Four Lemma(proof) I am proving four lemma: 
I want to show that 
if the rows are in the commutative diagram are exact and m and p are surjective, and q is injective, then n is surjective. 
See the following link.
http://en.wikipedia.org/wiki/Five_lemma
When they are proving (1)
they say $t(n(c)) = p(h(c)) = t(c′)$, why can't they say that $n(c)=c'$ because $t$ is injective which follows by exactness at D'. Then everything is much easier.
But instead they say $t(c'-n(c))=0$ and so on.
Please help. 
 A: This comes immediately from a diagram chase as follows (in the notation of wikipedia): Suppose you take a $c' \in C'$ and want to know what in $C$ maps to it. Now $t(c') \in D'$ and by surjectivity of $P$ there is some $d \in D$ such that $P(d) = t(c')$. Now by exactness of the lower row, $u(t(c')) = 0$ and so by commutativity of the last square, $q(j(d)) = 0$. However $q$ injective implies that $j(d) = 0$ and so by exactness of the top row, $d = h(c)$ for some $c \in C$.
Now $t(n(c)) = P(d) = t(c')$ from which it follows by exactness of the bottom row again that $n(c) - c' = s(b')$ for some $b' \in B$. By surjectivity of $m$ we can write that $m(b) = b'$ from which it follows that $n(g(b)) = n(c) - c'$. It follows that $c' = n(c - g(b))$ from which it follows that $n$ is surjective.
Edit: You can only say that $t$ is injective if on the bottom row you have $$  B'  \stackrel{s}{\longrightarrow} C' \stackrel{t}{\longrightarrow} D' $$
with $B'= 0$ or $s$ being the zero map. Remember, for $t$ to be injective you need the image of $s$ to be zero. $s$ is any map at the moment and $B'$ and $R$ - module so how do you $t$ is injective?
