2
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

It works exactly the same way, right?

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

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.