Prove that there exists $T \in \mathcal{L}(V, W)$ such that $\operatorname{null}(T) = U$ if and only if $\dim (U) \ge \dim (V) - \dim (W)$ 
Suppose that $V$, and $W$ are finite dimensional vector spaces and that $U$ is a subspace of $V$. Prove that there exists $T \in \mathcal{L}(V, W)$ such that $\operatorname{null}(T) = U$ if and only if $\dim(U) \ge \dim(V) - \dim(W)$.

The answer to this question partially makes sense, particularly the forward direction in which we assume $\operatorname{null}(T) = U$. However, the other direction does not, here is the answer given:

Suppose that $\dim(U) \ge \dim(V) - \dim(W)$. Let $(u_1, \ldots, u_m)$ be a basis of $U$. Extend to a basis $(u_1, \ldots, u_m, v_1, \ldots, v_n)$ of $V$. Let $w_1, \ldots, w_p$ be a basis of $W$. For $a_1, \ldots, a_m, b_1, \ldots, b_n \in \mathbf{F}$ define $T(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n)$ by:
$$
T(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n) = b_1w_1 + \ldots + b_nw_n
$$
Clearly $T \in \mathcal{L}(V, W)$ and $\operatorname{null}(T) = U$.

I don't see why $\operatorname{null}(T) = U$. Additionally, I do not see the reasoning for defining the linear mapping as it is - what is the thought process behind choosing it to be that? Why does it map to $b_1w_1 + \ldots + b_nw_n$ and not use another constant (i.e. $c_1w_1 + \ldots + c_nw_n$ where $c_1, \ldots, c_n \in \mathbf{F}$)?
 A: Another way to write it:

Define $T : V \to W$ by $T(u_i) = 0_W$ for $i \in \{1,\dots,m\}$, $T(v_j) = w_j$ for $j \in \{1,\dots,n\}$ and extend it by linearity.

Note that the hypotheses implies that
$$p = \dim(W) \geq \dim(V) - \dim(U) = (m+n)-m = n,$$
so choosing $w_1,\dots,w_n$ of $w_1,\dots,w_p$ makes sense.
Also, note that this satisfies your definition, for if $a_1,\dots,a_m,b_1,\dots,b_n \in \mathbf F$, then
\begin{align}
T(a_1u_1 + & \cdots + a_mu_m + b_1v_1 + \cdots + b_nv_n) \\
&= a_1T(u_1) + \cdots + a_mT(u_m) + b_1T(v_1) + \cdots + b_nT(v_n) \\
&= b_1w_1 + \cdots + b_nw_n.
\end{align}
Now, in one hand, it is easy to see that $U \subseteq \operatorname{null}(T)$ since every $u \in U$ can be written as a linear combination of $u_1,\dots,u_m$. On the other hand, let $v \in V$ and write it as
$$v = c_1u_1 + \cdots + c_mu_m + d_1v_1 + \cdots + d_nv_n$$
for some $c_1,\dots,c_m,d_1,\dots,d_n \in \mathbf F$. If $v \in \operatorname{null}(T)$, then
$$0_W = T(v) = d_1w_1 + \cdots + d_nw_n$$
and since $w_1,\dots,w_n$ are linearly independent, $d_1 = \cdots = d_n = 0$. So
$$v = c_1u_1 + \cdots + c_mu_m \in U.$$
A: Let's first consider this definition:
$$T(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n) = b_1w_1 + \ldots + b_nw_n.$$
This definition only really makes sense because $(u_1, \ldots, u_m, v_1, \ldots v_n)$ is a basis for $V$. So, any vector $x \in V$ can be expressed uniquely in the form
$$x = a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n.$$
Now, if $x \in U$, then $x$ must uniquely take the form
$$x = a_1u_1 + \ldots + a_mu_m = a_1u_1 + \ldots + a_mu_m + 0v_1 + \ldots + 0v_n,$$
since $(u_1, \ldots, u_m)$ is a basis for $U$. So, according to our definition of $T$, for $x \in U$, we have
\begin{align*}
T(x) &= T(a_1u_1 + \ldots + a_mu_m) = T(a_1u_1 + \ldots + a_mu_m + 0v_1 + \ldots + 0v_n) \\
&= 0w_1 + \ldots + 0w_n = 0.
\end{align*}
So, $U \subseteq \operatorname{Null} T$.
Conversely, suppose $x \in \operatorname{Null} T$. We still know $x$ is in the form
$$x = a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n,$$
but this time we know that
$$0 = T(x) = b_1w_1 + \ldots + b_nw_n.$$
Therefore,
$$x = a_1u_1 + \ldots + a_mu_m + 0 \in U,$$
completing the proof that $U = \operatorname{Null} T$.

Why define it with $b_1, \ldots, b_n$ instead of $c_1, \ldots, c_n$? Well, remember that $b_1, \ldots, b_n$ are not constants, they are placeholder variables. $T$ is defined by expansion with respect to the basis $(u_1, \ldots, u_m, v_1, \ldots, v_n)$, and the way that the author has chosen to denote such an expansion is by
$$x = a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n.$$
Thus, $b_1, \ldots, b_n$ are defined implicitly as (linear) functions of $x$, taking the vector $x$, and returning the coordinate of the corresponding basis vector $v_i$.
To simply replace them with $c_1, \ldots, c_n$, without defining them somehow, would mean the transformation is ill-defined. What are $c_1, \ldots, c_n$ in this context? How does the value of $x$ change them?
You could replace $b_1, \ldots, b_n$ with certain functions of $b_1, \ldots, b_n$ to obtain an equally valid construction $T'$ such that $\operatorname{Null} T' = U$. For example, the following $T'$ will also work:
$$T'(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n) = b_nw_1 + \ldots + b_1w_n.$$
Such functions are usually not unique!
A: First of all your proof does not emphasis where the assumption is used.
To define a linear map $T$, it is  sufficient to assign the images of members of a basis. Now we want $\text{null}(T)=U$, so first we take a basis $\mathcal{B}_0$ of $U$. Then extend $\mathcal{B}_0$ to a basis $\mathcal{B}$ of $V$. Now to have a linear map $T$ with $\text{null}(T)=U$ we have to assign each member of $\mathcal{B}_0$ to $0$ and rest of the members of the $\mathcal{B}$ have to assign so that $\{T(v):v\in\mathcal{B}\smallsetminus\mathcal{B}_0\}$ is linearly independent in $W$. Now the assumption $\dim(U)\geq\dim(V)-\dim(W)$ $\implies \dim(W)\geq\dim(V)-\dim(U)$ $\implies \dim(W)\geq|\mathcal{B}\smallsetminus\mathcal{B}_0|$. Thus this assures there are $|\mathcal{B}\smallsetminus\mathcal{B}_0|$ many linearly independent vectors in $W$. Which guarantees the existence of required $T$.
