# Continuity of a path in Hatcher's Algebraic Topology at page 65

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.

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 $$0=t_0 such that $$\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.