Proposition. every map $\alpha: [s_0,s_1] \to \mathbb{R}^n$ is homotopic rel $\{s_0,s_1\}$ to the linear map

$$\beta=\frac{(s_1-s_0)\alpha(s_0)+(s-s_0)\alpha(s_1)}{s_1-s_0}:[s_0,s_1] \to \mathbb{R}^n$$

The proof says

Proof. Define a rel $\{0,1\}$ homotopy $\gamma:\alpha \simeq\beta$ by

$$\gamma(s,t)=(1-t)\alpha(s)+\beta(s):I \times I \to \mathbb{R}^n$$

Question marks all over this one.

Firstly, in this context of the problem, I understand homotopic rel $\{0,1\}$ to be some homotopy $\gamma$ that satisfies

$$\gamma(0,t)=\alpha(0)=\beta(0) \text{ and } \gamma(1,t)=\alpha(1)=\beta(1)$$

for any $t \in I$. Am I right so far?

Then if I check with the $\gamma$ given in the proof, I am not convinced it's a homotopy rel $\{0,1\}$. I mean, let us try $s=0$.

$\gamma(0,t)=(1-t)\alpha(0)+t\beta(0)$. I do not know what $\alpha(0)$ is since $\alpha(s)$ is not specified, so leave that as it is. $\beta(0)$ is computable so substitute $s=0$ and $s_0=0,s_1=1$ in the above given $\beta(s)$ and I get $\beta(0)=\alpha(0)+\alpha(1)$.

Therefore, $\gamma(0,t)=(1-t)\alpha(0)+t(\alpha(0)+\alpha(1))=\alpha(0)-t\alpha(0)+t\alpha(0)+t\alpha(1)=\alpha(0)+t\alpha(1)$


Well, is $\alpha(0)+t\alpha(1)=\alpha(0)=\beta(0)$ for all $t \in I$? It doesn't look like it, and even if it is, why is it equal??

This is extremely obscure to me, I simply don't understand it, how is $\gamma$ a homotopy rel $\{0,1\}$ of $\alpha, \beta$?


This should be a comment, not an answer, but I'm new here.

There are two typos that seem to be causing you problems. First, your formula for $\beta$ should read $$\beta(s) = \frac{(s_1-s)\alpha(s_0)+(s-s_0)\alpha(s_1)}{s_1-s_0} : [s_0, s_1] \to \mathbb{R}^n.$$ I've changed an $s_0$ to an $s$, which should fix your calculation for $\beta(s_0)$.

Second, and more fundamentally - this may be a typo in your book - your homotopy should be rel $\{s_0, s_1\}$, not rel $\{0, 1\}$.

Then you should be able to check $\beta(s_0) = \alpha(s_0)$, so that $$\gamma(s_0, t) = (1-t)\alpha(s_0) + t\beta(s_0) = \alpha(s_0) = \beta(s_0)$$ as desired, and likewise for $s_1$. But that first typo seems to be screwing you up.

  • $\begingroup$ Hi there, thanks for answering(actually, this did answer me haha). First off, the typo is indeed something that had to be sorted for me and I managed now. And the notes....seems very unreliable doesn't it, I was wondering why they specified to $0,1$ instead of a general subspace $s_0,s_1$ as in the proposition. Anyways, it did help thanks so much! $\endgroup$ – John Trail May 3 '16 at 15:40
  • $\begingroup$ No problem - the typo is one of those things that if you've seen often enough it stands out, but if you're learning the subject it's not so obvious there's a mistake! $\endgroup$ – Thurmond May 3 '16 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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