0
$\begingroup$

Consider the following diagram which lives in the category of $R$-modules.

$$ \begin{array}{ccccccccc} 0 & \xrightarrow{i} & A & \xrightarrow{f} & B & \xrightarrow{q} & C & \xrightarrow{d} & 0 \newline & & \downarrow & & \downarrow & & \downarrow & & \newline 0 & \xrightarrow[j]{} & A & \xrightarrow[g]{} & E & \xrightarrow[r]{} & F & \xrightarrow[e]{} & 0 \end{array} $$

Let the first down arrow be an equivalence and the other down arrows be both epimorphisms.

Can we prove that the right square is a push out diagram? If not what can be said about the kernels of these two maps? Are they isomorphic?

What can be said about the diagram which is mirror revers of this diagram?

$\endgroup$
3
  • 1
    $\begingroup$ I'm afraid you can't load packages here. Try \array $\endgroup$ May 3, 2013 at 6:15
  • 1
    $\begingroup$ You've got end{array} instead of \end{array}. Also, for some reason its necessary to use \newline instead of \\ to end a line. $\endgroup$
    – Mark Grant
    May 3, 2013 at 6:46
  • 4
    $\begingroup$ There is substantial evidence that Hamid and all the answerers intended that the rows in the question's diagram are exact. Nevertheless, it might be good to explicitly say so. $\endgroup$ May 3, 2013 at 13:25

2 Answers 2

3
$\begingroup$

The right square is a pushout square as soon as the first downward arrow is an epimorphism, and we do not even need $f$ and $g$ to be monomorphisms. This is true in any abelian category.

Since diagrams are not easy to draw, let me label the downward arrows as $a : A \to A'$ (which we assume is an epimorphism but not necessarily an isomorphism), $b : B \to E$ and $c : C \to F$. Suppose $h_1 : E \to G$ and $h_2 : C \to G$ are morphisms such that $h_1 \circ b = h_2 \circ q$. Then, $h_1 \circ b \circ f = h_2 \circ q \circ f = 0$, so $h_1 \circ g \circ a = 0$ (because $g \circ a = b \circ f$). Since $a$ is an epimorphism, we must have $h_1 \circ b = 0$, thus $h_1 = h \circ r$ for a unique $h$, by the universal property of cokernels. Moreover, $h \circ c \circ q = h \circ r \circ b = h_1 \circ b = h_2 \circ q$, so $h \circ c = h_2$, because $q$ is an epimorphism. So we indeed have a pushout square.

$\endgroup$
3
  • $\begingroup$ Is this square also a pull back in any abelian category? $\endgroup$
    – HHH
    May 8, 2013 at 9:53
  • $\begingroup$ What is the minimal condition to impose to a square to deduce that it is a pull back or a push out? $\endgroup$
    – HHH
    May 8, 2013 at 9:55
  • $\begingroup$ See Fernando Muro's answer. For my answer, you only need an additive category and the existence of kernels and cokernels. $\endgroup$
    – Zhen Lin
    May 8, 2013 at 15:23
5
$\begingroup$

I guess you mean 'isomorphism' where you say 'equivalence' of $R$-modules. With your assumptions it is very easy to check (diagram chasing) that the folded sequence $$0\rightarrow B\stackrel{\binom{q}{?}}\longrightarrow C\oplus E\stackrel{(?,-r)}\longrightarrow F\rightarrow 0$$ is short exact. Hence the square on the right is both a pull-back and a push-out, essentially by definition. In particular, since it is a pull-back, the kernels of the two vertical maps on the right are the same.

$\endgroup$
2
  • 1
    $\begingroup$ Thank you for your answer. But I can't justify what do you mean by $R$. You may mean $E$. Am I right? $\endgroup$
    – HHH
    May 3, 2013 at 13:37
  • $\begingroup$ Yes, indeed, corrected. $\endgroup$ May 3, 2013 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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