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Fixed a connected topological space $X$ it's an exercise to show that, if $X$ admits a universal covering $Y \rightarrow X$, then the category $C$ of finite covering spaces of $X$ is small.

I'm interested in the converse proposition, that is: "if $C$ is small then $X$ admits a universal covering?". I strongly suspect this is true, someway I can think about the universal covering as a "limitator" for the category. But I have no idea on how use the smallness of $C$.

I thought about the problem studying Galois Categories, where our $C$ is a classical example. This means that it has a list of good properties:

  • Initial and final object
  • There exists finite sums, fibre products, finite projective limits, quotients by finite groups of automorphisms.
  • Every morphism factorize as the composition of an epimorphism and a monomorphism.
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1 Answer 1

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You probably want to say that the category of finite covering spaces of $X$ is equivalent to a small category. This is however true for any topological space $X$. For simplicity suppose that $X$ is connected. If $p:Y\to X$ is an $n$-sheet covering then there is a bijection $Y\to X\times\{1,\dots,n\}$ such that $p$ becomes the projection $X\times\{1,\dots,n\}\to X$. All topologies on $X\times\{1,\dots,n\}$ form a set (a subset of $P(X)$), those making the projection $X\times\{1,\dots,n\}\to X$ to a covering form a subset. So we get a small category.

There are certainly $X$'s without universal covering - e.g. if $X$ is locally path connected but not semi-locally $1$-connected. For example, $\mathbb{R}^2$ without the points $1/n$ ($n=1,2,\dots$) on the $x$-axis.

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I think I get it, thank you. I didn't know about the difference of "being small" and "being equivalent to a small c.", but I will go study it. Thank you again. –  Giovanni De Gaetano Apr 22 '11 at 7:54

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