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In Hatcher's Algebraic Topology, while constructing a simply-connected covering space $\tilde{X}$ for a semilocally simply-connected, locally path-connected, path-connected space $X$, Hatcher makes the following claim:

For a point $[\gamma]\in X$ let $\gamma_t$ be the path in $X$ that equals $\gamma$ on $[0,t]$ and is stationary at $\gamma(t)$ on $[t,1]$. Then the function $t\mapsto [\gamma_t]$ is a path in $\tilde{X}$ lifting $\gamma$ [...].

where $$\tilde X = \{[\gamma]\mid \gamma\text{ is a path in $X$ starting at $x_0$}\}$$ is endowed with the topology generated by the basis consisting of the sets $$U_{[\gamma]} = \{[\gamma \cdot \eta ] \mid \text{$\eta$ is a path in $U$ with $\eta(0)=\gamma(1)$}\}$$ for all $\gamma$ starting at $x_0$ and $U$ open in $X$.

I just can't figure out why the function $t\mapsto [\gamma_t]$ is continuous. Any help would be much appreciated.

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1 Answer 1

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Define the function $$\begin{align*} \varphi:&I \to \tilde X\\ &t\mapsto [\gamma_t] \end{align*}$$ Let $\gamma:I \to X$ be a path. Then, by compactness, we can find $0=t_0<t_1<\ldots < t_n =1$ such that $\gamma\left([t_i-1,t_i]\right) \subseteq U_i$ for some $U_i\in \mathcal U$, for all $i=1,\ldots,n$ where $\mathcal U$ is the basis of $X$ described at page 64.

Let $p_i$ be the homeomorphism $p:{U_i}_{[\gamma_{t_i}]}\to U_i$, the existence of which follows from the fact that $\gamma_{t_i}(1)\in U_i$ and the preceding paragraphs in Hatcher.

Claim: $\varphi=p_i^{-1}\circ \gamma$ on each $[t_{i-1},t_i]$.

Proof of claim: Let $t\in [t_{i-1},t_i]$. Then $\gamma_t(1)=\gamma(t)\in U_i$ and so ${U_i}_{[\gamma_t]}$ is well-defined. Now, $\gamma_t$ is homotopic to $\gamma_{t_i} \cdot \alpha$, where $\alpha$ is the path $s\mapsto \gamma((1-s)\cdot t_i + s\cdot t)$ which lies entirely in $U_i$. This shows that $[\gamma_t]=[\gamma_{t_i}\cdot\alpha]\in {U_i}_{[\gamma_{t_i}]}$, and thus $ {U_i}_{[\gamma_{t_i}]}={U_i}_{[\gamma_{t}]}$ by a preceding remark. Thus,

$$\varphi(t)=[\gamma_t]\in {U_i}_{[\gamma_{t_i}]}\Rightarrow (p_i\circ \varphi)(t) = \gamma_t(1)=\gamma(t)\\ \Rightarrow \varphi(t)=(p_i^{-1}\circ\gamma)(t)$$

This proves the claim. $\square$

Then the result follows by the pasting lemma.

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