I saw in an exercise that if $T$ is a linear transformation $T: V\rightarrow W$ and $T_2: W\rightarrow Z$ and $T_1: X\rightarrow V$ are invertible then the rank of the composition doesn't change. So, my question is regarding this: is necessary the hypothesis that $T_1,T2$ are invertible in order to hold the conclusion, I think just assuming $T_2$ to be $1:1$ and $T_1$ to be onto is sufficient.
Proposition: Let $T: V\rightarrow W$ be a linear transformation and let $T_1: X\rightarrow V$ be a surjective linear transformation and $T_2: W\rightarrow Z$ be an injective linear transformation. Then
$$\dim(T(V))=\dim(TT_1(X))=\dim(T_2T(V))=\dim (T_2TT_1(X))$$
Proof: Since $T_1$ is surjective then $T_1(X)=V$, so $T(V)=T(T_1(X))$ and then $\dim(T(V))= \dim(TT_1(X))$. A similar argument shows that $\dim(T_2T(V))=\dim (T_2TT_1(X))$.
Now we will show that $\dim(T_2T(V)) =\dim(T(V))$. Since $T_2$ is one-to-one, then is injective when is restricted to $T(V)$. So $T_2\restriction _{T(V)}: T(V)\rightarrow T_2(T(V))$, is $1:1$ and onto. Hence $\dim(T_2T(V)) =\dim(T(V))$.
Is this correct?
Thanks in advance.
$\DeclareMathOperator\rank{rank} $You are right. For linear maps $f:Y\to Z$, $g:X\to Y$, we have $\rank(f\circ g) = \rank(g)$ when $f$ is injective and $\rank(f\circ g)=\rank(f)$ when $g$ is surjective.
In fact we only need $f$ to be injective on $g(X)$ for the first statement and we only need $\widetilde g:X\to Y/\ker(f)$ to be surjective for the second statement.