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 $f\otimes g$?
 A: 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.
