I am reading Hatcher's proof of the homotopy lifting property of the covering map $p: \mathbb{R}\to S^1$. Starting with a homotopy $F: Y \times I \to S^1$ and a map $\tilde{F}:Y \times \{0\} \to \mathbb{R}$, first he shows existence of the lift for a neighborhood $N$ of a point $y_0 \in Y$. Letting $U_i$ be the open cover of $S^1$ such that $p^{-1}(U_i)$ is a disjoint union of $\tilde{U}_i^j$ mapped homeomorphically onto $U_i$ by $p$, he then finds a neighborhood $N$ of $y_0$ and a partition of $[0,1]$ such that $F(N \times [t_i, t_{i+1}]) \subset U_i$. We build $\tilde{F}:N \times [0,t_i]$ inductively.

Near the end of this paragraph, he makes a comment I don't understand. Having constructed $\tilde{F}:N \times [0,t_i]$, we then find $\tilde{U}_i$ such that $\tilde{F}(y_i, t_i) \in \tilde{U}_i$. He says that we may have to shrink $N$ in order to assume that $\tilde{F}(N \times \{t_i\})\subseteq \tilde{U_i}$.

What I don't understand is: why would we ever have to shrink $N$? $\tilde{F}\left.\right| N \times [t_{i-1}, t_i]$ is defined to be $p^{-1} \circ F$, where $p^{-1}$ refers to the homeomorphism $p: \tilde{U}_{i-1} \to U_{i-1}$. But then $\tilde{F}(N \times \{t_i\}) \subseteq \tilde{U}_{i-1} \cap \tilde{U}_i$, because $F(N \times [t_{i-1}, t_i]) \subseteq U_{i-1}$ and $F(N \times [t_i, t_{i+1}]) \subset U_i$, and $p$ is a homeomorphism on $\tilde{U}_{i-1} \cap \tilde{U}_i$.

What am I missing?

  • $\begingroup$ I don't think you have to shrink $N$. You are basically already shrinking $N$ enough when you require that $F(N\times [t_i,t_{i+1}]) \subset U_i$ for every subinterval in your partition. $\endgroup$ – William Stagner May 19 '15 at 19:20
  • $\begingroup$ @WilliamStagner That's what I thought too! $\endgroup$ – Eric Auld May 19 '15 at 19:26

The problem here is that the two maps "$p^{-1}$" for $U_{i-1}$ and $U_i$ may not agree on all of $U_{i-1}\cap U_i$, although they do agree at $F(y_0,t_i)$. For example, for the covering space $p:{\mathbb R}\to S^1$, if we use a cover of $S^1$ by two open intervals $U_\alpha$ and $U_\beta$ then it is impossible to choose lifts $p^{-1}:U_\alpha \to \tilde U_\alpha$ and $p^{-1}:U_\beta \to \tilde U_\beta$ that agree on both arcs of $U_\alpha \cap U_\beta$.

In the notation of the question, the map $p:\tilde U_{i-1}\cap \tilde U_i\to U_{i-1}\cap U_i$ is a homeomorphism onto its image, but this image might not be all of $U_{i-1}\cap U_i$.

If the neighborhood $N$ was connected there would be no problem, but to take advantage of this one would have to impose some restrictions on the space $Y$, such as being locally connected. This would be OK for the applications at hand where $Y$ is either a point or an interval, but eventually one wants to allow more general spaces $Y$.

Incidentally, the online version of the book has been mildly edited here to make explicit the fact that the arguments in this proof apply to arbitrary covering spaces, not just the covering space ${\mathbb R}\to S^1$. The essential content is still the same, however.


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