How does it follow in this proof of the Rank-Nullity Theorem that $\mathcal{N}(A_0)=\mathcal{N}(A)\cap L=0$? I'm reading the following proof of the Rank-Nullity Theorem:

Theorem: Assume $V$ and $W$ are vector spaces, $V$ finite dimensional, and $A:V\to W$ a linear map. Then $\dim\mathcal N(A)+\dim\mathcal R(A)=\dim V$.
Proof: Let $\{w_1,\ldots,w_\ell\}$ be a basis of $\mathcal{N}(A)\subset V$, and complete it to a basis $\{w_1,\ldots,w_\ell,u_1,\ldots,u_m\}$ of $V$. Set $L=\text{Span}\{u_1,\ldots,u_m\}$, and consider $A_0:L\to W, A_0=A\mid_L$. Clearly: $$w\in\mathcal R(A)\implies w=A(a_1w_1+\cdots+a_\ell w_\ell+b_1 u_1+\cdots+b_mu_m)=A_0(b_1u_1+\cdots+b_mu_m),$$so $\mathcal R(A_0)=\mathcal R(A)$. Furthermore, $\mathcal N(A_0)=\mathcal N(A)\cap L=0$. Hence $A_0:L\to\mathcal R(A_0)$ is an isomorphism. Thus $\dim\mathcal R(A)=\dim\mathcal R(A_0)=\dim L=m$, and the result follows.

I can follow this proof up until the statement that $\mathcal N(A_0)=\mathcal N(A)\cap L=0$, which I'm sure is a trivial fact but I can't see how it follows. I can see that $\mathcal N(A_0)=\mathcal N(A)\cap L$, since: $$\{v\in V:Av=0\}\cap L=\{v\in L:Av=0\}=\{v\in L:A_0v=0\}=\mathcal N(A_0).$$
I am not seeing how the conclusion that $\mathcal N(A)\cap L=0$ follows from this, though. Any hints?
 A: It doesn't follow from this: you use the fact that this intersection is $0$ and that equality to deduce that $A_0$ is an isomorphism. 
To show that this intersection has just the element $0$ in it, take an element in the intersection, and write it following the basis in two ways using the fact that it is in the intersection. What can you deduce from that?
A: This is a general fact about (not necessarily linear) functions and restrictions to subsets. I will state this more generally, using the notion of inverse images. Specifically, let's suppose that we have a function $f : X \to Y$ and $B \subseteq Y$. Then, even if $f$ is not invertible, we denote:
$$f^{-1}(B) = \{x \in X : f(x) \in B\} \subseteq X.$$
In particular, if $f$ is linear, then
$$\mathcal{N}(f) = f^{-1}(\{0\}).$$
Now I'll state the more general form:

Suppose $X, Y$ are sets, $A \subseteq X$, $B \subseteq Y$, and $f : X \to Y$. Let $g : A \to Y$ be $f|_A$. Then,
  $$g^{-1}(B) = f^{-1}(B) \cap A.$$

Again, substituting $B = \{0\}$ gives the claim in the quoted proof.
To prove this more general claim, suppose $x \in g^{-1}(B)$. Note that this implies that $x$ lies in the domain of $g$, which is $A$, and hence also $X$. Further, we have $g(x) \in B$. As $g$ is a simple restriction, this implies $f(x) = g(x) \in B$. Thus, $x \in f^{-1}(B)$, but also $x \in A$, hence $x \in f^{-1}(B) \cap A$.
Conversely, suppose $x \in f^{-1}(B) \cap A$. Then $x \in A$, so $g(x)$ at least makes sense, and $f \in f^{-1}(B)$, so $f(x) \in B$. But again, by definition of $g$, this means $g(x) = f(x) \in B$, so $x \in g^{-1}(B)$, completing the proof.
