# $\ker(f\otimes g)=\langle\{a\otimes b:(a\in \ker f \text{ or }(b\in\ker g)\}\rangle$

There's a theorem that if $f:A\to B$ and $g:A'\to B'$ are epimorphisms, then their tensor product's kernel is given as $$\ker(f\otimes g)=\langle\{a\otimes b:(a\in \ker f )\ or\ (b\in\ker g)\}\rangle$$ where $<A>$ denotes the module generated by the set $A$. But if I define $\widetilde{f}:A\to Im(f)$ as $\widetilde{f}(a)=f(a)$ for all $a$ in $A$, $\widetilde{f}$ is an epimorphism. And also, $\ker(\widetilde{f}\otimes g)=\ker(f\otimes g)$ since if $$t\in A\otimes B$$ then there exist $a_i$and $b_i$ such that $$t=\sum_{i} a_i\otimes b_i$$ so $$(f\otimes g)(t)=\sum_{i} f(a_i)\otimes g(b_i)$$ $$=\sum_{i} \widetilde{f}(a_i)\otimes g(b_i)$$ $$=(\widetilde{f}\otimes g)(t)$$ which yields $$(\widetilde{f}\otimes g)(t)=0 \iff (f\otimes g)(t)=0$$ Then I can say $$\ker(f\otimes g)=\langle\{a\otimes b:(a\in \ker f )\ or\ (b\in\ker g)\}\rangle$$ for any $f$ even if it is not an epimorphism. At this point it is apparent that I made mistake, but I cannot see it. I made this mistake in proving that $-\otimes F$ preserves every short exact sequence, where F is an arbitrary free module. Can anyone point out what I did wrong? Thank you.

• This is a common misconception, say, I have $A, B$ are both $R-$modules (one on the right, and the other on the left), and $A'$ is a submodule of $A$, then $A' \otimes B$ needs not be a subgroup of $A \otimes B$. If you wants to know why this happens, you should trace back to how the tensor product of $A \otimes B$ is built. What I mean is, the "structure" of $\mbox{Im} f\otimes B'$, and $A' \otimes B'$ has nothing to do with each other. They are completely different. Hence $f \otimes g$, and $\tilde{f} \otimes g$ are 2 different morphisms. – user49685 Dec 6 '13 at 8:44

Let's take a closer look: when you write $(f\otimes g)(t)=(\widetilde{f}\otimes g)(t)$ you are assuming that LHS and RHS lie in the same module — i.e. that the natural map $\operatorname{Im}f\otimes B\to A\otimes B$ is a monomorphism. Of course the map $\operatorname{Im}f\to A$ is a monomorphism, but it's tensor product with $B$ needn't be.
• Thank you, but I have a minor additional question; does $f$'s being monomorphism change the result? – generic properties Dec 6 '13 at 8:42
• @dielectric I'm not sure I understand the question correctly. Is $Kef(f\otimes g)=\langle...\rangle$ if $f$ is mono (and $g$ is epi)? No, take, for example, $f:x\mapsto ax,\mathbb Z\to\mathbb Z$, $g:\mathbb Z\to\mathbb Z/b$ – Grigory M Dec 6 '13 at 13:38