Showing a linear transformation is onto Let $T:V\to W$ be a linear transformation from V to W. Is it sufficient to show that rank(T)=dim(W) to show that T is an onto function? I know that T is onto if R(T)=W. So the question is does dim(R(T))=dim(W) implies that R(T)=W but it doesn't seem right.
 A: That is true precisely when $\dim W$ is finite. Indeed, suppose we have a subspace $U \subseteq W$ with $\dim U = \dim W$. If $\dim W$ is finite then we have that $U = W$. Take a basis $\{u_1, \dots, u_n\}$ for $U$ where $n = \dim U = \dim W$. If there is a $w \in W - U$ then $\{u_1, \dots, u_n, w\}$ is linearly independent so $\dim W \geq n + 1 = \dim W + 1$, a contradiction. Finiteness of $\dim W$ here was really used in the final step - that $\dim W \geq \dim W + 1$ is a contradiction. This is obvious for $\dim W$ finite, but it's not contradictory for $\dim W$ infinite. Indeed, the key fact here is that any infinite set $S$ contains a proper subset $T < S$ with the same cardinality. This is called Dedekind infiniteness and is (under the axiom of choice) equivalent to infinitude. The argument can then proceed as follows: we claim that for any infinite dimensional vector space $W$, there exists a proper subspace $U < W$ with $\dim U = \dim W$. Take a basis $S$ of $W$ and take a proper subset $R < S$ of the same cardinality. Then let $U = span(R)$, which is proper in $W$ as $R < S$. Hence, $\dim U = |R| = |S| = \dim W$ but $U < W$, in contrast to the finite case.
Apply the above to $U = im(T)$.
