Let $f$ : $\Bbb Z_2$ $\rightarrow$ $\Bbb Z_4$ given as $f$($a$ + $2$$\Bbb Z$) = $2a$ + $\Bbb Z_4$ is a monomorphism . And knowing that <0,2> as subgroup of $\Bbb Z_4$ is isomorphic to $\Bbb Z_2$ how can I prove that the morphism $f$ $\otimes $ $1$ : $\Bbb Z_2$$\otimes $ $\Bbb Z_2$ $\rightarrow$ $\Bbb Z_4 \otimes $ $\Bbb Z_2$ is the zero morphism
Any help by sketching a proof or proving this statement will be apreciated.
If I understood you correctly, we have (as $\;\Bbb Z$- modules = abelian groups) for any basic tensor $\;a\otimes b\in \Bbb Z_2\otimes\Bbb Z_2\;$:
$$f\otimes1(a\otimes b):=f(a)\otimes b:=(2a)\times b=2(a\otimes b)=a\otimes (2b)=\ldots$$