On the swapping map of $\mathbb{G}_m$ On Voevodsky's paper 'Cancellation theorem', Lemma 4.8, he stated in the proof that the map
$$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x^{-1})\end{array}$$
is $\mathbb{A}^1$-homotopic to the identity, i.e., one could find an explicit homotopy (finite correspondence)
$$h:\mathbb{G}_m\times\mathbb{G}_m\times\mathbb{A}^1\longrightarrow\mathbb{G}_m\times\mathbb{G}_m$$
such that $h(x,y,0)=(x,y), h(x,y,1)=(y,x^{-1})$.
But I don't see how to find such a map.
Alternatively, he wants to show that the map
$$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x)\end{array}$$
is $\mathbb{A}^1$-homotopy to
$$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(x, y^{-1})\end{array}$$.
 A: Recall that $S_t^1$ is the reduced motive $\tilde M(\mathbf G_m)$ of $\mathbf G_m$, obtained most explicitly as the kernel of the projector $M(\mathbf G_m) \to M(\mathbf G_m)$ given by $\operatorname{id}-1_*$ where $1 \colon \mathbf G_m \to \mathbf G_m$ is the constant map $1$. In particular, it is a direct summand of $M(\mathbf G_m)$ such that $H^*(\tilde M(\mathbf G_m)) = H^1(\mathbf G_m)$ for any Weil cohomology theory $H$.
For a scheme $X$ and morphisms $f_1, \ldots, f_n \colon X \to \mathbf G_m$, write $\tilde f_1, \ldots, \tilde f_n \colon X \to \tilde M(\mathbf G_m)$ for the compositions with the projection $M(\mathbf G_m) \to \tilde M(\mathbf G_m)$, and for simplicity write $[f_1,\ldots,f_n]$ for the morphism $\tilde f_1 \otimes \ldots \otimes \tilde f_n \colon X \to \tilde M(\mathbf G_m)^{\otimes n}$. The claim in Lemma 4.8 is now:
Lemma. Let $X$ and $f, g \colon X \to \mathbf G_m$ be as above. Then the association $(f,g) \mapsto [f,g]$ is a skew-symmetric bilinear pairing in $f$ and $g$.
In particular, this implies that $[f,g] = -[g,f] = [g,f^{-1}]$, as claimed. As cited in the proof of Lemma 4.8, this is really a relative version of (the easy part of) Proposition 3.4.3 of [Suslin–Voevodsky].
Proof of Lemma. For simplicity, we will assume the base scheme $S$ is the spectrum of a field $k$, and $X$ is an integral smooth affine $k$-variety. (The ultimate application is to $X = \mathbf G_m \times \mathbf G_m$, so this hypothesis is harmless. These hypotheses are only used so that I know what the correct definitions are.)
If $f_1,f_2,g \colon X \to \mathbf G_m$ are morphisms, define the correspondence $Z \subseteq (X \times \mathbf A^1) \times \mathbf G_m$ by
$$y^2 - \big(t(f_1(x)+f_2(x))+(1-t)(1+f_1(x)f_2(x))\big)y + f_1(x)f_2(x) = 0$$
for $(x,t,y) \in X \times \mathbf A^1 \times \mathbf G_m$. The projection to $X \times \mathbf G_m$ is an isomorphism: the equation is linear in $t$, so we can eliminate $t$. Thus, $Z$ is integral. The projection $Z \to X \times \mathbf A^1$ is finite flat of rank $2$, so $Z$ is a finite correspondence $\phi \colon X \times \mathbf A^1 \to \mathbf G_m$. Its fibre at $0$ is $[1] + [f_1f_2]$ and its fibre at $1$ is $[f_1]+[f_2]$, so tensoring with $g$ gives
$$[f_1,g] + [f_2,g] = [1,g] + [f_1f_2,g] \in \operatorname{Hom}\big(M(X),\tilde M(\mathbf G_m)\otimes \tilde M(\mathbf G_m)\big).$$
But $[1,g] = 0$ since $1 \colon X \to \mathbf G_m$ induces the zero map when composed with the projector $\operatorname{id}-1_*$ on $\mathbf G_m$. Thus, we see that $[f_1f_2,g] = [f_1,g]+[f_2,g]$, and likewise $[f,g_1g_2] = [f,g_1]+[f,g_2]$. This shows that the map $(f,g) \mapsto [f,g]$ is bilinear.
To see that it is alternating, consider the correspondence $Z \subseteq (X \times \mathbf A^1) \times (\mathbf G_m \times \mathbf G_m)$ given by
$$\left\{\begin{array}{l}
y_1^2 - \big(t(f(x)+g(x))+(1-t)(1+f(x)g(x))\big)y_1 + f(x)g(x) = 0, \\
y_1 = y_2,\end{array}\right.$$
for $(x,t,y_1,y_2) \in X \times \mathbf A^1 \times \mathbf G_m \times \mathbf G_m$. This is again an integral subscheme that is finite flat of rank $2$ over $X \times \mathbf A^1$, and it gives a correspondence $X \times \mathbf A^1 \to \mathbf G_m \times \mathbf G_m$ whose restrictions to $t=0$ and $t=1$ witness the relation
$$[fg,fg] = [f,f] + [g,g] \in \operatorname{Hom}\big(M(X),\tilde M(\mathbf G_m) \otimes \tilde M(\mathbf G_m)\big).$$
By bilinearity, we can rewrite the left hand side as $[f,f]+[f,g]+[g,f]+[g,g]$, so we conclude that $[f,g] + [g,f] = 0$, i.e. the pairing is skew-symmetric. $\square$
Remark. To actually produce a homotopy from $[f,g]$ to $[g,f^{-1}]$, we really have multiple components:

