# Finding unique morphism in a diagram of exact sequences such the whole diagram commute.

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 $$f$$?

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

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 $$f=f'$$.