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On page 64 of Hatcher's book on Algebraic Topology, he writes the following:

...if the map $\pi_1(U) \to \pi_1 (X)$ is trivial for one choice of basepoint in $U$, it is trivial for all choices of basepoint since $U$ is path-connected.

I interpret this as the following claim:

Let $\iota : U \to X$ be the canonical embedding, let $x \in U$, and let $\iota_{\ast,x} : \pi_1(U,x) \to \pi_1(X,x)$ be the induced homomorphism. If $\iota_{\ast ,x}$ is trivial, then $\iota_{\ast, y}$ is trivial for every $y \in U$.

Is this a correct interpretation? How does one prove this? I am having trouble proving it. Let $\alpha$ be a path in $U$ from $x$ to $y$. Then $\hat{\alpha} : \pi_1(U,x) \to \pi_1(U,y)$ given by $\hat{\alpha}([\gamma]) = [\overline{\alpha} \ast \gamma \ast \alpha]$ is an isomorphism. Initially, I thought that $i_{\ast ,y} = \iota_{\ast ,x} \hat{\alpha}^{-1}$, but this doesn't appear to be true (domains/codomains don't match up). I have tried other things, but to no avail...

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  • $\begingroup$ The equality you wrote isn't true indeed, but what happens if you make the domains/codomains match by adding another $\hat{\alpha}$ ? Think about it this way : changing the basepoint in $U$ should correspond to changing the basepoint in $X$ $\endgroup$ – Max Mar 13 at 20:42
  • $\begingroup$ @Max Would I then view $\alpha$ as a loop in $X$ and then write $\hat{\alpha} \circ \iota_{\ast , x} \circ \hat{\alpha}^{-1}$? Would this equal $\iota_{\ast , y}$? It seems that it does. $\endgroup$ – user193319 Mar 13 at 20:53
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    $\begingroup$ Stare really hard at the equation $\hat{\alpha}\circ \iota_{*,x} = \iota_{*,y}\circ\hat\alpha$ and things should become clear $\endgroup$ – Max Mar 13 at 20:58

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