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This question refers to an argument from Section 6, p. 22 of Freed–Hopkins, "Chern–Weil forms and abstract homotopy theory." There's something presumably straightforward I'm just not getting.

All presheaves considered are presheaves of sets on the site of smooth manifolds, although I don't think the domain category really figures in this argument.

Let $\mathscr{F}_\bullet$ be a simplicial presheaf and $\mathscr G$ a sheaf, viewed as a simplicial sheaf with all face and degeneracy maps the identity. Associated to $\mathscr{F}_\bullet$ is the coequalizer presheaf ${\pi}_0\mathscr F$ of the two face maps $d_0$ and $d_1$

$$\mathscr F_1 \to \mathscr F_0,$$

and the sheafification of this coequalizer is denoted $\underline{\pi}_0\mathscr F$.

Now, the claim is that the set of simplicial presheaf maps

$$\mathscr F_\bullet \to \mathscr G$$

is in natural bijection with the set of sheaf maps

$$\underline{\pi}_0\mathscr F \to \mathscr G.$$

It's clear enough that sheaf maps $\underline{\pi}_0\mathscr F \to \mathscr G$ and presheaf maps $\pi_0\mathscr F \to \mathscr G$ are in natural bijection—this is the universal property of sheafification. It also seems clear that a simplicial presheaf map $\varphi_\bullet\colon\mathscr F_\bullet \to \mathscr G$ descends to a map $\pi_0\mathscr F \to \mathscr G$, precisely because the "simplicial map" condition means that

$$\varphi_1 = \varphi_0 \circ d_0 = \varphi_0 \circ d_1\colon \mathscr F_1 \to \mathscr G.$$

What I don't get is the converse; it seems like a map $\pi_0(\mathscr F_\bullet) \to \mathscr G$ suffices to define $\varphi_0$ and $\varphi_1$, but not that it should be possible to define $\varphi_2\colon\mathscr F_2 \to \mathscr G$. It seems that one would need the maps

$$\varphi_1 \circ d_j\colon\mathscr F_2 \to \mathscr G \qquad (0 \leq j \leq 2)$$

to be equal in order to do so, and I don't understand how this follows from $\pi_0$, which only knows things at levels 0 and 1 in the simplicial structure.

Given, it seems like the key application will have to do with groupoids, for which all data is determined in levels 0 and 1, but I want to know why this works in general.

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  • $\begingroup$ Do you know that $\pi_0$ is left adjoint to the inclusion of sets into simplicial sets? $\endgroup$ – Qiaochu Yuan Jan 18 '16 at 23:49
  • $\begingroup$ I didn't, but I see that's enough. I still don't really understand why this is true, though. I see the formulas seem to check out, but do you have some intuition why this really works? $\endgroup$ – jdc Jan 19 '16 at 0:30
  • $\begingroup$ Sets, as simplicial sets, are discrete, so functions into them are locally constant, hence determined by what they do on connected components. $\endgroup$ – Qiaochu Yuan Jan 19 '16 at 0:39
  • $\begingroup$ That's really helpful. Thanks! $\endgroup$ – jdc Jan 19 '16 at 0:45
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Pursuant to Qiaochu's hint, presheaves aren't essential to this part of the argument, which follows from the same argument for simplicial sets. Specifically, given a simplicial set $S_\bullet$ and a set $T$, maps $\psi\colon\pi_0 S \to T$ correspond to maps $S_\bullet \to T$ in a way following from the simplicial identities, specifically that

$$d_i d_j = d_{j-1} d_i \qquad (i < j).$$

Define $\phi_0$ as the composition $S_0 \overset{p}\to \pi_0 S \overset\psi\to T$, and define $\varphi_1 = \varphi_0 d_0 = \varphi_0 d_1$, which works because by definition the compositions $pd_0$ and $pd_1$ are equal.

We want to see $\varphi_2 = \varphi_1 d_i$ is independent of $i \in \{0,1,2\}$ and hence well-defined, and so on. Inductively suppose $\varphi_n$ is well-defined as $\varphi_{n-1} d_i$ for $0 \leq i \leq n$. Then for $0 \leq i < j \leq n+1$, we have

$$\varphi_n d_j = \varphi_{n-1} d_i d_j = \varphi_{n-1} d_{j-1} d_i = \varphi_n d_i,$$

completing the induction.

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