# Monomorphisms and epimorphisms in the category of finite coverings of a topological space

I'm working my way through Lenstra's Galois Theory for Schemes, and I've run into a bit of a conundrum with Exercise 3.14(b).

In this exercise, we consider the category $\textbf{FC}_X$ of finite coverings $Y\xrightarrow{\pi_Y} X$ of a fixed topological space $X$, with morphisms given by continuous maps $Y\xrightarrow{h} Z$ such that the appropriate diagram commutes, i.e. $\pi_Z\circ h=\pi_Y$. [Note: I would put the diagram itself, but \xymatrix isn't supported in MathJaX: is there something analogous that can be used here to make commutative diagrams?]

Part (b) of the problem is to show that $h:Y\rightarrow Z \$ being a monomorphism in $\textbf{FC}_X$ is equivalent to $h$ being injective and $h$ being an epimorphism in $\textbf{FC}_X$ is equivalent to $h$ being surjective. The implications "$h$ surjective" $\Rightarrow$ "$h$ epic" and "$h$ injective" $\Rightarrow$ "$h$ monic" are both almost immediate, so there's no trouble there.

In order to show that monic $h$ are injective, I supposed that $h(y)=h(y')$ for some $y,y'\in Y$, and considered the covering $\pi_Y':Y\rightarrow Z$ given by $\pi_Y'(y'')\equiv\pi_Y(y)$, for $y''\in Y$, and the maps $f,g:Y\rightarrow Y$ defined by $f(y'')\equiv y$ and $g(y'')\equiv y'$. These are constant, so they are continuous, and some simple algebra shows that $(f\circ\pi_Y)(y'')=(g\circ\pi_Y)(y'')=\pi_Y'(y'')$; hence they are morphisms in $\textbf{FC}_X$. Moreover, some more simple playing with compositions yields $(h\circ f)(y'')=(h\circ g)(y'')$ for all $y''\in Y$, and so the fact that $h$ is a monomorphism implies that $y\equiv f=g\equiv y'$; therefore $h$ is injective, as desired.

The only possible problem I see with the above argument is that $\pi_Y'$ is, emphatically, not surjective, but Lenstra (much like Hatcher) doesn't seem to require coverings to be surjective, so I thought nothing of this point. Moving onto the "epic implies surjective" part -- in light of the fact that I decided coverings are not required to be surjective -- I figured a similar argument could easily be found.... but after MUCH effort, I can't seem to find appropriate choices of $f$ and $g$ to get the sort of diagram present in the definition of epimorphisms... at least not one which will give me, for every $z\in Z$, $h(y)=z$ for some $y\in Y$.

So now I'm wondering: are these coverings meant to be surjective (in which case this part of the problem is trivial)? If so, then does anyone know how I can adjust the argument regarding injectivity (since $\pi_Y'$ is not surjective, so it wouldn't be a covering)? If not, can anyone see what the right choice of a covering $Y'\xrightarrow{\pi_{Y'}}X$ and morphisms $f,g:Z\rightarrow Y'$ are to give me that $h$ is surjective? Am I just missing some small, stupid, obvious detail?

Any help with this would be greatly appreciated... I keep trying, but cannot seem to finish this one bit of the problem up...

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What definition of covering are you using? Is it merely a local homeomorphism, or does it have the stronger property of being (locally) a fibre bundle? – Zhen Lin Nov 2 '13 at 19:35
The definition Lenstra gives is: a continuous map $f:Y\rightarrow X$ "is said to be a covering of $X$ if it is locally a trivial covering, i.e., if $X$ can be covered by open sets $U$ for which $f:f^{-1}(U)\rightarrow U$ is a trivial covering," meaning $f^{-1}(U)\cong U\times E$ for some finite discrete set $E$. – user101616 Nov 3 '13 at 11:35

Ah, of course! And then passing to $\textbf{sets}$ this pushout diagram would be preserved (since the forgetful functor in this case has a right adjoint). Therefore $h$ is epic in $\textbf{sets}$, which is equivalent to it being surjective. Thanks! And yeah, I figured forcing coverings to be surjective would cause problems. :D – user101616 Nov 3 '13 at 11:28