A star-shaped region is homotopic equivalent to a single point set

Question

I am having trouble solving a problem that is probably quite simple.

I want to show that an open star-shaped convex set is equivalent to a single point set $\{x_0\}$, with $x_0$ the one in the definition of star-shaped set.
I know that two smooth manifolds $M$ and $N$ are said to be homotopy equivalent if there exist continuous maps $f : M \rightarrow N$ and $g : N \rightarrow M$ so that $f \circ g$ is homotopic to $id_N$ and $g \circ f$ is homotopic to $id_M$.
At the same time, star-shaped region in $R^n$ is a region $U\subset R^n$ satisfying the following property: $$\exists x_0 \in U \text{ so that } \forall x ∈ U \text{ and } ∀t:0 \le t \le 1, \text{ one has }tx_0 + (1 − t)x ∈ U.$$ My problem is finding $g$ and $f$.

I think that one of them might be the inclusion map $i:x_0$ $\hookrightarrow$ $X$, but in this case I have no clue about what could be the other function. Another option is something involving $x \mapsto tx_0 + (1 − t)x $.

It is probably trivial, as I have already said but I have been thinking about a solution all day and right now I am pretty confused.

Thanks in advance

Answer 1

As you've said, there's a map $i : \{ x_0 \} \hookrightarrow X$, which will be half of our homotopy equivalence. For the other half, we'll need a map $r : X \to \{ x_0 \}$, but as is pointed out in the comments there's only a single choice for $r$! It must be a constant function.

So now we have our maps $i$ and $r$. We want to show they're a homotopy equivalence. As you've mentioned, this means we want to show that $\text{id}_X \sim ir$ and $\text{id}_{\{x_0\}} \sim ri$.

To show $\text{id}_{\{x_0\}} \sim ri$ we want a function $H : \{x_0\} \times [0,1]$ so that

1. $H(x_0,0) = \text{id}_{\{x_0\}}(x_0) = x_0$
2. $H(x_0,1) = r(i(x_0)) = x_0$

Can you find a function $H$ that meets these criteria? I'll give you a hint -- you should think of a very simple function! I'll leave the answer under the fold

The constant function $H(x_0, t) = x_0$ works

Now for the other homotopy, $\text{id}_X \sim ir$. Here we'll want to use your other idea, the function $x \mapsto t x_0 + (1-t)x$. Indeed, a homotopy here is a function $K : X \times [0,1] \to X$ so that

1. $K(x, 0) = \text{id}_X(x) = x$
2. $K(x, 1) = i(r(x)) = x_0$

But notice your idea $H(x,t) = t x_0 + (1-t)x$ does the job perfectly!

So now $H$ and $K$ are our desired homotopies, and we see that $i$ and $r$ are a homotopy equivalence!

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As a quick aside, I'll say a word about intuitively why we expect this to be true (though you likely already know this). Two spaces are homotopy equivalent if we can deform one space into the other, so we want to know why a star-shaped space can be deformed into a point. The insight is that a star-shaped space is one where every point admits a (linear) path to some fixed $x_0$. So by squishing every point along its path to $x_0$, we can deform the whole space down to $\{ x_0 \}$. The argument that we've given above is just a formalization of this intuition. Indeed, your map $K$ is literally this squishing map (and if this isn't obvious, I suggest thinking about it, perhaps drawing a picture of the situation in case $X$ is literally the interior of a star).

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I hope this helps ^_^