I know that $\pi_1(\Sigma_g) = \langle a_1,b_1,\cdots,a_g,b_g | [a_1,b_1]\cdots[a_g,b_g]\rangle$, where $\Sigma_g$ is the orientable surface of genus $g$, and $[a_i,b_i]$ is their commutator. The idea of my argument for why no surface with $g \geq1$ could work is as follows (let me know if this seems fishy):

We know that the induced homomorphism from the fundamental group of a covering space, $p_*: \pi_1(\tilde X) \to \pi_1(X)$ must be injective. $\pi_1(\Sigma_g)$ has $2g$ generators of infinite order. In an injective homomorphism, they must be sent to distinct generators of infinite order in $\pi_1(X)$. However, $\pi_1(X) = \langle a,b |a^2\rangle \cong \mathbb{Z} * \mathbb{Z}_2$ only has one generator of infinite order. Therefore $g =0$ is the only case which might work.

However, this is just $S^2$, which is the universal cover for $\mathbb{R}P^2$. It seems "obvious" that this (nor a space homeomorphic to it) couldn't also be the universal cover for $\mathbb{R}P^2 \vee S^1$, but I'm having a hard time formalizing that.

I know that a covering space of a wedge sum should restrict to a covering space of each of the summands, but I could imagine a CW-construction of $S^2$ that includes a copy of $S^1$ (with a 2-cell on each side, for example), so it isn't totally clear to me why this fact makes it clear $S^2$ doesn't work.

Thanks in advance!


1 Answer 1


Hint: a covering projection is a local homeomorphism. Every point of a surface (oriented or not) has a neighbourhood homeomorphic to $\Bbb{R}^2$. No point of the copy of $S^1$ in $\Bbb{R}P^2 \vee S^1$ has a neighbourhood homeomorphic to $\Bbb{R}^2$. (Since the connected component of a neighbourhood of any such point $x$ that contains $x$ becomes disconnected if you remove $x$.)

  • $\begingroup$ isn't a cover a manifold if and only if the base space is? $\endgroup$ Apr 22, 2018 at 20:38
  • $\begingroup$ I suppose by this answer, yes (modulo some second countability concerns maybe). $\endgroup$ Apr 22, 2018 at 20:40
  • 1
    $\begingroup$ @qbert: there may be some concerns about paracompactness too, However, all that matters here is that given a covering projection, if the total space is a manifold, then so is the base space. $\endgroup$
    – Rob Arthan
    Apr 22, 2018 at 20:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .