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Suppose $X = \mathbb{R}^n$. Let $\gamma, \alpha : [0,1] \to X $ be to paths such that $\gamma(0) = \alpha(0) = x_0 , \; \; \gamma(1) = \alpha(1) = x_1$. We want to show $\gamma$ and $\alpha$ are homotopic.

My try: Take $F(s,t) = f_t(s) = (1-t)\gamma(s) + t\alpha(s)$

Clearly, $F(0,t) = x_0$ and $F(1,t) = x_1$.

Since $\alpha, \gamma, t, 1$ are continuous, then $F$ must be continuous function. Hence it is a homotopy. Therefore $\alpha$ and $\gamma$ are homotopic

IS this enough to show they are homotopic?

Cameron Williams
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ILoveMath
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1 Answers1

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Yes, it's perfect. Note that this proof would also work if $\mathbb{R}^n$ were replaced by any convex subset.

Najib Idrissi
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  • Would it also be true for simply connected subsets? – Not Euler Apr 12 '19 at 14:00
  • Well, it's also true that any two paths in a simply connected space with the same start & end points are homotopic (relative to the endpoints). But this argument no longer works, because the proposed homotopy $F$ may not be well defined. In order for $F$ to be well defined, we need to know that $(1-t)\gamma(s) + t \alpha(s) \in X$ for all $t,s \in [0,1]$. So we need $X$ to be convex for this particular idea to work. – diracdeltafunk Mar 26 '22 at 23:15
  • But of course the point of building $F$ is to show that the set $X$ has trivial fundamental group! If you already know that $\pi_1(X,x_0)$ is trivial, you can use this directly to get that any two paths starting at $x_0$ with the same endpoints are homotopic. Take $\alpha, \beta : [0,1] \to X$ such that $\alpha(0) = \beta(0) = x_0$ and $\alpha(1) = \beta(1)$. Then $\alpha \overline{\beta}$ is homotopic to a constant path (relative endpoints) by assumption, so $\alpha \sim \alpha \overline{\beta} \beta \sim \beta$. – diracdeltafunk Mar 26 '22 at 23:18