Proving an isomorphism $T_1T_2$ knowing $T_1$ and $T_2$ are 1-1 and onto, respectively Say V is a finite dimensional vector space, and $T_1: V \to\ V$ and $T_2: V \to\ V$ are linear transformations. Assume that $T_1$ is one to one and $T_2$ is onto.
I am uncertain if my proof that $T_1T_2$ is an isomorphism is valid. 
If you don't want to look at the picture - Is it necessary to show that $T_1$ and $T_2$ are each isomorphisms themselves by Rank-Nullity?
Here is an image of my work:

 A: I think it is easier to show the following easy claim which follows at once from the dimension (or rank-nullity, if we talk matrices language) theorem:
Claim: An operator $\,T:V\to V\,$ , with $\,\dim V<\infty\,$ is bijective iff it is injective iff it is surjective.
With the above, it follows at once that both $\,T_1,T_2\,$ are bijections (isomorphisms of vector spaces)...
A: Thanks to a discussion with @David_Mitra, I have an answer that I understand. In short, my picture answer is not sufficient, and I do need to show that $T_1$ and $T_2$ are each isomorphisms. 
More specifically, we just need to make sure that $T_1$ and $T_2$ are each one-to-one:


*

*By Rank-Nullity, we can show that $T_2$ is not only onto, but also 1-1.

*Observe $T_2(v) = 0$ if $T_1(T_2(v))=0$ because $Ker(T_1) = {0} $ (because $T_1$ is 1-1)

*Observe $v = 0$ if $T_2(v)=0$ because $Ker(T_2) = {0} $ (because $T_2$ is 1-1)

*Therefore, $Ker(T_1T_2)=Ker(T_1(T_2(v)))=0$ - so $T_1T_2$ is 1-1

*By Rank-Nullity, we can show that $T_1T_2$ is onto

*$T_1T_2$ is an isomorphism!