*

*A homotopy from $[f,f]+[f,g]+[g,f]+[g,g]$ to $[fg,fg]$ (even this requires multiple steps a priori, but see below for a simplification);

*A homotopy from $[fg,fg]$ to $[f,f]+[g,g]$;

*A homotopy from $[g,f] + [g,f^{-1}]$ to $0$.

Then some things cancel, some things don't, and we crucially use multiple times that constant maps $1 \colon X \to \mathbf G_m$ induce the zero map $M(X) \to \tilde M(\mathbf G_m)$.
In fact the first and second item on the list above can be simplified by using some cancellation, as the closed subschemes witnessing these homotopies are actually almost the same. The closed subscheme
$$\left\{\begin{array}{l}
y_1^2 - \big(t(f(x)+g(x))+(1-t)(1+f(x)g(x))\big)y_1 + f(x)g(x) = 0,\\
y_2^2 - \big(t(f(x)+g(x))+(1-t)(1+f(x)g(x))\big)y_2 + f(x)g(x) = 0,
\end{array}\right.$$
of $X \times \mathbf A^1 \times \mathbf G_m \times \mathbf G_m$ has two components: one where $y_1 = y_2$ that we used for $[fg,fg]$, and a second component $W$ witnessing $[f,g] + [g,f] = [1,fg]+[fg,1]$, which again is zero on the reduced motive $\tilde M(\mathbf G_m) \otimes \tilde M(\mathbf G_m)$. Using $W$ avoids redundancies, leaving only the need for something witnessing $[g,f] + [g,f^{-1}] = 0$. We ultimately end up with two subschemes of $X \times \mathbf A^1 \times \mathbf G_m \times \mathbf G_m$ that are both finite flat of degree $2$ over $X \times \mathbf A^1$.
If you want to consider the entire motive $M(\mathbf G_m)$, then the same argument provides a homotopy from
$$(f \otimes g) + (g \otimes f) + (g \otimes 1) + (g \otimes 1)$$
to
$$(1 \otimes fg) + (fg \otimes 1) + (g \otimes f) + (g \otimes f^{-1}),$$
witnessed by the two degree $2$ correspondences explained above. The projector $(\operatorname{id}-1_*) \otimes (\operatorname{id}-1_*)$ on $M(\mathbf G_m) \otimes M(\mathbf G_m)$ sends $f \otimes g$ to
$$[f,g] := (f \otimes g) - (f \otimes 1) - (1 \otimes g) + (1 \otimes 1),$$
and one immediately sees that terms of the form $[f,1]$ or $[1,g]$ vanish. This gives a homotopy from $[f,g]+[g,f]$ to $[g,f]+[g,f^{-1}]$, and the result follows by cancelling $[g,f]$ on both sides.
A: Regarding your precise question, the map $\mathbb G_m \times \mathbb G_m \to \mathbb G_m \times \mathbb G_m, (x,y) \mapsto (y,x^{-1})$ is not $\mathbb A^1$-homotopic to the identity. This can be seen by taking Betti realization and looking at the fundamental group.
But I have in my notes that the swap map $(x,y) \mapsto (y,x)$ on $\mathbb P^1 \wedge \mathbb P^1$ is $\mathbb A^1$-homotopic to the map $(x,y) \mapsto (x,-y)$ via a sequence of two homotopies, induced by the homotopies $\begin{pmatrix} t & 1-t \\ 1+t & -t\end{pmatrix}$ and $\begin{pmatrix}1 & 0 \\ 2t & -1 \end{pmatrix}$ via the functoriality of Thom spaces. If you only care about the stable story, that's enough. I think I convinced myself you can't do it in one homotopy.
