Failure of proof for $\operatorname{null}T_1 \subset \operatorname{null}T_2 \implies \exists S~\text{s.t.}~T_2 = ST_1$ In S.Axler's "Linear Algebra Done Right (3ed) an exercise asks of the following:

Suppose $W$ is finite dimensional and $T_1, T_2 \in \mathcal{L}(V,W)$.
  Prove that $\operatorname{null} T_1 \subset \operatorname{null} T_2$
  if and only if there exists $S \in \mathcal{L}(W,W)$ such that $T_2 = ST_1$.

I am having troubles proving in the direction $\operatorname{null}T_1 \subset \operatorname{null}T_2 \implies \exists S~\text{s.t.}~T_2 = ST_2$. I have tried several things but most of them assume that $V$ is finite dimensional, which is not stated in this case.
Most of them are similar to the solution stated here, the gist of which is to extend a basis of $\operatorname{null} T_1$ to a basis of $V$.
However both $\operatorname{null}T_1$ and $V$ are not necessarily finite. Why then is it valid to assume that a basis of $\operatorname{null}T_1$ is extendable
to a basis of $V$? Is there any implication on their dimensions from the question which I am missing out?
The relevant proofs in the book only show that any basis of a subspace $U$ of a finite space $V$ can be extended to a basis of $V$, but do not assume anything for infinite-dimensional subspaces.
 A: Let $\{T_1(v_1),T_1(v_2),\ldots,T_1(v_k)\}$ be a basis for $\operatorname{range}T_1$, extend it to the basis 
$\{T_1(v_1),T_1(v_2),\ldots,T_1(v_k),w_{k+1},\ldots,w_m\}$ for $W$.
Since $T_2(v_j)\in W$ for each $j$, it is well-known that (see the Lemma.) there exists
$S\in\mathcal{L}(W,W)$ such that
$$ST_1(v_j)=T_2(v_j)\quad\mbox{for }j=1,2,\ldots,k,$$
and $S(w_{k+1})=S(w_{k+2})=\cdots=S(w_m)={\it 0}$.
Finally, we check that $T_2=ST_1$. Given $x\in V$, and write $T_1(x)$ as
a linear combination of vectors in $\{T_1(v_1),T_1(v_2),\ldots,T_1(v_k)\}$,
that is,
$$T_1(x)=\sum_{j=1}^ka_jT_1(v_j)\quad\mbox{for some scalars }a_1,a_2,\ldots,a_k.$$
Then $T_1(x)=\displaystyle T_1\left(\sum_{j=1}^ka_jv_j\right)$ and
$${\it 0}=T_1(x)-\sum_{j=1}^ka_jT_1(v_j)
=T_1\left(x-\sum_{j=1}^ka_jv_j\right),$$
which implies $x-\displaystyle\sum_{j=1}^ka_jv_j\in\operatorname{null}T_1\subset\operatorname{null}T_2$. Therefore
\begin{align}
T_2(x)
&=T_2\left(\sum_{j=1}^ka_jv_j\right)
=\sum_{j=1}^ka_jT_2(v_j)
=\sum_{j=1}^ka_jST_1(v_j)
=ST_1\left(\sum_{j=1}^ka_jv_j\right)
=ST_1(x),
\end{align}
and hence $T_2=ST_1$.
A: This can also be proved using the First Isomorphism Theorem,which is one of the most important theorems (at least in my point of view) for abstract algebra. We have that 
$$V/null\ T_{1}\cong image\ T_{1}$$,let the isomorphism be $\phi_{1}$,and
$$V/null\ T_{2}\cong image\ T_{2}$$,let the isomorphism be $\phi_{2}$.
Define a map:
$$\rho\ :\ v+null\ T_{1}\ \rightarrow\ v+null\ T_{2}$$
Since $null\ T_{1} \subseteq null\ T_{2}$,the map $\rho$ is well-defined and linear,maps the space $V/null\ T_{1}$ to a subspace of $V/null\ T_{2}$. Because we have the two isomorphisms above,there will be a linear map $\phi'=\phi_{2}\rho \phi^{-1}_{1}$ that maps $image\ T_{1}$ to a subspace of $image\ T_{2}$ in the vector space $W$. Calculate: 
$$\phi'\ T_{1}v=\phi_{2}\rho \phi^{-1}_{1}T_{1}v=\phi_{2}\rho (v+null\ T_{1})=\phi_{2}(v+null\ T_{2})=T_{2}v$$
we therefore see that $\phi' T_{1}= T_{2}$ and this is what we want to get. Vector​ space $W$ is of finite-dimensional,we can easily expand the map $\phi'$ into $S\in \mathcal{L}(V,W)$ where $S|_{image\ T_{1}}=\phi'$,and then we obtain $ST_{1}=T_{2}$,as desired.
