Homotopy equivalence between two basepoints of the etale homotopy type of the one-torus Let $T = \mathbb{G}_m$ be the torus, and let $\tilde{T}$ be its étale universal cover (a pro-object in schemes of finite type). Then both $T$ and $\tilde{T}$ have a well-defined étale homotopy type. Explicitly, the homotopy types are $ét(T) = B\hat{\mathbb{Z}}$ and $ét(\tilde{T}) = *,$ for $*$ the point. In particular, the natural covering map $$\pi:\tilde{T}\to T$$ gives a basepoint (in a suitable homotopy sense) $$ét(\pi)$$ of $ét(T)$. Now group structure on $T$ lets us define a new point $\pi^2 := \pi*\pi,$ which is the composition of $\pi$ with the squaring map $[2]:T\to T$. While both $\pi, \pi^2:\tilde{T}\to T$ realize $\tilde{T}$ as a universal cover of $T$, they are not equal (as can be seen e.g. by looking at the map on tangent spaces at $1$). On the other hand, since $B\hat{\mathbb{Z}}$ is (I think?) connected, the maps $$ét(\pi), ét(\pi^2):ét(\tilde{T})\to ét(T)$$ should be homotopy equivalent in a suitable homotopy category. 
Question is there a way to see the homotopy equivalence between $ét(\pi)$ and $ét(\pi^2)$ explicitly? Here by "explicitly", I mean as an interval in the mapping space between natural topological models, or an interval in some other model category. Edit: I'd also like for the functor from varieties to etale types to take etale maps to fibrations, so that in particular the etale types of $\pi, \pi^2$ are not a priori equal.
 A: There is a completely explicit way to do it in this case I think. The Etale homotopy type of $T$ is represented by the pro-object $\{B \mathbb{Z}/n\}_{n \in \mathbb{N}^\times}$, ignoring for a moment the Galois action and assumeing that the field is of characteristic $0$ for simplicity. The object 
$\tilde{T}$ is very similar in nature. It is represented by the pro-object 
$(T,n)_{n \in \mathbb{N}^\times}$ with structure maps 
$(T,nm) \stackrel{[m]}{\to}(T,n)$. The map $\pi: \tilde{T} \to T$ is given by 
$(T,n) \stackrel{[n]}{\to} T$. Here $[n]$ is rising to the $n$-th power. 
The map $\pi * \pi$ is given by $(T,n) \stackrel{[2n]}{\to} T$. 
We want to show that the two maps are homotopic. For this we need to pull back the pro-cover $(T \stackrel{[n]}{\to} T)_{n \in \mathbb{N}^\times}$ along the two maps $\pi$ and $\pi * \pi$. For both you just get the same pro-cover over every copy $(T,n)$ because the pull-back $[n]$ along $[m]$ is just $[n]$.
After passing to the corresponding simplicial sets, we end up with the following problem in topology: 
Consider the pro-space indexed by $(\mathbb{N}^\times)^2$ given by 
$X_{m,n} = B\mathbb{Z}/n$, with structure maps given by the reduction maps 
$\mathbb{Z}/nk \to \mathbb{Z}/n$ in the second coordinate and by 
the multiplication my $k$ maps $\mathbb{Z}/n \to \mathbb{Z}/n$ in the first coordinate. We need to show that the map $X_{n,m} \to X_{n,m}$ given by multiplication by $2$ is homotopic to the identity, in an explicit way. 
Now, you can note that the maps in this story are all naturally based. So the connectivity of the target is somehow irrelevant, we only need to know explicitly how to contract the source. But there is a null-homotopy 
$(X_{n,m}) \to (X_{n,m})$ given by the map 
$(X_{n,m} \times \Delta^1) \to (X_{n,m})$ given by sending the edge of $\Delta^1$ to the coherent collection of compositions $X_{n,nm} \to X_{n,m} \to X_{n,m}$ where the first map is the structure map (i.e. multiplication by $n=0$) and the second map is the identity (the fact that this gives an edge in the hom space between the pro-objects follows from the identity $Pro-Hom(X_i,Y_j) = holim_j hocolim_i Him(X_i,Y_j)$). Call this simplex $\sigma$, we get that 
$\partial_0 \sigma = Id_X$ while $\partial_1 \sigma$ factor through a point-system. This gives an explicit contraction of $X$ and by composing with $\pi$ and $[2]\pi$ we get a homotopy between them.
