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

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

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 $\alpha, \beta$?

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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.

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  • $\begingroup$ 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

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