$\DeclareMathOperator\Mor{Mor}\DeclareMathOperator\Hom{Hom}\newcommand{\et}{\mathrm{et}}\newcommand{\top}{\mathrm{top}}$In Voevodsky's paper "Étale topologies of schemes over fields of finite type over $\mathbf{Q}$" (link), there's a proposition in part of the proof of his main theorem (an anabelian result for normal schemes). It's as follows:
$\mathcal{O}^*(X)$ denotes the units in the structure sheaf. The object at the top of the diagram is the inverse limit $$\widehat{\mathcal{O}}^*(\overline{X})=\lim \mathcal{O}^*(\overline{X})/\mathcal{O}^*(\overline{X})^n$$
The object at the bottom denotes morphisms of etale sites over $(\operatorname{Spec}(K))_{\et}$ (where $K$ is a field of finite type over $\mathbb{Q}$)
$$\mathcal{O}_{\top}^*(X)=\operatorname{Mor}^0_K(X_{\et},(\mathbb{G}_m)_{\et})$$
(the superscript $0$ is not relevant here). And the identification (3.1.) that he cites is the isomorphism coming from the identification of torsors (principal bundles) and homs from the fundamental group $$\lim_nH^1_{\et}(\overline{X},\mu_n)=H^1_{\et}(\overline{X},\widehat{\mathbb{Z}}(1))\cong \operatorname{Hom}(\pi_1^{\et}(\overline{X},\overline{x}),\pi_1^{\et}(\mathbb{G_m},1))$$ where of course $\pi_1^{\et}(\mathbb{G_m},1)\cong \widehat{\mathbb{Z}}$ (We write $\pi_1^{\et}(\mathbb{G_m},1)$ instead of $\widehat{\mathbb{Z}}$ to make it clear how we get a map from maps of etale sites into this $H^1$). I'm struggling to explicitly compare the maps at finite level. The map $\chi_n$ referred to in the proof comes from the Kummer sequence $$1\longrightarrow \mathcal{O}^*(\overline{X})/\mathcal{O}^*(\overline{X})^n\overset{\chi_n}{\longrightarrow}H^1_{\et}(\overline{X},\mu_n)\longrightarrow \operatorname{Pic}(\overline{X})_n\longrightarrow 1$$ Then the map on the top right is the map you get by passing to the inverse limit. The map on the bottom left sends a unit to the corresponding map of etale sites (thinking of $\mathcal{O}^*_X$ via the functor of points) given by pullback (i.e., $\phi\in \Hom(X,\mathbb{G}_m)$ maps to $\phi_{\et}\in \Mor(X_{\et},(\mathbb{G}_m)_{\et})$ where $\phi_{\et}^{-1}(U)=U\times_\phi X$ for any etale $\mathbb{G}_m$-scheme $U$). Then the map on the bottom right comes from the functoriality of $\pi_1$ and the fact that a map on etale sites takes etale covers to etale covers.
At finite level on the sites, maybe we take the composition
$$\Mor(X_{\et},(\mathbb{G_m})_{\et})\longrightarrow \Hom(\pi_1^{\et}(\overline{X},\overline{x}),\pi_1^{\et}(\mathbb{G_m},1))\longrightarrow \Hom(\pi_1^{\et}(\overline{X},\overline{x}),\mu_n)$$ where we consider the last $\mu_n$ as the group of roots of unity that gives rise to the constant sheaf $\mathbf{\mu}_n$ (we've based changed to the algebraic closure). I think the first map sends a morphism $\phi$ of etale sites to a map $\phi_\pi$ given by automorphisms on pullbacks of etale covers of $\mathbb{G}_m$ (I guess we're assuming $\overline{x}\mapsto 1$). But maybe we want to consider $(\mu_n)_{\et}$. I'm a bit lost here. And then of course the $\chi_n$'s at finite level are mysterious as well.
Any comments would be appreciated!
Best Answer
The key problem here is something so subtle I didn't notice it the first time I read your question - apologies.
Of course, "of course" is a dangerous phrase in mathematics. Usually this is because things that are "of course" true are not always actually true. That's not a problem here (assuming you're working over an algebraically closed field of characteristic zero).
The problem in this case is that saying "of course" hides the construction of the isomorphism $\pi_1^{\et}(\mathbb{G_m},1) \to \widehat{\mathbb{Z}}$. To see why this is crucial, note that, assuming the diagram commutes, if you just replace the isomorphism with a different isomorphism, say after multiplication by $-1$, the diagram will certainly not commute, because the two paths around the diagram will differ by $-1$.
So how do you construct the isomorphism? There are multiple approaches, with the comparison to the topological fundamental group being one (since we are working over an algebraically closed field of characteristic zero). However, for our purposes, the best one is Kummer theory. Kummer theory applied to the identity function on $\mathbb G_m$ produces for each $n$ a homomorphism $\pi_1^{\et}(\mathbb{G_m},1) \to \mu_n$. Taking the inverse limit of these gives a homomorphism $\pi_1^{\et}(\mathbb{G_m},1) \to \widehat{\mathbb Z}(1)$, and then one checks this is an isomorphism.
To check this agrees with the isomorphism coming from the comparison with the topological fundamental group, one just needs to calculate the Kummer covering explicitly as adjoining the $n$th' root of $x$ and see that the generator of the topological $\pi_1$, a loop around zero, acts on this covering by sending the $n$th root of $x$ to $e^{ 2\pi i/n}$ times the $n$th root of $x$. In particular, it is necessary to fix the isomorphism $\mu_n \to \mathbb Z/n$ to send $e^{ 2\pi i /n}$ to $1$, since the generator of the topological $1$ corresponds to $1 \in \mathbb Z \subset \widehat{\mathbb Z} \to \mathbb Z/n$.
Now that we have fixed this, we can understand what the b bottom-right arrow is doing at finite level. We know the bottom-right arrow involves composing this isomorphism $\pi_1^{\et}(\mathbb{G_m},1) \to \widehat{\mathbb{Z}}$ with the map of étale fundamental groups arising from a map $X \to \mathbb G_m$. At finite level, this would be he composition of a map $\pi_1^{\et}(\mathbb{G_m},1) \to \mathbb Z/n$ with the map of étale fundamental groups arising from a map $X \to \mathbb G_m$.
Now here is the crucial point. Homomorphism from the étale fundamental group to a fixed finite group $G$ correspond to (pointed) $G$-coverings, i.e. finite étale coverings with an action of $G$ that is simply transitive on the fibers. Under this correspondence, composition with the induced map of fundamental groups of a map of spaces is equivalent to pullback of étale coverings.
Why? This is the definition of the induced map of fundamental groups via Grothendieck's Galois theory, as the unique map of groups whose induced map on categories of representations agrees with the pullback functor.
So at finite level, the bottom-right arrow sends a map $ X \to \mathbb G_m$ to the pullback of the Kummer class in $H^1(X, \mu_n)$ (arising from the identity function on $\mathbb G_m$) along that map.
So you shouldn't think of this as some mysterious equivalence between maps and fundamental groups induced by and Kummer theory, which seem totally different, but rather the plausible statement that the pullback of a Kummer class is a Kummer class.
How do you check that? One way is to use the definition of the connecting map of Kummer theory in terms of cocycles - Cech cohomology suffices for this. Each step of the process commutes with pullback in exactly the way you would expect.
Concretely, at level $n$, you can use that the Kummer map applied to $f$ is the class of the étale covering $y^n=f$. So this is the statement that the pullback of the cover $y^n=x$ along a map sending $x$ to $f$ is the cover $y^n=f$, which is clear from the definition of pullback.