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Show that the change-of-basepoint homomorphism $\beta_h$ depends only on the homotopy class of $h$.

The change-of-basepoint homomorphism is defined as $\beta_h:\pi_1(X, x_1) \to \pi_1(X,x_0)$ sending $[f] \mapsto [h \cdot f \cdot \overline{h}]$, where $\overline{h}$ is the inverse path of $h$.

Now in order for this to depend only on the homotopy class of $h$ if I take some $g$ such that $h \simeq g$, then I should have that $\beta_h=\beta_g$. It would satisfy to show that $$\beta_h[f]\beta_{\overline{g}}[f]= [e]$$ where $e$ is the constant loop staying at the base point of $\pi(X,x_0)$ i.e. $x_0$. So what I have is that $$\beta_h[f]\beta_{\overline{g}}[f]=[h \cdot f \cdot \overline{h}][\overline{g} \cdot f \cdot g] = [h \cdot f \cdot \overline{h} \cdot \overline{g} \cdot f \cdot g]$$

now since $h \simeq g$ we have that $\overline{h} \simeq \overline{g}$ but how can I use this here? My issue is that I don't know how to relate these homotopies with the homotopy class $[h \cdot f \cdot \overline{h} \cdot \overline{g} \cdot f \cdot g]$.

Do I just remove everything except $f$ as they're homotopic or how should I do this?

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In order to show that $\beta_h=\beta_g$ you need to show that the inverse of $\beta_g$ ($\beta_{\overline g}$) is iverse to $\beta_h$ (as you denoted).
But what you need to show is that $(\beta_h \circ \beta_{\overline g})([f])=[f]$ namely that $\beta_h \circ \beta_{\overline g}$ is the identity.

So $$(\beta_h \circ \beta_{\overline g})([f])=\beta_h(\beta_{\overline g}[f])=\beta_h[\overline g \cdot f \cdot g]=[h\cdot\overline g\cdot f\cdot g \cdot \overline h]=[f]$$ the last eqaily is since $g \simeq h$ we have $[h\cdot \overline g]=[g\cdot \overline h]=[e].$
you also need to show that $(\beta_{\overline g} \circ \beta_h)([f])=[f]$ but it is a very similar calculation.

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