# Two covering spaces covering each other are equivalent?

Let $X_1,X_2,X$ be connected and locally path-connected spaces and $p_1:X_1\to X$ and $p_2:X_2\to X$ be covering maps.

Suppose that there are maps $f:X_1\to X_2$ and $g:X_2\to X_1$ such that $p_1=p_2f$ and $p_2=p_1g$.

My Question: Is it true that $p_1$ and $p_2$ are equivalent?

In other words, does there exists a homeomorphism $k:X_1\to X_2$ such that $p_1=p_2k$?

Thanks.

## 1 Answer

Yes, I think so. The existence of $f$ means $(p_1)_* \pi_1 X_1 \subseteq (p_2)_* \pi_1 X_2$ where $(p_1)_*$ and $(p_2)_*$ are just the inclusion of subgroups of $\pi_1 X$. Similarly, the existence of $g$ means $(p_2)_* \pi_1 X_2 \subseteq (p_1)_* \pi_1 X_1$. Therefore $\pi_1 X_1 = \pi_2 X_2$ as subgroups of $\pi_1 X$, and the classification of covering spaces implies $X_1$ and $X_2$ are equivalent covers.

• Nice (+1) and thanks for correcting my wrong answer. – Tim kinsella Aug 5 '16 at 17:02
• But how do you resolve the issue of base points? We actually have $p_{1*}\pi(X_1,x_1)$ is conjugate to some subgroup of $p_{2*}\pi_1(X_2,x_2)$ provided $p_1(x_1)=p_2(x_2)$ and $p_{2*}\pi_1(X_2,x_2)$ is conjugate to some subgroup of $p_{1*}\pi_1(X_1,x_1)$. So, is it true that given two subgroups $H_1,H_2\subset G$ such that$gH_1g^{-1}\subset H_2$ and $kH_2k^{-1}\subset H_1$ we have $H_1$ and $H_2$ are conjugate? – YYF Aug 5 '16 at 18:26
• If the maps $f, g$ of covering spaces do not respect basepoints (as in $f(x_1) \neq x_2$, etc.), then you're right that we only have "inclusion up to conjugacy" and this is not enough to conclude that the covering spaces are the same, but only merely "conjugate" (i.e., change of base points). For a concrete example of two inequivalent covering spaces satisfying all the conditions in your question except pointedness of $f$ and $g$, take $f = g$ to be the identity map in examples (3) and (4) in Hatcher's AT book, page 58. – JHF Aug 5 '16 at 18:55
• I guess that even if $f(x_1)=x_2$ but $g(x_2)\neq x_1$ we still don't have the actual equivalence of two covering spaces. Right? – YYF Aug 5 '16 at 19:20
• Right. In fact there is no difference in assuming one of the maps preserves basepoints, since we can always factor $f: X_1 \to X_2$ into a map of covering spaces that does preserve basepoints followed by the "identity" $X_2 \to X_2$ that "changes" basepoints, and then we can combine this second map into $g$. – JHF Aug 5 '16 at 20:43