What is the kernel of the tensor product of two maps?

Solution 1:

Yes, that's true. Let $f_i : V_i \to W_i$ be two linear maps. Since $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $W_1 \otimes W_2$, we may assume that $f_1,f_2$ are surjective. But then they are split, so that we can assume that $V_i = W_i \oplus U_i$ and that $f_i$ equals the projection $V_i \to W_i$, with kernel $U_i$. Then $V_1 \otimes V_2 = W_1 \otimes W_2 \oplus W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2$ and $f_1 \otimes f_2$ equals the projection of $V_1 \otimes V_2$ onto $W_1 \otimes W_2$. Hence the kernel is $W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2 = U_1 \otimes V_2 + V_1 \otimes U_2$.

This shows even more: The kernel is the pushout $(\ker(f_1) \otimes V_2) \cup_{\ker(f_1) \otimes \ker(f_2)} (V_1 \otimes \ker(f_2))$.

By the way, this argument is purely formal and works in every semisimple abelian $\otimes$-category. What happens when we drop semisimplicity, for example when we consider modules over some commutative ring $R$? Then we only need some flatness assumptions:

Let $f_1 : V_1 \to W_1$ and $f_2 : V_2 \to W_2$ be two morphisms in an abelian $\otimes$-category (for example $R$-linear maps between $R$-modules). If $f_1,f_2$ are epimorphisms, then we have exact sequences $\ker(f_1) \to V_1 \to W_1 \to 0$ and $\ker(f_2) \to V_2 \to W_2 \to 0$. Applying the right exactness of the tensor product twice(!), we get that then also

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2 \to 0$

is exact. If $f_1,f_2$ are not epi, we can still apply the above to their images and get the exactness of

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to \mathrm{im}(f_1) \otimes \mathrm{im}(f_2) \to 0.$

Now assume that $\mathrm{im}(f_1)$ and $W_2$ are flat. Then $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $\mathrm{im}(f_1) \otimes W_2$ which embeds into $W_1 \otimes W_2$. Hence, we have still that the sequence

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2$

is exact. In other words, we have a sum decomposition $$\ker(f_1 \otimes f_2) = \alpha(\ker(f_1) \otimes V_2) + \beta(V_1 \otimes \ker(f_2)),$$ where $\alpha : \ker(f_1) \otimes V_2 \to V_1 \otimes V_2$ and $\beta : V_1 \otimes \ker(f_2) \to V_1 \otimes V_2$ are the canonical morphisms. In general, these are not monic! However, this is the case, by definition, when $V_1$ and $V_2$ are flat. So in this case we can safely treat $\alpha$ and $\beta$ as inclusions and write $$\ker(f_1 \otimes f_2) = V_1 \otimes \ker(f_2) + \ker(f_1) \otimes V_2.$$