Let $$\begin{array}{c} 0 & \xrightarrow{} &M' & \xrightarrow{i} & M & \xrightarrow{p} & M'' \\ & & & & \downarrow{g} & & \downarrow{h}\\ 0 & \xrightarrow{} &N' & \xrightarrow{j} & N & \xrightarrow{k} & N'' \\ \end{array}$$

be a commutative diagram ,$hp=kg$, in some abelian category like modules or abelian groups where both rows are exact, then prove there is an unique morphism $f:M' \to N'$ such $gi=jf$. My attemp to prove this was by diagram chasing trying to emulate proposition 2.70 of Rotmans´s Homological Algebra. My natural candidate for $f:M' \to N'$ is as $f:=j^{-1}gi$ where $j^{-1}:N \to N'$ is the preimage of $j$ but there may be some $n \in N$ such $n \notin Im(j)$ so how I can properly define this $f$?

  • 2
    $\begingroup$ The image of $j$ is the kernel of $k$, and the image of $g\circ i$ is contained in the kernel of $k$? $\endgroup$ Mar 1, 2020 at 20:04
  • $\begingroup$ That´s right beacuse the commutative of the right square. So candidate for$f$ is right? @CharlieFrohman $\endgroup$
    – Cos
    Mar 1, 2020 at 20:28
  • $\begingroup$ The candidate for $f$ is right. $\endgroup$ Mar 2, 2020 at 1:25

1 Answer 1


By diagram chasing: Let $m'\in M'$, then $i(m')\in \ker(p)$ by exactness, so we have $$ k(gi(m')) = h(pi(m')) = h(0) = 0 $$ so $gi(m')\in \ker(k)=j(N')$, then there exists $n'\in N'$ such that $gi(m')=j(n')$. Define $f(m')=n'$. This is well defined, because if $gi(m')=j(n'')$ for some $n''\in N'$ then $j(n')=j(n'')$, but by exactness, $j$ is injective, thus $n'=n''$.

Now note that $$ gi(m') = j(n') = jf(m') $$ and thus the diagram is commutative.

Finally, let $f':M'\to N'$ be another morphism such that the diagram is commutative. Note that $$ jf = gi = jf', $$ but $j$ is a monomorphism, so $jf=jf'$ implies that $f=f'$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .