# Homotopy of paths

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 $$s$$?)
• 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. – Alfdav Mar 10 at 2:46