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
$$\begin{array}{ccc}\mathbb{G}\_m\times\mathbb{G}\_m&\longrightarrow&\mathbb{G}\_m\times\mathbb{G}\_m\\(x,y)&\longmapsto&(x, y^{-1})\end{array}$$.



On the swapping map of $\mathbb{G}_m$
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](https://www.math.ias.edu/vladimir/sites/math.ias.edu.vladimir/files/susvoenew.pdf)].


*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.


