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If there is a continuous function from the closed unit disk to itself such that it is identity map on boundary, must it be onto?

David
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2 Answers2

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Yes.

Suppose $f: D^2 \to D^2$ did not have $p \in D^2$ in the image, and such that $f$ restricts to the identity on the boundary. One may pick a homeomorphism $g: D^2 \to D^2$ that restricts to the identity on the boundary, with $g(p) = 0$, so $gf: D^2 \to D^2$ misses $0$.

Now compose with the map $D^2 \to S^1, x \mapsto x/\|x\|$. This defines a retraction $D^2 \to S^1$. But that's silly, as if there were such a retraction, the map $\pi_1(S^1) \to \pi_1(D^2)$ induced by the inclusion would be an injection, and it's not.

  • Thank you. Also, "I'm wondering if there exists a fixed point in the D's interior" – David Mar 02 '15 at 06:57
  • @David I think there should be, but I don't have a proof in mind yet. You should post this as a separate question. –  Mar 02 '15 at 07:07
  • I've revised it. – David Mar 02 '15 at 07:11
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    @David: The new question is fundamentally a different question than the first one. Generally, if you have multiple (but only a little related) questions, you should post them each separately. I'm suggesting you post this one as a new question. –  Mar 02 '15 at 07:12
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    Ok!thanks for your suggestion and I'm wondering how to construct the function g in your proof. – David Mar 02 '15 at 07:15
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    Pick a self-homeomorphism $g'$ of the unit disc that takes $p$ to $0$. (You can do this very explicitly; here are the ones you want.) On the boundary, say $g'$ restricts to $h$; extend $h^{-1}$ to an automorphism of the unit disc using the Alexander trick. Call this, uh, $h'$? Then $h'g'$ is the desired homeomorphism. –  Mar 02 '15 at 07:42
  • Got it thank you very much! – David Mar 02 '15 at 08:05
  • If you want to do this in any dimension, it's a bit harder, because you can't resort to picking an easy self-homeomorphism like up there. But it's still true; one can prove it with some smooth manifold theory. –  Mar 02 '15 at 08:06
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My proof (hopefully bug-free) for arbitrary dimension. If $f$ is not surjective, let $p\in \mathbf{D}^n-f(\mathbf{D}^n)$. Since $f$ is identity on the boundary sphere, $p$ is in the interior of $\mathbf{D}^n$. Thus the map $g: \mathbf{D}^n\rightarrow \mathbf{S}^{n-1}$ that sends $x$ to the unique intersection of ray $pf(x)$ with $\mathbf{S}^{n-1}$, is well-defined and continuous. We have $gi=\mathbf{1}_{\mathbf{S}^{n-1}}$ where $i: \mathbf{S}^{n-1}\hookrightarrow\mathbf{D}^n$ is inclusion, inducing identity composition on cohomology $$ i^*g^*: H^{n-1}(\mathbf{S}^{n-1})\rightarrow H^{n-1}(\mathbf{D}^{n}) \rightarrow H^{n-1}(\mathbf{S}^{n-1}). $$

But this is impossible: the middle group is $0$ since $\mathbf{D}^n$ is contractible while the first and third groups are $\mathbf{Z}$ by fundamental algebraic topology.

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