Let $\alpha:I\rightarrow X$ be a path such that $\alpha(0)=x_0,\alpha(1)=x_1$ let $h:I\rightarrow I$ continuous such that $h=0,h=1$. Show that $\alpha$ is homotopic to $\alpha \circ h $.
I have noted that since $$\alpha:I\rightarrow X$$ and $$h:I\rightarrow I$$ then
$$\alpha \circ h:I \rightarrow X$$ and also $(\alpha \circ h)(0)=x_0, (\alpha \circ h)(1)=x_1$ therefore both are paths in $X$ and have the same begin and end points.
I must show that there exist a function $F:I\times I \rightarrow X$ s.t.
$F(t,0)=\alpha(t),F(t,1)=(\alpha \circ h)(t)$
$F(0,s)=x_0,F(1,s)=x_1$
any hints? thanks.
It might be easier to build the homotopy in $I$ and then compose with $\alpha$ to get the desired homotopy in $X$.
In other words, first find a function $G : I \times I \to I$ which is a homotopy from $h : I \to I$ to the identity function $Id : I \to I$ and also $G(0,t)=0$, $G(1,t)=1$. Then you can define $F : I \times I \to X$ by the formula $F(s,t) = \alpha(G(s,t))$.
Can you find a formula for $G(s,t)$?
(Hint: Can you parameterize the line segment from $h(s)$ to $s$?)