Clever proof for showing that if a graph $G$ is critically $k-$colorable then $\delta(G) \geq k - 1$ While reading for my graph theory class, I came across a short - yet curious - proof for the following theorem: if a graph $G$ is critically $k-$colorable then $\delta(G) \geq k - 1$. Here is the proof to the claim:

Suppose (for a contradiction) that $G$ is $k-$critical and that $v \in V(G)$ satisfies $\text{deg}(G) < k-1$. Then $G - v$ has a $(k-1)-$coloring, and this coloring extends to a $k-1$-coloring of $G$. This yields a contradiction$._{\Box}$

Everything in this proof makes sense besides one particular item: in the second line, what does the author mean by extending a coloring from a subgraph of $G$ to the whole graph? In addition, why does $G-v$ have a $(k-1)$-coloring that extends to all of $G$ (if that makes any sense)? I understand this may seem like an easy Google search but to be honest I can't find anything helpful and figured someone could provide some insight.
Note that this is not a homework question but simply for going beyond what I am learning in class.

Source:
https://www.sfu.ca/~mdevos/notes/graph/445_colouring0.pdf
 A: 
in the second line, what does the author mean by extending a coloring from a subgraph of $G$ to the whole graph?

Recall that a coloring formally is a function $f : V(G) \to \{1,2,\cdots,k\}$ for some integer $k$. So the notion of extension is basically the same as it would be for functions in other cases (extending a function from some substructure to a larger one).

In addition, why does $G-v$ have a $(k-1)$-coloring that extends to all of $G$ (if that makes any sense)?

We have a coloring $f : V(G-v) \to \{1,2,\cdots,k-1\}$ from $k$-criticality. We then define a coloring $\widetilde{f} : V(G) \to \{1,2,\cdots,k-1\}$ where $\widetilde{f} = f$ on $V(G-v)$. To complete the extension, we need to define $\widetilde{f}(v)$.
Notice that $\deg_G(v) < k-1$ by hypothesis (we can choose such a $v$ where $\deg_G(v) = \delta(v) < k-1$ in this approach by contradiction). Hence, there are at most $k-2$ nodes in $V(G)$ adjacent to $v$, which does not exhaust all of the possible $k-1$ colors in $f,\widetilde{f}$. Consequently, let $\widetilde{f}(v)$ be any of the colors not taken up by a neighbor of $v$.
This ensures that $\widetilde{f}$ is indeed a $k-1$ vertex-coloring. However, this contradicts the $k$-criticality of $G$, in the sense that deleting $v$ does not change the chromatic number of $G$.
