$D$-Linear Map Surjective Map Implies Injective

Problem: Let $V$ and $W$ be finite dimensional vector spaces over a division ring $D$ with the same dimension. Let $f : V \to W$ be a $D$-linear map. Then if $f$ is surjective, it is also injective.

I already showed the converse. Basically, the converse comes down to showing that $0 \rightarrow V \stackrel{f}{\rightarrow} W \stackrel{\pi}{\rightarrow} C \rightarrow 0$ is an exact sequence, where $C$ is the cokernel of $f$ and $\pi : W \to W/f(V)$ is the canonical projection, and using the fact that $\dim W = \dim V + \dim C$ to show $C=0$. But I am having trouble showing surjectivity implies injectivity. My book says that the "surjectivity implies injectivity" implication is similar to the one I already proved, but I don't see it. What relevant exact sequence will show that the kernel of $f$ is trivial? E.g.,I tried to find some set $X$ and map $h : X \to V$ such that $h(X)=0$ and $0 \rightarrow X \stackrel{h}{\rightarrow} V \stackrel{f}{\rightarrow} W \rightarrow 0$, but I couldn't identify the relevant pair.

If $f$ is surjective, $0\to\ker(f)\to V\xrightarrow f W\to 0$ is an exact sequence, and $\dim(\ker f)+\dim(W)=\dim(V)$.