When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:

n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $\pi_1(X,x_0) \to \Sigma_n$,where $\Sigma_n$ is n-symmetric group.

Then does it imply that given a homomorphism $\pi_1(X,x_0) \to \Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.


Consider a representation $h:\pi_1(M)\rightarrow \Sigma_n$ and $U_n=\{1,...,n\}$ the following $n$-cover can be associated: the quotient of $\hat X\times U_n$ by the diagonal action of $\pi_1(M)$ where $\hat X$ is the universal cover of $X$.

Conversely given a $n$-cover $p:N\rightarrow M$ you can associate to it its holonomy obtained by the action of $\pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.

  • $\begingroup$ It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $X\times Y/\sim$ where $(x,y)\sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $\pi_1M$ bundle is $\hat{X}$ and the new fibres are $Y = U_n$. $\endgroup$
    – William
    Jan 27 '19 at 17:23

One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.

"Intuitively" the idea is that since $B\Sigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $B\Sigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $B\pi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.

An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $\Sigma_n$; therefore they are classified by homotopy classes of maps $[X, B\Sigma_n]$ where $B\colon Grp \to Top$ is a classifying space functor. Since $\Sigma_n$ is a discrete group it follows that $B\Sigma_n \sim K(\Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $\pi_1K(\Sigma_n, 1)\cong \Sigma_n$ and $\pi_iK(\Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $B\Sigma_n$ all vanish, a result from obstruction theory is that

$[X,B\Sigma_n] \cong [X^{(2)}, B\Sigma_n]$

where $X^{(2)}$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.

Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $B\pi_1 X$. This is because since $\pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $\pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $\pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get

$[X,B\Sigma_n] \cong [X^{(2)}, B\Sigma_n] = [(B\pi_1X)^{(2)}, B\Sigma_n]\cong [B\pi_1 X, B\Sigma_n]$

Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $G\to H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.


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