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Here is a quote from Hatcher's Algebraic Topology:

Given a set $U \in \mathcal{U}$ and a path $\gamma$ in $X$ from $x_0$ to a point in $U$, let $$U_{[\gamma]} = \{[ \gamma \cdot \eta ] \mid \eta \text{ is a path in } U \text{ with } \eta (0) = \gamma (1) \}$$ As the notation indicates, $U_{[\gamma]}$ depends only on the homotopy class $[\gamma ]$. Observe that $p : U_{[\gamma]} \to U$ is surjective since $U$ is path-connected and injective since different choices of $\eta$ joining $\gamma (1)$ to a fixed $x \in U$ are all homotopic in $X$, the mapping $\pi_1(U) \to \pi_1(X)$ being trivial.

See page 64 for more context. Note that $p : \widetilde{X} \to X$ is defined as $p([\gamma]) = \gamma (1)$, where $\gamma$ is a path in $X$ starting at $x_0$. I can see that the map is surjective. However, I am having a little troubling working out the injectivity of the map.

If $\iota : U \to X$ denotes the canonical embedding, then the above quote says that $\iota_* : \pi_1 (U) \to \pi_1(X)$ is trivial (in fact, it is trivial for any base point). Suppose that $\eta_1$ and $\eta_2$ are paths in $U$ with the same starting and ending point such that $p([\gamma \cdot \eta_1]) = p([\gamma \cdot \eta_2])$. This implies $\eta_1 (1) =\eta_2 (1)$. Then $[\eta_1 \cdot \overline{\eta}_2] \in \pi_1(U)$, so $\iota_*([\eta_1 \cdot \overline{\eta}_2]) = [\iota \circ (\eta_1 \cdot \overline{\eta}_2]$ is trivial in $\pi_1(X)$, which implies $\iota \circ \eta_1 \simeq \iota \circ \eta_2$. Hatcher seems to be claiming that this implies $\eta_1 \simeq \eta_2$. If this is what he is asserting, I don't see how to deduce that conclusion.

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    $\begingroup$ No that's not what he was asserting at all, he did say "homotopic in $X$". So $\iota\circ \eta_1 \simeq \iota\circ \eta_2 \implies [\gamma\eta_1]=[\gamma\eta_2]$, where in this second equality everything is in $X$ $\endgroup$ – Max Mar 17 at 21:14
  • $\begingroup$ @Max Oh, okay. It was a slight abuse in notation that was confusing me. $\endgroup$ – user193319 Mar 18 at 0:21

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