Homomorphism Induced by Inclusion is Trivial

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...

• 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 $X$ – Max Mar 13 at 20:42
• @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. – user193319 Mar 13 at 20:53
• Stare really hard at the equation $\hat{\alpha}\circ \iota_{*,x} = \iota_{*,y}\circ\hat\alpha$ and things should become clear – Max Mar 13 at 20:58