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Let $S$ be a commutative ring with unity, and let $A,B,A',B'$ be $S$-modules. If $\phi:A\rightarrow A'$ and $\psi:B\rightarrow B'$ are $S$-module homomorphisms, is it true that

$$\operatorname{im}(\phi\otimes\psi)=\operatorname{im}(\phi)\otimes_S \operatorname{im} (\psi)?$$

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up vote 11 down vote accepted

No, because the map $im(\phi)\otimes_S im(\psi) \to A'\otimes_S B'$ may not be injective.

E.g. consider the case $S = \mathbb Z$, $A = A' = B =\mathbb Z/2\mathbb Z,$ $B' = \mathbb Q/\mathbb Z$, $\phi = id_A$, and $\psi: \mathbb Z/2\mathbb Z \hookrightarrow \mathbb Q/\mathbb Z$ is the unique injection.

Then the map $A \otimes_S B \to im(\phi) \otimes_S im(\psi)$ is an isomorphism, while the map $\phi \otimes_S \psi$ is the zero map, since $A'\otimes_S B' = 0$.

More generally, if you restrict to the case when $\phi$ is the identity, and $\psi$ is injective, you are asking whether the injection $\psi: B \hookrightarrow B'$ induces an injection $A\otimes_S B \hookrightarrow A \otimes_S B'$. The answer is no in general because tensoring is not left exact. (The preceding example is one illustration of this.)

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Thanks a lot for great answer. I don't quite see what the unique injection $\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Q}/\mathbb{Z}$ is though...could you please tell me? And what if all the modules $A,A',B,B'$ are free $S$- modules (even finitely generated), do we then have $$\operatorname{im}(\phi\otimes\psi)=\operatorname{im}(\phi)\otimes_{S}\operator‌​name{im}(\psi)?$$ – Bart Patzer May 16 '12 at 13:16
$\psi : A \rightarrow \psi(A)$ and $\phi: B \rightarrow \phi(B)$ are surjective. Thus $\psi\otimes 1_B:A \otimes B \rightarrow \psi(A) \otimes B$ and $1_{\psi(A)} \otimes \phi : \psi(A) \otimes B \rightarrow \psi(A) \otimes \phi(B)$ are surjective. Thus $\psi \otimes \phi = (1_{\psi(A)} \otimes g) \circ (\psi \otimes 1_B)$ is surjective. What am i missing? – Manos May 16 '12 at 13:51
Thus $im(\psi \otimes \phi) = \psi(A) \otimes \phi(B)=im(\psi) \otimes im(\phi)$. – Manos May 16 '12 at 13:59
@Manos: Dear Manos, The map from the tensor product of the images to the image of the tensor product is surjective, but not injective in general (e.g. as in the example I give). Regards, – Matt E May 16 '12 at 14:42
@BartPatzer: Dear Bart, The group $\mathbb Q/Z$ has a unique subgroup of order two; let $\psi$ be the unique isomorphism of $\mathbb Z/2\mathbb Z$ with this subgroup. Regards, – Matt E May 16 '12 at 14:43

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