let $h:X→Y$ be continuous, with $h(x_0)=y_0$ and $h(x_1)=y_1$. Let $α$ be a path in $X$ from $x_0$ to $x_1$ , and let $β=h∘α$ . Show that $\widehat{\beta}∘(h_{x0})_∗=(h_{x1})_∗∘\widehat{\alpha}$.
Already answer is the following:
$\widehat{\beta}∘(h_{x_0})∗([f])=[\bar{β}]⋆(h_{x_0})_∗([f])⋆[β]=[h∘\bar{α}]⋆[h∘f]⋆[h∘α]=[h∘(\bar{α}⋆f⋆α)]=( h_{x_1})∗([\bar{α}⋆f⋆α])=(h_{x_1})_∗∘\widehat{\alpha}([f]).$
But I do not understand why $[\bar{\beta}]=[h∘\bar{\alpha}]$ and why $[h∘(\bar{α}⋆f⋆α)]=( h_{x_1})∗([\bar{α}⋆f⋆α])$? Could someone help me please? Thanks
Edit: As someone mentioned in the comments, here are some specifications. $\widehat{α}:π_1(X,x_0)\to π_1(X,x_1), (h_{x_0})_∗:π_1(X,x_0)→π_1(Y,y_0), (h_{x_1})_∗:π_1(X,x_1)\to π_1(Y,y_1), \widehat{β}:π_1(Y,y_0)\to π_1(Y,y_1)$