Constructing a Universal Cover--Proving Injectivity

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.

• 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 $X$ – Max Mar 17 at 21:14
• @Max Oh, okay. It was a slight abuse in notation that was confusing me. – user193319 Mar 18 at 0:21