Homomorphisms induced by $h$

Question

Let $X$ and $Y$ be spaces. Let $x_0 \in X$ and let there be a continuous function $h$ such that $h(x_0)=y_0$. Then there is a homomorphism $h_{*}: \pi_1(X,x_0)\to \pi_1(Y,y_0)$ by the rule $h_{*}\circ [a]=[h\circ a].$

I'm not sure how to show this is a well defined homomorphism.

I was told to take two classes of paths $[a]$ and $[b]$ in $\pi_1(X, x_0)$ and suppose $H$ is a homotopy between $a$ and $b$. Then $f\circ H$ is a homotopy between $f \circ a$ and $f\circ b$. I don't understand why this shows it is well defined.

Answer 1

To show $h_*$ is well-defined say we have $[a]=[b] \in \pi_1(X,x_0)$. We need to show $h_*([a])=h_*([b])$ that is show that $[h\circ a]=[h\circ b]$. As $[a]=[b]$ there exists a path homotopy $H:I\times I \rightarrow X$ from $a$ to $b$. Define $F:I \times I \rightarrow Y$ by $F(s,t)=h(H(s,t))$ and $F$ will be a path homotopy from $h\circ a$ to $h\circ b$ which implies that $[h\circ a]=[h\circ b]$ as needed. Let me know if you want me to explain more.

Answer 2

the map $h_{*}$ is well-defined, for if $F$ is a path homotopy between the paths $a$ and $a'$, then $h\circ F$ is a path homotopy between the paths $h\circ a$ and $h\circ a'$.