OK, so I asked a similar question before; $B{\mathbb G}_m$ is a simplicial presheaf over number field $k$. I see that there is some $A^1$-homotopy equivalence between the sheaf represented by ind-scheme ${\mathbb P}^\infty$ and that by $B{\mathbb G}_m$. As in the paper $A^1$-homotopy theory [MV], the construction is through Nisnevich site. I would like to know if some description of this map exists that stays within the etale site. MV gives a general construction good for any algebraic group; but I feel like something explicit can be said for the simplest case, namely ${\mathbb G}_m={\mbox{Spec}}\ k[t,t^{-1}]$.
2 Answers
This is more like an extended comment to David Roberts' answer. There (at least) two ways to see that $\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible.
Using Morel's unstable connectivity theorem which can be found in his book "$\mathbb{A}^1$-algebraic topology over a field", in the chapter on $\mathbb{A}^1$-homotopy and $\mathbb{A}^1$-homology sheaves. The result holds for the $\mathbb{A}^1$-homotopy theory over an infinite perfect field, and states that any $n$-connected simplicial presheaf is also $\mathbb{A}^1$-$n$-connected. It is claimed in Example 3.2.20 of [MV] and explained in Examples 2.11 of Dugger-Isaksen "Motivic cell structures" that there is an $\mathbb{A}^1$-weak equivalence $\mathbb{A}^n\setminus\{0\}\cong S^{n-1}\wedge\mathbb{G}_m^{\wedge n}$. In particular, $\mathbb{A}^n\setminus\{0\}$ is simplicially $(n-2)$-connected, and by Morel's unstable connectivity theorem also $\mathbb{A}^1$-$(n-2)$-connected. As mentioned in S. Carnahan's comment, taking the colimit of $\mathbb{A}^n\setminus\{0\}$ via the obvious inclusions therefore shows that $\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible.
The alternative is the classical topological argument that $S^\infty$ is contractible (see the MO-discussion How do you show that $S^{\infty}$ is contractible?), made algebraic: the shift-by-1 map is $$ S:\mathbb{A}^\infty\setminus\{0\}\to\mathbb{A}^\infty\setminus\{0\}: (x_1,x_2,\dots)\mapsto (0,x_1,x_2,\dots) $$ which is homotopic to the identity via $$ f_T:\mathbb{A}^\infty\setminus\{0\}\times\mathbb{A}^1\to \mathbb{A}^\infty\setminus\{0\}: (x_1,x_2,\dots)\mapsto (1-T)(x_1,x_2,\dots)+TS(x_1,x_2,\dots). $$ This works since the straight line through $(x_1,x_2,\dots)$ and $(0,x_1,\dots)$ does not go through $0$. Then the image of the shift map $S$ can be contracted to a point via $$ g_T:\mathbb{A}^\infty\setminus\{0\}\times\mathbb{A}^1\to \mathbb{A}^\infty\setminus\{0\}: (x_1,x_2,\dots)=(1-T)(0,x_1,x_2,\dots)+T(1,0,0,\dots). $$ This is just the same argument as in topology. In topology, the above maps must be renormalized to induce maps on $S^\infty$, but that is not necessary here. As the above maps are polynomial, they induce even naive $\mathbb{A}^1$-homotopies from the identity to the constant map, so $\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible.
The explicit map is then given as in David Robert's answer. Formulated slightly differently, take $\mathbb{A}^\infty\setminus\{0\}\to\mathbb{P}^\infty$ to be the universal $\mathbb{G}_m$-bundle, and take the standard covering $\mathcal{U}$ of $\mathbb{P}^\infty$ by the $\mathbb{A}^\infty$ with coordinates $x_i\neq 0$, $i\in\mathbb{N}$. The universal $\mathbb{G}_m$-bundle induces a map $\check{C}(\mathcal{U})\to B\mathbb{G}_m$. Here, $\check{C}(\mathcal{U})$ is the Cech nerve of the cover $\mathcal{U}$ which is simplicially equivalent to the action groupoid. The map itself is the $\mathbb{G}_m$-cocycle for the universal bundle in degree $1$, and can be extended to a map of the whole Cech nerve since the cocycle condition is satisfied. The diagram $\mathbb{P}^\infty\leftarrow\check{C}(\mathcal{U})\to B\mathbb{G}_m$ gives the required weak equivalence in $\mathbb{A}^1$-homotopy category. Note that the map itself is already defined in the simplicial model category because $\check{C}(\mathcal{U})\to \mathbb{P}^\infty$ is a simplicial weak equivalence.
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$\begingroup$ Thanks, Matthias! That algebraisation of the topological proof is really nice, and immediately graspable without any deep A^1 homotopy theory. $\endgroup$– David Roberts ♦Commented Jun 30, 2014 at 10:57
OK, here's a first attempt, with a small gap.
The ind-scheme $\mathbb{P}^\infty$ is the quotient of the ind-scheme $\mathbb{A}^\infty \setminus\lbrace 0\rbrace = \lim_{n\to \infty}\mathbb{A}^n\setminus\lbrace0\rbrace$ by the free action of $\mathbb{G}_m$, so here is a weak equivalence of simplicial presheaves $N(\mathbb{G}_m \curvearrowright\mathbb{A}^\infty \setminus\lbrace 0\rbrace) \to \mathbb{P}^\infty$, where the source is the nerve of the action groupoid and the target is the constant simplicial presheaf. I say 'weak equivalence', but more properly this is I guess a weak equivalence in the local projective model structure on simplicial presheaves, where 'local means whatever topology the $\mathbb{G}_m$ bundle over $\mathbb{P}^\infty$ is trivial in (I'm guessing étale, or possibly even Zariski). Next, there is a map of simplicial sheaves $N(\mathbb{G}_m \curvearrowright\mathbb{A}^\infty \setminus\lbrace 0\rbrace) \to B\mathbb{G}_m$, which I conjecture is a weak equivalence in the $\mathbb{A}^1$-homotopy sense. This should be true as long as $\mathbb{A}^\infty \setminus\lbrace 0\rbrace \to \ast$ is a $\mathbb{A}^1$-weak equivalence. This last statement is the one that I feel is a gap, in that I have no clue how to prove this, only argue by analogy with the differential geometric or topological case.
The span $$ \mathbb{P}^\infty \leftarrow N(\mathbb{G}_m \curvearrowright\mathbb{A}^\infty \setminus\lbrace 0\rbrace) \to B\mathbb{G}_m, $$ assuming that one can fill in the gap, should represent an $\mathbb{A}^1$-weak equivalence, and my guess, based on the assumption that the $\mathbb{G}_m$ bundle over $\mathbb{P}^\infty$ trivialises in the étale (resp. Zariski) topology, is that this construction uses the étale (resp. Zariski) site.
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1$\begingroup$ For your gap, could you show that the closed immersion $\mathbb{A}^n \setminus \{0\} \to \mathbb{A}^{n+1} \setminus \{ 0 \}$ is something like a suspension, and take a colimit? $\endgroup$– S. Carnahan ♦Commented Jun 12, 2014 at 23:45
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$\begingroup$ @S.Carnahan, maybe, but I don't know the A^1 homotopy machinery so as to make that rigorous. $\endgroup$– David Roberts ♦Commented Jun 13, 2014 at 0:58