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.
$F(t,0)=\alpha(t),F(t,1)=(\alpha \circ h)(t)$
any hints? thanks.