# 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.

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$t$ is not injective. – Norbert Oct 20 '12 at 11:36
Thanks Norbert. – Reader Oct 20 '12 at 12:23

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?

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Thanks but my question was: is $t$ injective? – Reader Oct 20 '12 at 12:21
@Reader You cannot say that $t$ is injective because there is no zero to the left of the bottom row. End of story and your proof does not work. – user38268 Oct 20 '12 at 12:21
Thanks BenjaLim – Reader Oct 20 '12 at 12:23
you mean $C'=0$ and not $B'=0$? – Reader Oct 20 '12 at 12:43
@Reader No I meant $B'= 0$. – user38268 Oct 20 '12 at 12:58