Proof.
Denote $U$ in $\mathcal{C}$ the invertible object which is assumed to exist in the statement of the lemma. We extend $F$ to motives by setting
\[ F(X, p, m) = \left(\text{the image of the projector }F(p)\text{ in }F(X)\right) \otimes U^{\otimes -m} \]
which makes sense because $U$ is invertible and because $\mathcal{C}$ is Karoubian. An important feature of this choice is that $F(X, c_2, 0) = U$. Observe that
\begin{align*} F((X, p, m) \otimes (Y, q, n)) & = F(X \times Y, p \otimes q, m + n) \\ & = \left(\text{the image of }F(p \otimes q)\text{ in }F(X \times Y)\right) \otimes U^{\otimes -m - n} \\ & = F(X, p, m) \otimes F(Y, q, n) \end{align*}
Thus we see that our rule is compatible with tensor products on the level of objects (details omitted).
Next, we extend $F$ to morphisms of motives. Suppose that
\[ a \in \mathop{\mathrm{Hom}}\nolimits ((Y, p, m), (Z, q, n)) = q \circ \text{Corr}^{n - m}(Y, Z) \circ p \subset \text{Corr}^{n - m}(Y, Z) \]
is a morphism. If $n = m$, then $a$ is a correspondence of degree $0$ and we can use $F(a) : F(Y) \to F(Z)$ to get the desired map $F(Y, p, m) \to F(Z, q, n)$. If $n < m$ we get canonical identifications
\begin{align*} s : F((Z, q, n)) & \to F(Z, q, m) \otimes U^{m - n} \\ & \to F(Z, q, m) \otimes F(X, c_2, 0) \otimes \ldots \otimes F(X, c_2, 0) \\ & \to F((Z, q, m) \otimes (X, c_2, 0) \otimes \ldots \otimes (X, c_2, 0)) \\ & \to F((Z \times X^{m - n}, q \otimes c_2 \otimes \ldots \otimes c_2, m)) \end{align*}
Namely, for the first isomorphism we use the definition of $F$ on motives above. For the second, we use the choice of $U$. For the third we use the compatibility of $F$ on tensor products of motives. The fourth is the definition of tensor products on motives. On the other hand, since we similarly have an isomorphism
\[ \sigma : (Z, q, n) \to (Z \times X^{m - n}, q \otimes c_2 \otimes \ldots \otimes c_2, m) \]
(see proof of Lemma 45.4.6). Composing $a$ with this isomorphism gives
\[ \sigma \circ a \in \mathop{\mathrm{Hom}}\nolimits ((Y, p, m), (Z \times X^{m - n}, q \otimes c_2 \otimes \ldots \otimes c_2, m)) \]
Putting everything together we obtain
\[ s^{-1} \circ F(\sigma \circ a) : F(Y, p, m) \to F(Z, q, n) \]
If $n > m$ we similarly define isomorphisms
\[ t : F((Y, p, m)) \to F((Y \times X^{n - m}, p \otimes c_2 \otimes \ldots \otimes c_2, n)) \]
and
\[ \tau : (Y, p, m)) \to (Y \times X^{n - m}, p \otimes c_2 \otimes \ldots \otimes c_2, n) \]
and we set $F(a) = F(a \circ \tau ^{-1}) \circ t$. We omit the verification that this construction defines a functor of symmetric monoidal categories.
$\square$
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