Maps involving direct sums of modules This might seem like a strange question... I am curious: 
If $A$ is a ring and $M_1, M_2, M_3, M_4$ are $A$-modules, and we have a map $f: M_1\oplus M_2 \rightarrow M_3\oplus M_4$, then is there some way to extract maps $f_1:M_1 \rightarrow M_3$ and $f_2: M_2 \rightarrow M_4$ from $f$? If so, can we say that $f$ is injective if and only if $f_1$ and $f_2$ are? 
 A: The composition of $f$ with the canonical inclusions and projections yields maps $$f_1:M_1 \hookrightarrow M_1 \oplus M_2 \xrightarrow{f} M_3 \oplus M_4 \twoheadrightarrow M_3 $$ and $$f_2:M_2 \hookrightarrow M_1 \oplus M_2 \xrightarrow{f} M_3 \oplus M_4 \twoheadrightarrow M_4. $$
In other words, we have $f_1(m) = \operatorname{pr}_1(f(m,0))$ and $f_2(m) = \operatorname{pr}_2(f(0,m))$. We don't quite have $f = f_1 \oplus f_2$, as generally $f$ will intertwine parts of $M_1$ and parts of $M_2$. (That is to say, generally the first component of $f$ will depend on its second input and the second component of $f$ will depend on its first input.) For example, if all the modules involved are the same and $f(a,b) = (a+b,0)$, then $f_1$ is the identity and $f_2$ is the zero map, and $(f_1 \oplus f_2)(a,b) = (a,0)$.
Also, with these maps it is not true that $f$ is injective iff $f_1$ and $f_2$ are. For instance if all the modules involved are the same and $f(a,b) = (b,a)$, then clearly $f$ is injective, but $f_1$ and $f_2$ are both zero.
