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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.

  1. 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.

  2. 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?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.

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.

  1. 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.

  2. 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.

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.

  1. 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.

  2. 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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Matthias Wendt
Matthias Wendt
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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.

  1. 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.

  2. 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.