The full question is here: Let f be a correspondence between vector spaces V & W (that is, a map that is one to one and onto). Show that the spaces V & W are isomorphic via f if and only if there are bases B $\subset$ V and D $\subset$ W such that corresponding vectors have the same coordinates: $Rep_B(\vec{v}) = Rep_D(f(\vec{v}))$.
I understand the proof to the first part of the iff statement (if V& W are isomorphic, then etc, etc.), but I'm struggling to understand the proof for the second part of the statement, as seen below by the solution.
The solution states this: " ... For the other half, assume that there are bases such that corresponding vectors have the same coordinates with respect to those bases. Because f is a correspondence, to show that it is an isomorphism, we need only show that it preserves structure. Because $Rep_B\vec{v}$ = $Rep_Df(\vec{v})$, the map f preserves structure if and only if representations preserve addition: $Rep_B(\vec{v}_1 + \vec{v}_2) = Rep_B(\vec{v}_1) + Rep_B(\vec{v}_2)$ and scalar multiplication: $Rep_B(r*\vec{v}) = r*Rep_B(\vec{v})$. The addition calculation is this: $( c_1 + d_1)\vec{\beta_1} + ... + (c_n+d_n)\vec{\beta_n} = c_1\vec{\beta_1} + ... + c_n\vec{\beta_n} + d_1\vec{\beta_1} + ... + d_n\vec{\beta_n}$, and the scalar multiplication calculation is similar.
My problem with this solution is that it doesn't seem to show an isomorphism between V & W at all, because the preservation of addition calculation only shows preservation between $Rep_B(\vec{v})$ and the expanded basis calculation, rather than the corresponding $Rep_D(f(\vec{v}))$. Wouldn't the proper equation look something more like this:
$Rep_B(\vec{v_1}+\vec{v_2}) = Rep_D(f(\vec{v_1} + \vec{v_2})) =>$ $Rep_B(\vec{v_1})+Rep_B(\vec{v_2}) = Rep_D(f(\vec{v_1})) + Rep_D(f(\vec{v_2})) =>$ $(c_1+d_1){\beta_1} + ... + (c_n + d_n){\beta_n} = (c_1 + d_1)D_1 + ... + (c_n+d_n)D_n$
(I know the last imiplication doesn't make sense, but it's just to illustrate where my thought process is at with regards to this question).
Thanks for any insight regarding this specific solution or any better proofs that you may have!
Let $B=\{\vec{b}_1,\vec{b}_2,\ldots, \vec{b}_n\}$ and $D=\{\vec{d}_1,\vec{d}_2,\ldots, \vec{d}_n\}.$ According to the assumptions for any $\vec{v}$ we have $$\vec{v} =\sum_{k=1}^n \alpha_k\vec{b}_k,\qquad f(\vec{v})=\sum_{k=1}^n \alpha_k\vec{d}_k$$ Therefore th mapping $\vec{v}\mapsto f(\vec{v})$ acts by $$\sum_{k=1}^n \alpha_k\vec{b}_k\longmapsto \sum_{k=1}^n \alpha_k\vec{d}_k$$ This mapping is an isomorphism. Indeed as was proved, it is linear. Moreover it is one-to-one as$$\sum_{k=1}^n \alpha_k\vec{d}_k =0\quad \implies \alpha_1=\ldots=\alpha_n=0 $$ due to linear indepedence of $\vec{d}_k.$ It is surjective automatically (because one-to-one linear mappings preserve the dimension), or by inspection, as all the vectors $\vec{d}_k$ belong to the range of the mapping, so do all their linear combinations.