# Question about the first proof in Hatcher's Algebraic Topology

I have a question about Hatcher's proof that the fundamental group of a circle is Z. Specifically, halfway through, ( http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf , page 30), he proves an important a lemma stated rather generally:

Given a map $Y\times I\rightarrow S^1$ (Y is any space, and I is an interval), there is a unique lifting to $Y\times I\rightarrow \mathbb{R}$ once we've specified an initial condition $Y\times \{0\}\rightarrow \mathbb{R}$.

In the proof of this, he breaks $Y\times I$ into smaller pieces $N\times \{y_0\}$, and keeps adjusting the size of the $N$. In the corresponding proofs in Munkres or Fulton, $Y$ is fixed as just another copy of $I$, and there is no need to modify the pieces we're looking at as the proof progresses.

My question: what is different in Hatcher's more general proof? Specifically, why does he keep modifying the $N$? My feeling is that this is because $Y$ could be disconnected or bad in some other sense. My question seems a little vague, but (a) if you look at Hatcher's proof, you'll see what I mean (near the bottom of p. 30), and (b) I'm hoping that maybe this situation, and the difficulties with it, are well known enough for someone to answer anyway.

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in the initial condition part should we be mapping into $\mathbb{R}$? –  Sean Tilson Dec 16 '10 at 14:08
Yes, we should. Thanks. –  Lost In Math Dec 19 '10 at 21:15

This steps seems to be unnecessary. In what follows, I'll assume you have p. 30 in front of you (since it's available freely online).

Start with the part where he has constructed neighborhood $N$ of $y_0 \in Y$ and a partition $0 = t_0 < \dots t_n = 1$ such that $F(N \times [t_i,t_{i+1}]) \subset U_i$ for each $i = 0, \dots, n-1$. The lift is constructed by defining $\widetilde{F}:N \times I \to \mathbb{R}$ to be $h_{i} \circ F$, where $h_{i}$ is the homeomorphism from the evenly covered neighborhood $U_i$ to the sheet $\widetilde{U_i}$ which contains $h_{i-1} \circ F(y_0,t_i)$. (The homeomorphism $h_0$ is determined by the given lift $\widetilde{F}$ at time $t=0$.)

The only question is whether this is well-defined. One might worry that $N \times \{t_i\}$ is mapped by the $(i-1)$-st lift into the sheet $\widetilde{U}_{i-1}$ but that the neighborhood $N$ needs to be modified so that the $i$-th lift glues to this. In other words, could it be that we need to replace $N$ with a smaller neighborhood $N'$ so that $h_{(i-1)} \circ F(N' \times \{t_i\}) \subset \widetilde{U}_{i-1} \cap \widetilde{U}_{i}$?

This is not the case because, by design, $F(N \times \{t_i\}) \subset U_{i-1} \cap U_i$. Therefore, the map above is well-defined. The lift is continuous by the glueing lemma for open sets.

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The modification step is necessary. In the notation set by the other answer to this question:

By construction, $F(N \times \{t_i\}) \subset U_{i-1} \cap U_i$. By construction, and for all $k$, the "inverse" $h_{k}$ is the homeomorphism from the evenly covered neighborhood $U_k$ to the sheet $\tilde{U_k}$. So $\tilde{U_k}$ is homeomorphic to $U_k$. Now, at stage $i$ of the construction $\tilde{U_i}$ and $\tilde{U}_{i-1}$ intersect - both of them contain $h_{i-1}(F(y_0,t_i))$.

However, even after all this, the set $\tilde{U}_{i-1} \cap \tilde{U}_{i}$ may not be homeomorphic to $U_{i-1} \cap U_i$ (for example it could have fewer connected components). However, the set $\tilde{U}_{i-1} \cap \tilde{U}_{i}$ does contain a neighborhood of $h_{i-1}(F(y_0,t_i))$; this smaller neighborhood suffices for the proof.

Spanier (Chapter 2, Theorem 3) deals with this subtle point in a slightly different fashion - he doesn't shrink the set $N$, but instead "shuffles" the collection of sheets $\{\tilde{U}_\alpha\}$ for a fixed $U$.

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