Recall the following proposition in Hatcher's Algebraic Topology:

Proposition 1.31 The map $p_* : \pi_1(\widetilde{X},\widetilde{x}_0) \to \pi_1(X,x_0)$ induced by a covering space $p : (\widetilde{X}, \widetilde{x}_0) \to (X,x_0)$ is injective. The image subgroup $p_*(\pi_1(\widetilde{X},\widetilde{x}_0))$ in $\pi_1(X,x_0)$ consists of the homotopy classes of loops in $X$ based at $x_0$ whose lifts to $\widetilde{X}$ starting at $\widetilde{x}_0$ are loops.

I am currently working through the proof of following proposition:

Proposition 1.33 Suppose a given covering space $p : (\widetilde{X},\widetilde{x}_0) \to (X,x_0)$ and a map $f : (Y,y_0) \to (X,x_0)$ with $Y$ path-connected and locally path-connected. Then a lift $\widetilde{f} : (Y,y_0) \to (\widetilde{X},\widetilde{x}_0)$ of $f$ exists iff $f_*(\pi_1(Y,y_0)) \subseteq p_*(\pi_1(\widetilde{X},\widetilde{x}_0))$

Here is the proof Hatcher gives in his book:

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In the first paragraph, why does Hatcher say that $h_0$ is path homotopic to a loop at $x_0$ which lifts to a loop in $\widetilde{X}$ at $\widetilde{x}_0$? This seems unnecessary. If $[h_0] \in p_*(\pi_1(\widetilde{X},\widetilde{x}_0)$, then $h_0$ itself lifts to a loop in $\widetilde{X}$ at $\widetilde{x}_0$. It seems unnecessary to talk of a homotopy between $h_0$ and $h_1$.

My next two questions concern the second paragraph. I'm having trouble following the proof that $\widetilde{f}$ is continuous. It appears that he does this by proving continuity at an arbitrary point $y \in Y$. Usually these sorts of proof start out as follows:

Let $y \in Y$ be arbitrary, and let $\mathcal{O}_1$ be an open set about $\widetilde{f}(y)$. We need to find an open set $\mathcal{O}_2$ about $y$ such that $\widetilde{f}(\mathcal{O}_2) \subseteq \mathcal{O}_1$

As you may see, he shows that we can find an open set $V$ about $y$ such that $\widetilde{f}(V) \subseteq \widetilde{U}$. But $\widetilde{U}$ was not arbitrarily selected at the beginning. How do we know we can fit it in an arbitrarily chosen neighborhood about $y$?

My next question is, why is $\widetilde{(f \gamma )} \cdot \widetilde{(f \eta)}$ a lift of $(f \gamma ) \cdot(f \eta)$? Is he implicitly using the following: if $\alpha$ is a path in $X$ from $x_0$ to $x_1$, $\beta$ a path from $x_1$ to $x_2$, then $\widetilde{\alpha \cdot \beta} = \widetilde{\alpha} \cdot \widetilde{\beta}$? I tried proving this, but I don't think it is true since the end point of $\widetilde{\alpha}$ may be different from the starting point of $\widetilde{\beta}$.


For the first question : $[h_0]=p_*[\gamma]$ doesn't tell you that there is $\delta$ such that $h_0 = p\circ \delta$, it tells you that $h_0$ and $p\circ \gamma$ are homotopic. A priori there is no reason to believe that $h_0$ should itself be the projection of a loop.

The sentence that follows explains why a posteriori it is actually the case that $h_0$, too, is the projection of a loop.

For the second question, $\tilde{U}$ is not arbitrary but it is picked among a basis of neighbourhoods, so it is enough. That it is enough is a point-set topology basic fact, that we may for instance state as

Let $f:X\to Y$ be a function between two topological spaces; let $x\in X$ and let $\mathcal{V}$ be a basis of neighbourhoods of $f(x)$. Then, if for each $U\in \mathcal{V}$ there is $V\subset X$ a neighbourhood of $x$ such that $f(V)\subset U$, $f$ is continuous at $x$.

For your last question, Hatcher is not being very precise here, in the sense that $\widetilde{\gamma}$ in this last section is not the same as the one at the beginning which specifically denoted a lift with starting point $\widetilde{x_0}$; here it denotes some lift, with the starting point "obvious from context", as is often the case.

In particular, a claim that is true is that if $\widetilde{\alpha},\widetilde{\beta}$ are lifts of $\alpha,\beta$ that can be put after another, then so can $\alpha, \beta$, and $\widetilde{\alpha}\widetilde\beta$ is a lift of $\alpha\beta$. It is then your job here to figure out which specific lifts are meant here, but that shouldn't be a problem.

  • $\begingroup$ But $[h_0] \in p_*(\pi_1(\widetilde{X},\widetilde{x}_0))$, so why doesn't proposition 1.31 tell us directly that $h_0$ lifts to a loop? $\endgroup$ – user193319 Mar 4 at 19:07
  • $\begingroup$ Proposition 1.31 tells us that $[h_0]$ is the class of a loop that lifts to a loop, the sentence that comes afterwards tells you why it follows that all elements of that class actually lift to a loop $\endgroup$ – Max Mar 4 at 19:13

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