# Are monics and epics in the category of finite dimensional vector spaces actually injective and surjective linear transformations?

Consider the category of finite dimensional vector spaces with morphisms being linear transformations.

Is it still true that monics and epics are actually injective and surjective linear maps, respectively? The converse is surely true since the category is concrete.

I know this monics and epics are precisely the injective and surjective maps in the category of Sets and in the category of groups, but it is not necessarily true in the category of topological spaces, so I'm just curious if it is true or not in the category of f.d. vector spaces, and if so why?

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If $f:V\to W$ is not injective, then there are two different maps $\ker(f) \to V$ yielding the same composition with $f$, namely the zero map and the inclusion map, so $f$ is not monic.

If $f:V\to W$ is not surjective, then there are two different maps $W \to W$ yielding the same composition with $f$, namely the identity map and any projection onto $\operatorname{ran}(f)$, so $f$ is not epic.

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Thanks for the clear answer! –  YN Chew Nov 17 '13 at 3:22

They key to answering these questions is to try and pick out pieces of your objects using nice, easy-to-understand things. In Set, we have the single point that allows us to "test" maps. In Vect, I would imagine that the 1-d space works. So imagine that $f:V \rightarrow W$ were monic, but not injective. The the kernel is nontrivial, and so if $k$ is the one-dimensional space, we get a nonzero map $v:k \rightarrow V$ that picks out a nonzero vector in the kernel. But there is also $0:k \rightarrow V$ that sends all of $k$ to zero, and $f\circ 0 = f \circ v = 0$, and so since is monic, $0 = v$. But this is impossible for a nontrivial field. So monics are the same as injections. We "test" monics with $k$ just as we test them in Set with the single point.

If $f$ is epi, then using the quotient space $W/f(V)$, comparing $\pi$ to $0$, you can also reason that $W/f(V) = 0$, so $W = f(V)$.

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@TreverWilson's solution is spot on. I like to use rank-nullity whenever I get the chance to though.

Let $f:V\rightarrow W$ be monic. Seeking a contradiction, suppose $f$ is not injective so $\dim\ker f\neq0$. This gives an exact sequence $$0\rightarrow\ker f\xrightarrow{i}V\xrightarrow{f}W$$ That is, $i$ is injective so $\dim\ker i=0$. Moreover, $f\circ i=0=f\circ 0$ so $i=0$. Thus $\dim\mathrm{im}\,i=0$. Now, applying rank-nullity to $i$ gives $$\dim\ker f=\dim\ker i+\dim\mathrm{im}\,i=0+0=0$$ a contradiction. Hence $f$ is injective.

Proving that $f$ epi implies $f$ surjective is similar using the exact sequence $$V\xrightarrow{f}W\rightarrow\mathrm{cok}\,f\rightarrow 0$$

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