I'm working on some exam prep questions, and I am having a bit of trouble with this one:

If $f:M \rightarrow N$ and $g:N \rightarrow V$ are maps of connected and locally path connected spaces, if $g:N \rightarrow V$ and the composition $g \circ f: M \rightarrow V$ are both covering spaces, show that $M \rightarrow N$ is a covering space

I first show that for every $n \in N$ there exists an open $U$ with $n \in U$ such that $f^{-1}(U) = \cup_i S_i$ disjoint open subsets in $M$.

I fix $n_0 \in N$, and since $g:N \rightarrow V $ is a covering, $g$ maps $N$ surjectively onto $V$, and so there exists $v_0 \in V$, such that $g(n_0) = v_0$, with $n_0 \in S_j $ for some $S_j \in g^{-1}(v_0) = \cup_i S_i$.

Now $g \circ f: M \rightarrow V$ is also a covering, so:

$(g \circ f)^{-1}(v_0) = f^{-1} \circ g^{-1}(v_0) = f^{-1}(\cup_i S_i) = \cup_k S_k$, disjoint open subsets in $M$.

Now since $n_0 \in S_j \subseteq \cup_i S_i $, this means that $f^{-1}(S_j) = \cup_l S_l$ are disjoint subsets in $M$.

I think what I have done above is correct, but now I am stuck on proving that $f$ maps each $S_l$ homeomorphically to $S_j$.

I am not too sure how I would go about showing this, perhaps I need to use the fact of connectedness or local path connectedness? Any help would be greatly appreciated!


  • $\begingroup$ What is hard to prove? That $f$ restricted to $S_t$ is injective, or that it is an open map? (we supposed, it is continuous) $\endgroup$ – Édes István Gergely Apr 13 '16 at 14:54
  • $\begingroup$ Why did you edit your question making it unintelligible? $\endgroup$ – user228113 Apr 14 '16 at 0:36

This is Lemma 80.2 in Munkres, Topology, 2nd edition.


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