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In order to prove that a convex subpace $X \subseteq \mathbb{R^n}$ is homotopy equivalent to a point I did the following:

Let $X \subseteq \mathbb{R^n}$ be a convex subspace in Euclidean space and $Y=\{0\}$ be a one point space. Let $x \in X$ and define $f(x) = 0$ and $g(0) = x_0$ for some $x_0 \in X$. Then we have $(f \circ g)(0) = 0$ and $(g \circ f)(x) = x_0$.

However this seems wrong to me since $(f \circ g) = 1_Y$ but $(g \circ f) \neq 1_X$ since $x_0 \neq x$. Where is my reasoning flawed and how can I fix this?

mathripper
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2 Answers2

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You have to exhibit a homotopy between $h = g \circ f : X \to X$ and $1_X$, i.e., a continuous function $H : X \times I \to X$ such that $h(x) = H(x, 0)$ and $1_x(x) = H(x, 1)$ for every $x \in X$. To do this define $H(x, t) = tx + (1-t)x_0$, then $H : X \times I \to X$ because $X$ is convex and clearly $H(x, 0) = x_0 = h(x)$ and $H(x, 1) = x = 1_X(x)$ for every $x \in X$.

In more complex cases, you might also have to exhibit a homotopy between $f \circ g$ and $1_Y$, but, in this case, $f \circ g = 1_Y$ as you observed, so there is nothing to do.

Rob Arthan
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  • Oh ok so I only have to find maps homotopic to the identities but not necessarily equal to the identity right? Edit: never mind I just read the other answer again. – mathripper Feb 03 '21 at 23:55
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    Exactly: that's the difference between the idea that $X$ and $Y$ are homotopy equivalent and the idea that they are homeomorphic. – Rob Arthan Feb 04 '21 at 00:02
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Your reasoning is actually correct. A homotopy equivalence is one in which $g \circ f$ is homotopic to $1_X$, not on the nose equal.

How can you use the convexity assumption to homotope $g \circ f$ to the identity?

Sam Freedman
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