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.


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 $s$?)

  • $\begingroup$ I understand your approach, however I cannot find a way to do the parametrization, I dont think i quite got that when the teacher explained it $\endgroup$ – Alfdav Mar 10 at 2:37
  • $\begingroup$ Wait a second, thinking a bit about it, i defined $$G(s,t)=(1-t)h(s)+st$$ I think that it should work because $G(s,0)=h(s), G(s,1)=s, G(0,t)=0, G(1,t)=1$ can you confirm it? then your original suggestion of the composition should work. $\endgroup$ – Alfdav Mar 10 at 2:46
  • 1
    $\begingroup$ That looks good. $\endgroup$ – Lee Mosher Mar 10 at 3:40

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