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

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?

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.