Isomorphism by restricting surjective linear mapping between finite dimensional spaces, and quotient space by null space In Axler's book Linear Algebra Done Right, Section 3D, Exercise 8:
Suppose $V$ is finite-dimensional and $T:V\rightarrow W$ is a surjective linear map of $V$ onto $W$. Prove that there is a subspace $U$ of $V$ such that $\left.T\right|_U$ is an isomorphism of $U$ onto $W$. (Here $\left.T\right|_U$ means the function $T$ restricted to $U$. In other words, $\left.T\right|_U$ is the function whose domain is $U$, with $\left.T\right|_U$ defined by $\left.T\right|_U (u)=Tu$ for every $u\in U$.)
An immediate solution would be considering a basis $w_1, \dots, w_m$of $W$. Then there exists $v_1, \dots, v_m \in V$, linearly independent, and $Tv_i=w_i$. Let $U =$ span($v_1, \dots, v_m$), then we get an isomorphism $\left.T\right|_U : U\rightarrow W$.
My question is that do we have $U = V/(\operatorname{null}T)$, where $V/(\operatorname{null}T)$ is the quotient space of $V$ by null$T$? And do we have  $\left.T\right|_U = \tilde{T}$, where $\tilde{T}: V/(\operatorname{null}\to W)$ is defined by
$\tilde{T}(v+\operatorname{null}T)=Tv$?
 A: Given finite-dimensional $k$-vector spaces $V$ and $W$ with a surjective linear transformation $T : V \to W,$ by the First Isomorphism Theorem, there exists a unique linear transformation $S : V / \ker T \to W$ such that $T = S \circ P,$ where $P : V \to V / \ker (T)$ is the canonical surjection defined by $P(v) = v + \ker T.$ Consequently, your intuition is essentially correct; however, it is not technically true that $U = V / \ker T.$
We will need to do things in a slightly different order than your proof, but we will ultimately establish that $V = U \oplus \ker T$ so that $U$ can be identified (by the First Isomorphism Theorem) with the quotient space $V / \ker T.$ Recall that any linearly independent set of fewer than $\dim V = n$ vectors can be extended to a basis for $V.$ Consequently, we may extend a basis $\{v_{m + 1}, \dots, v_n \}$ of $\ker T$ to a basis $\{v_1, \dots, v_m, v_{m + 1}, \dots, v_n \}$ of $V,$ where we denote $\dim W = m.$ We may define (as you have) the subspace $U = \operatorname{span}_k \{v_1, \dots, v_m \}.$ We claim that $T(U) = W$ so that $T|_U : U \to W$ is surjective and therefore an isomorphism. Given any vector $w$ in $W,$ there exists a vector $v$ in $V$ such that $T(v) = w$ by hypothesis that $T$ is surjective. Using our basis, we may write $v = a_1 v_1 + \cdots + a_m v_m + a_{m + 1} v_{m + 1} + \cdots + a_n v_n,$ from which it follows that $$\begin{align} w = T(v) &= T(a_1 v_1 + \cdots + a_m v_m + a_{m + 1} v_{m + 1} + \cdots + a_n v_n) \\ \\ &= T(a_1 v_1 + \cdots + a_m v_m) + a_{m + 1} T(v_{m + 1}) + \cdots + a_n T(v_n) \\ \\ &= T(a_1 v_1 + \cdots + a_m v_m) \end{align}$$ by hypothesis that $v_{m + 1}, \dots, v_n$ are basis elements for $\ker T.$ We have therefore that every vector $w$ of $W$ can be written as $T(u)$ for some vector $u$ in $U$ so that $T(U) = W,$ as desired.
Last, in order to prove that $V = U \oplus \ker T,$ we must show two things: (1.) $V = U + \ker T$ and (2.) $U \cap \ker T = \{0\}.$ Considering that $\{v_1, \dots, v_m, v_{m + 1}, \dots, v_n \}$ is a basis for $V$ such that $\{v_1, \dots, v_m \}$ is a basis for $U$ and $\{v_{m + 1}, \dots, v_n \}$ is a basis for $\ker T,$ the first condition holds. Given a vector $v \in U \cap \ker T,$ we have that $v = a_1 v_1 + \cdots + a_m v_m$ and $v = a_{m + 1} v_{m + 1} + \cdots + a_n v_n,$ from which it follows that $$a_1 v_1 + \cdots + a_m v_m - a_{m + 1} v_{m + 1} - \cdots - a_n v_n = 0.$$ But this implies that $a_1 = \cdots = a_m = a_{m + 1} = \cdots = a_n = 0$ so that $v = 0,$ as desired.

Perhaps this means nothing to you at the moment, but this is a consequence of a more general notion: we say that a short exact sequence of $R$-modules $$0 \to L \xrightarrow \varphi M \xrightarrow \psi N \to 0$$ splits if there exists an isomorphism $\gamma : M \to L \oplus N$ such that $\gamma \circ \varphi : L \to L \oplus N$ is the inclusion map and $\psi \circ \gamma^{-1} : L \oplus N \to N$ is the canonical surjection into the second coordinate. Given that $R = k$ is a field, we refer to an $R$-module (or a $k$-module, as it were) as a $k$-vector space. Every $k$-vector space is free as an $R$-module, and every free $R$-module is projective. Every short exact sequence of $R$-modules that ends with a projective $R$-module splits. By the First Isomorphism Theorem, we obtain the short exact sequence $$0 \to \ker T \xrightarrow i V \xrightarrow T W \to 0,$$ where $i : \ker T \to V$ is the inclusion map. By identifying $W$ and $V / \ker T,$ we obtain the short exact sequence $$0 \to \ker T \xrightarrow i V \xrightarrow P \frac V {\ker T} \to 0.$$ Ultimately, this is all to say that this behavior you intuitively detected is predictable and always works.
