morphism between short exact sequences of vector spaces

Question

I have recently been trying to learn some linear algebra, and have been following the book: Elements of Linear and Multilinear Algebra by John M. Erdman. The book states a theorem and has left the proof as an exercise to the reader. However, I don't even know how to start begining to prove the theorem. The theorem is as follows.

Consider the following diagram in the category of vector spaces and linear maps. $\require{AMScd}$ \begin{CD} 0@>>>U@>j>>V@>k>>W@>>>0\\ @.@VfVV@VVgV@VVhV\\ 0@>>>U'@>>j'>V'@>>k'>W'@>>>0 \end{CD} If the rows are exact and the left square commutes, then there exists a unique linear map $h:W\rightarrow W'$ that makes the right square commute.

I realise that $j$ and $j'$ are both injective and similarly that $k$ and $k'$ are both surjective, but I am unsure how to use this information to show that $h$ exists.

Answer 1

A map $h:W\to W'$ satisfies $h\circ k=k'\circ g$ iff for every $w\in W,$ written $k(v)$ for some $v\in V$ (by surjectivity of $k$), we have $h(w)=k'(g(v)).$ This proves the uniqueness of the solution $h$ if any.

Let us now prove this existence, i.e. prove that the previous condition $$w=k(v)\implies h(w)=k'(g(v))$$ defines $h$ non-ambiguously, and that such an $h$ is linear.

• the definition is non-ambiguous because if $k(v')=k(v)$ then $k'(g(v'))=k'(g(v)),$ since the difference is $(k'\circ g)(v'-v)=(h\circ k)(v'-v)=h(0)=0.$
• the linearity of $h$ follows easily from (again) the linearity of $k$ and $k'\circ g.$