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Let $p\colon(\widetilde{X},\widetilde{x_0})\to(X,x_0)$ be a covering space, where both $\widetilde{X}$ and $X$ are path-connected. Note that $p$ is not assumed to be surjective. In Prop. 1.32, Hatcher establishes a bijection between the right-cosets of $H=p_*(\pi_1(\widetilde{X},\widetilde{x_0}))$ in $\pi_1(X,x_0)$ and the fibre $p^{-1}(x_0)$ by sending at coset $H[g]$ to the endpoint $\widetilde{g}(1)$ of a lift $\widetilde{g}\colon[0,1]\to\widetilde{X}$ of $g$.

Since $\widetilde{X}$ is path-connected, this map is surjective. Why does $X$ need to be path-connected (equivalently, connected)? Is this only to ensure that we can speak of the number of sheets of the cover?

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    $\begingroup$ Yeah, I think that's just to ensure that you can say something reasonable about the covering. If your space is not even path connected, a covering can have totally different components... $\endgroup$
    – Danu
    Mar 9, 2017 at 11:49
  • $\begingroup$ @Danu: Sure, he later restricts himself to that case. Anyway, we still get this bijection even if $X$ is not connected. $\endgroup$
    – user363520
    Mar 9, 2017 at 13:07
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    $\begingroup$ I think the point is not that the theorems fail but rather that the situation is not interesting in this case $\endgroup$
    – Danu
    Mar 9, 2017 at 13:08

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If $X$ is not path-connected, we have $\pi_1(X,x_0)$ is isomorphic to $\pi_1(PC_{x_0}, x_0)$, where $PC_{x_0}$ is the path-connected component of $x_0$, via the map induced by inclusion. This reduces the discussion to the case where the base space is path-connected (since $p$ will also have its image lying in $PC_{x_0}$, due to path-connectedness of $\widetilde{X}$), so there is nothing interesting in letting $X$ be not-path-connected (just make sure that you are talking about the cardinality of sheets in the path-component, since it is not well-defined globally).

However, although $X$ not being path-connected is almost irrelevant, $\widetilde{X}$ being path-connected is very relevant, as you imply. This ensures (group-theoretically speaking) that the action of $\pi_1(X,x_0)$ in the fiber is transitive, which makes the bijection making the index being equal to the number of sheets work.

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