# tensor product of $R$-module homomorphisms

Let $M,M',N,N'$ be modules over a commutative ring $R$, and $f:M\to M'$ and $g:N\to N'$ are $R$-modules homomorphisms. Then prove or disprove the following statements.

a) If $f$ and $g$ are surjective, so is $f\otimes g:M\otimes N\to M'\otimes N'$.
b) If $f$ and $g$ are injective, so is $f\otimes g:M\otimes N\to M'\otimes N'$.

Can you help me please? solution or any hint? what is the kernel of $f\bigotimes g$

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Since the OP has indicated that he/she has solved it using my hints (see edit history and comments), I have edited to give a complete answer for future readers.

Statement a. While it is possible to take an arbitrary sum of tensors in $M'\otimes N'$ and show that it is in the image of $f\otimes g$, I will opt for using the universal property of tensor products and notion of epimorphisms as this will clarify the reason why Statement b is not true, let alone the dual of this one.

Let $h,k:M'\otimes N'\to L$ be arrows which satisfy $h\circ f\otimes g=k\circ f\otimes g$. Since the diagram below commutes note that we have $\bar h\circ f\times g=\bar k\circ f\times g$, where $\bar h:M'\times N'\to L$ is the composite of $h$ with the universal arrow. Since $f\times g$ is epi when both $f$ and $g$ are epi, $\bar h=\bar k$, and by the universal property of tensor products, $h=k$.

$$\require{AMScd}\begin{CD} M\times N @>f\times g>> M'\times N' \\ @VVV @VVV \\ M\otimes N @>f\otimes g>> M'\otimes N' \end{CD}$$

Statement b. If you attempt to argue in the same way here, you come to a halt right at the start, and it becomes clear that there is no way to work with an equality like $f\otimes g\circ h=f\otimes g\circ k$. This gives a moral justification for the fact that this statement is false.

To construct a counterexample all we have to do is find a non-flat module $A$ and an injection which does not remain injective by tensoring with $A$. E.g. let $R=\mathbb Z$, $f:\mathbb Z\to\mathbb Z$ be multiplication by $2$, and $g:\mathbb Z/2\mathbb Z\to \mathbb Z/2\mathbb Z$ be the identity. Then $f\otimes g:\mathbb Z/2\mathbb Z\to\mathbb Z/2\mathbb Z$ is the zero map.

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Comment if you need additional help, but please show what you have tried (in this case, showing what $g$ maps you have tried would be good). –  Karl Kronenfeld Jan 21 '14 at 22:04
I can't find such g.. Can you give me hint? –  user114952 Jan 21 '14 at 22:13
@user114952 If you feel stuck, try the simplest things you can think of. The simplest $\mathbb Z$-module is $\mathbb Z$ itself. –  Karl Kronenfeld Jan 21 '14 at 22:16
I can understand. thx –  user114952 Jan 21 '14 at 22:19
The careful reader will notice some ambiguity in the proof of Statement a with regard to which category we are using. The "commutative diagram" is, strictly speaking, an arrow in the category of bilinear maps, but it should be helpful to think of it as somehow transcending that status. –  Karl Kronenfeld Jan 23 '14 at 1:28