commented Jul 4, 2014 at 12:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .