Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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.

share|cite|improve this question
$t$ is not injective. – Norbert Oct 20 '12 at 11:36
Thanks Norbert. – Reader Oct 20 '12 at 12:23
up vote 0 down vote accepted

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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.