Advanced Modern Algebra, Ex 6.25 Been trying to figure this out for a while now (at least a week). The Problem Set was already handed in but I’m still trying to figure this out since I wasn’t able to answer this prior to handing it in. I really want to understand this better. Here’s the problem. 
Let $0\to A\to B\to C\to 0$ be a short exact sequence of Modules. If $M$ is any module, prove that there are exact sequences. 
$0\to A\oplus M\to B\oplus M\to C\to 0$
and 
$0\to A\to B\oplus M\to C\oplus M\to 0$. 
 A: Let $f:A\to B$ and $g:B\to C$ be the maps in your exact sequence above. Then let $\hat{f}:A\to B\oplus M$ be given by sending $a$ to $(f(a),0)$ and $\hat{g}:B\oplus M \to C\oplus M$ send $(b,m)$ to $(g(b),m)$. I will leave it to you now to show that $0\to A\to B\oplus M\to C\oplus M\to 0$ is exact.
The other problem is similar, so I will also leave it to you, but please comment if you get stuck.
A: Suppose
$$0\rightarrow A\stackrel{f}\rightarrow B\stackrel{g}\rightarrow C\rightarrow 0$$
is exact, and let
$$0\rightarrow A\oplus M\stackrel{f\oplus i}\rightarrow B\oplus M\stackrel{\overline g}\rightarrow C\rightarrow 0$$
with 
$$f\oplus i(a,m):=(f(a),m)\;\;,\;\;\overline g(b,m):=g(b)$$
then
$$\overline g\circ f(a,m)=\overline g(f(a),m):=gf(a)=0\Longrightarrow \operatorname {Im}f\subset \ker g$$
$$\overline g(b,m)=g(b)=0\Longrightarrow b\in\ker g=\operatorname {im}f\Longrightarrow \text{there exists}\;\; a\in A\,\,\text{s.t.}\,\,f(a)=b$$
$$\Longrightarrow (b,m)=(f(a),m)=f\oplus i(a,m)\Longrightarrow \ker g\subset\operatorname{Im}f$$
Since $\,f\oplus i\,$ is clearly $\,1-1\,$ and $\,\overline g\,$ is clearly onto, we're done.
A: @ Brett Frankel You wanted to write $\hat g: B\oplus M \to C $ defined as $(b,m)\mapsto g(b)$ and not $(b,m)\mapsto (g(b), m)$. I wanted to write this as a comment.  Sorry  I am posting it as an answer. 
Can somebody please tell me how can I write a comment?
