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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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    There is also an explanation in Keith Conrad's notes on tensor products (Example 2.14)-http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod2.pdf2016-12-28

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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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    @Wfpiggie For line$a$r maps between linear spaces they are equal. This follows (and it is equivalent) from the property that the rank of Kronecker product of two matrices (over an arbitrary field) equals the product of ranks. (Btw, this property deserves a proof as long as many classical textbooks in linear algebra, e.g. Horn & Johnson, give proofs only for real matrices via SVD.)2012-11-12