Hello. I have a question on covering spaces. Please let each space be paracompact, path-connected, locally path-connected and semi-locally simply connected.

Let $E\to B$ denote a normal covering space and $G$ its deck transformation group. It is the same as a $G$-principal bundle. On the other hand for a discrete group $G$ a $G$-principal bundle is a normal covering space.

$G$-principal bundles $E\to B$ are in one-to-one correspondence with pointed homotopy classes of maps $[B,BG]$. $BG$ denotes the classifying space construction.

$S^1$ is a classifying space for the integers. There is only one covering of $S^1$ with deck transformation group the integers, the universal covering $\mathbb{R}$. On the other hand $\pi_1(S^1)=[S^1,S^1]=\mathbb{Z}$. Can someone explain to me how this fits together? Especially with regard to $\pi_2(S^1)=0$ meaning that $S^2\times\mathbb{Z}$ is not recognized as a covering of $S^2$.

If $G$ is discrete the set $[B,BG]$ should be in a one-to-one correspondence with normal $|G|$-sheeted coverings of $B$. Can one say that $[B,B\pi_1(X)]$ is in a one-to-one correspondence with all normal coverings of $B$?



closed as too localized by S. Carnahan Mar 14 '11 at 15:19

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  • $\begingroup$ You should try this at math.stackexchange.com $\endgroup$ – Sean Tilson Mar 14 '11 at 5:26
  • $\begingroup$ You should consult your favorite algebraic topology book, and concentrate on the precise statements regarding the Galois correspondence between subgroups of $\pi_1(B,b)$ and connected coverings. As Sean Tilson remarked, math.stackexchange.com is a more appropriate site for this question. $\endgroup$ – S. Carnahan Mar 14 '11 at 15:19

The only oversight in your analysis is the statement that $S^1$ has only one covering with group $\mathbb{Z}$. It has only one CONNECTED covering with group $\mathbb Z$. It has plenty of others, such as $S^1\times \mathbb{Z}$.

  • $\begingroup$ But with this reasoning $S^2$ would have non-connected covers such as $S^2\times \mathbb{Z}$, too. How does this fit with $[S^2,S^1]=0$ but $[S^1,S^1]=\mathbb{Z}$? $\endgroup$ – user13624 Mar 14 '11 at 9:00
  • $\begingroup$ $S^2\times\mathbb{Z}$ is the only cover of $S^2$. That agrees with the fact that $[S^2,S^1]$ has only one element, namely 0. $\endgroup$ – Andreas Blass Mar 14 '11 at 14:19
  • $\begingroup$ This is exactly my problem. Aren't $S^2$, $S^2\times\{1,2\}$, $S^2\times\{1,2,3\}$ all covers of $S^2$? $\endgroup$ – user13624 Mar 14 '11 at 14:22
  • $\begingroup$ @gadwall: They're covers, but not $|\mathbb{Z}|$-sheeted ones. Remember that homotopy classes of maps to $BG$ classify principal $G$-bundles, not any old coverings. $\endgroup$ – Andreas Blass Mar 14 '11 at 14:52

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