Why is T one-to-one iff T is onto? In Friedberg's Linear Algebra, the following proof that T is one-to-one if and only if T is onto is presented. Here, $T:V\rightarrow F$ is linear.
"We have that T is one-to-one if and only if N(T)=0, if and only if nullity(T) = 0, if and only if rank(T) = dim(V), if and only if rank(T) = dim(W), and if and only if dim(R(T))=dim(W)."
My questions is how does it follow that rank(T) = dim(W) from knowing that rank(T) = dim(V)?

This holds only if V and W have equal (finite) dimensions. 
From the rank-nullity theorem one has:
$nullity(T)+rank(T) = dim(V)$
Since $T$ is injective then $nullity (T)=0$. Then this implies $rank (T)=dim (V)=dim (W) $ which suggests surjectivity.
