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Let $\alpha$ a path from $[0,1]$ to a topological space X. Let $\alpha(0)=\alpha(1)=c$, where $c\in X$.

The standard function to prove that $\alpha\cdot\bar \alpha$ is homotopic to the constant map $c_x$ (where $\bar \alpha =\alpha(1-t)$ is:

$F:I^2\rightarrow X$

such that

$F(s,t) = \alpha(2s)$, if $0\le s \le t/2$

$F(s,t) = \alpha(2s)$, if $t/2 \le s \le 1- (t/2)$

$F(s,t) = \alpha(2s)$, if $1-(t/2)\le s \le 1$

By the gluing lemma this function is continuous.

I'm wondering, why we can't use this function instead:

$H(s,t) = c$ , if $0\le t \lt 1$

$H(s,t) = \alpha \bar \alpha(s)$, if $t=1$

Yes, I know H is not continuous, but how can I prove this formally?

Thanks

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If $H$ is continuous and $H(s,t)=c$ when $0\le t<1$, then $H(s,1)$ is forced to be $c$. – 23rd Nov 5 '12 at 17:18

This question has been answered in a comment:

If $H$ is continuous and $H(s,t)=c$ when $0\leq t<1$, then $H(s,1)$ is forced to be $c$. – 23rd Nov 5 '12 at 17:18

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