Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.