In chapter 2.4 of Friedberg's text, no mention of subspaces is used in the theorems that he presents concerning isomorphisms. Unfortunately for me, exercise 17 of this chapter involves subspaces.
Here's the question and my vague outline of how I would like to attempt it:
Let $V$ and $W$ be finite-dimensional vector spaces and $T : V \to W$ be an isomorphism. Let $V_0$ be a subspace of $V$.
(a) Prove that $T(V_0)$ is a subspace of $W$ (I have already done this)
(b) Prove that $\dim(V_0) = \dim(T(V_0))$
Now what I would like to do is simply define a new linear transformation, $U$, that is equivalent to $T$, expect that it only operates on $V_0 \to U(V_0)$. Then I simply deduce that if $\beta$ is a basis for $V_0$, then $U(\beta)$ is a basis for $U(V_0)$ (I have proven this in a prior exercise). Finally, because $U$ is an isormorphism, and thus a bijection, $\beta$ and $U(\beta)$ have the same number of vectors, so $\dim(V_0) = \dim(U(V_0)) = \dim(T(V_0))$.
Is this legal? I'm very uncertain
The new linear transformation you are referring to is usually called the restriction of $T$ to $V_{0}$, denoted $T|_{V_{0}}$. Since $V_{0}$ is a subspace of $V$, you can check that $T|_{V_{0}}$ is a linear map from $V_{0}$ to $W$, so part (b) follows just as you wrote it.