Suppose $f: V \to W$ is linear and assigns a basis of $V$ to a basis of $W$ and let $\dim V = \dim W$. I want to show that $f$ is an isomorphism however I don't know what "$f$ assigns a basis of $V$ to a basis of $W$" means? Does it say that if $B = \{v_{1},...,v_{n}\}$ is a basis of $V$ then $f(B) = \{f(v_{1}),...,f(v_{n})\}$ is a basis of $W$?
 A: 
"$f$ assigns a basis of $V$ to a basis of $W$"

I would say it means:
There exists a basis $\{v_{1},...,v_{n}\}$ of $V$ and
a basis $\{w_{1},...,w_{n}\}$ of $W$ such that $f(v_j) = w_j$ for $j=1,\dots,n$.
A: If "$f$ assigns a basis of $V$ to a basis of $W$" means, as I think it should, that for some ordered basis $[b_1,\ldots,b_n]$, the family of vectors (in $W$) $[f(b_1),\ldots,f(b_n)]$ forms an ordered basis of $W$, then this immediately implies that $f$ is an isomorphism; no additional condition on the dimensions is required. Indeed, given that $[f(b_1),\ldots,f(b_n)]$ forms an ordered basis of $W$, there is for every family $[v_1,\ldots,v_n]$ in $V$ a linear map that sends $f(b_i)$ to $v_i$, for all$~i$. If $g$ is the linear map so obtained for $[v_1,\ldots,v_n]=[b_1,\ldots,b_n]$, then this means $g(f(b_i))=b_i$ for all$~i$, which implies by linearity that $g\circ f=\operatorname{Id}_V$, and similarly $f(g(f(b_i)))=f(b_i)$ for all$~i$ implies by linearity that $f\circ g=\operatorname{Id}_W$; thus $g$ is the inverse linear map of $f$. These even works for infinite dimensional spaces, after adapting the notation to cater for infinite basis and families of vectors.
