Proving that if two linear transformations are one-to-one and onto, then their composition is also.

Question

I am attempting to solve a problem with the following given conditions:

Let V, W. and Z be vector spaces, and let $T:V \longrightarrow W$ and $U: W\longrightarrow Z$ be linear.Prove that if U and T are one-to-one and onto, then UT is also.

Here is my attempt for the one-to-one part:

"Suppose that U and T are one-to-one and onto. By assumption, for $\forall \space T(x),T(y) \in W $ where T(x) = T(y), we have x = y. It also follows by assumption that $ \forall \space U(g),U(h) \in Z $, we have g = h. Now suppose g = T(x) and h = T(y). Then we have UT(x) = U(T(x)) = U(g) = U(h) = U(T(y)) = UT(y).

Thus we conclude UT is one-to-one."

Is this a valid proof?

Thanks!

Answer 1

$$\begin{align*}&UTv=UTv'\stackrel{U\,\text{is}\,1-1}\implies Tv=Tv'\stackrel{V\,\text{is}\,1-1}\implies v=v'\;,\;\;\text{so}\,\,UT\;\text{is}\;1-1\\{}\\ (i)\;&\forall\,z\in Z\;\exists\,w\in W\;\text{with}\;Uw=z\;\;\text{because surjectivity of}\;U\;,\;\;\text{and}\\ (ii)\;&\exists\,v\in V\;\text{with}\;Tv=w\;,\;\;\text{because surjectivity of}\;T\\ &\text{and from (i)-(ii) it follows}\;\;UTv=Uw=z\implies UT\;\text{is surjective}\end{align*}$$

The above is true for general functions and sets, not necessarily for linear maps and linear spaces.