Let $V$ be a vector space and $\alpha \in \operatorname{End}(V)$
(i) If $V$ is finite dimensional, then $\alpha$ is injective iff $\alpha$ is surjective.
(ii) Give example showing (i) is false if $V$ is not finite dimensional.
So on (i), since $V$ is finite dimensional, then $V$ has a basis with finite cardinality and hence $\dim(V)=n$. Also, the following holds, $\dim(V)=\dim(\ker(\alpha) + \dim(\operatorname{im}(\alpha))$ Since $\alpha$ is injective, then the $\ker(\alpha)$ is $0$ which implies that the $\dim(\ker(\alpha)$ is $0$ which implies that $\dim(\operatorname{im}(\alpha))=\dim(V)$ and that shows that $\alpha$ is surjective. (Is that right?)
On the second one, (ii), I just have no idea. I mean shouldn't Axiom of Choice be able to work here?