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Recently I came across the following problem that I cant's solve:

Let $f: (D^n, S^{n-1}) \rightarrow (D^n, S^{n-1})$ be a continuous map such that $f|_{S^{n-1}}$ has non-zero degree. Show that $f$ is surjective.

I found an answer to this question at Map Surjective on a Disk but unfortunately I currently don't have the privilege to leave a comment on the appropriate site.

The answer referred to above, makes use of something called the "Extension Theorem" which I don't know.

Is there a more direct way to solve the problem without using this theorem (basically only using more or less elementary homology theory)?

Clearly it follows from $deg(f|_{S^{n-1}}) \neq 0$ that $f|_{S^{n-1}}: S^{n-1} \rightarrow S^{n-1}$ is surjective, but where do I go from there?

Thanks in advance for any help.

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  • $\begingroup$ @TomBombadill I think you love $\text{surjective}$. +1 to you. $\endgroup$ – gaoxinge Jan 29 '14 at 13:16
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For $\mathbf{n=1}$ : a simple argument usig connectedness of the image $f(D^1)$, and the fact that it contains $\lbrace -1,+1\rbrace$ shows that the map is onto.


For $\mathbf{n\geq 2}$ : You can do this using homology. Suppose $f$ misses a point in $D^n$. As you pointed out, this point has to be in the interior of the disk. Using naturality of the long exact homology sequence, there is a commutative diagram $$ \begin{array}{ccc ccc ccc} H_n(D^n)&\rightarrow & H_n(D^n,S^{n-1})&\xrightarrow{\:\partial\:} & H_{n-1}(S^{n-1})&\rightarrow & H_{n-1}(D^n) & \\ \downarrow&&\downarrow&&\downarrow&&\downarrow\\ H_n(D^n\smallsetminus\lbrace\mathrm{pt}\rbrace)&\rightarrow & H_n(D^n\smallsetminus\lbrace\mathrm{pt}\rbrace,S^{n-1})&\xrightarrow{\:\partial\:} & H_{n-1}(S^{n-1})&\rightarrow & H_{n-1}(D^n\smallsetminus\lbrace\mathrm{pt}\rbrace) & \rightarrow &0 \end{array} $$ induced by the map of pairs $f:(D^n,S^{n-1})\to(D^n\smallsetminus\lbrace\mathrm{pt}\rbrace,S^{n-1})$. Since $D^n\smallsetminus\lbrace\mathrm{pt})$ deformation retracts onto $S^{n-1}$, $D^n$ is contractible, and the top homology of the sphere is $\Bbb Z$, we have a diagram \begin{array}{ccc ccc ccc} 0&\rightarrow & H_n(D^n,S^{n-1})&\rightarrow & H_{n-1}(S^{n-1})&\rightarrow & 0 & \\ &&\downarrow&&\downarrow&&\downarrow\\ & & 0&\rightarrow & H_{n-1}(S^{n-1})&\rightarrow & H_{n-1}(D^n\smallsetminus\lbrace\mathrm{pt}\rbrace) & \rightarrow &0 \end{array} (since $n\geq 2$.) The top map is an isomorphism (from exactness), yet by commutativity of the diagram, we must have that the vertical map $H_{n-1}(S^{n-1})\to H_{n-1}(S^{n-1})$ (which is multiplication by $\mathrm{deg}(f)\neq 0$) is $0$. This is a contradiction, and so $f$ is onto.

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  • $\begingroup$ What a great answer! Thank you very much for your wonderful explanations! $\endgroup$ – Tom Bombadil Jan 29 '14 at 13:52
  • $\begingroup$ Thanks for your kind reply :) $\endgroup$ – Olivier Bégassat Jan 29 '14 at 13:54
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Here's a less hi-tech level of proof.

Let $f:(D^n, S^{n-1})\rightarrow (D^n, S^{n-1})$ be any continuous function which is not surjective. We'll prove that $f|_{S^{n-1}}$ has degree $0$.

Suppose $y\in D^n$ is not in the image. Since $D^n$ is compact, $f(D^n)$ is closed, and therefore there is an open set around $y$ of points which are not in the image of $f$. In particular, we may assume wlog that $y$ is interior to $D^n$. Now, there is a homeomorphism of $D^n$ which maps $y$ to $0$, so we may assume wlog that $y=0$.

Now, let $\phi:D^n\setminus\{y\}\rightarrow S^{n-1}$ be the standard retraction. In polar coordinates, $\phi$ has the form $\phi(\omega ,t) = \omega$ for $\omega \in S^{n-1}$.

Now, consider the map $\phi\circ f:D^n \rightarrow S^{n-1}$. This is well defined precisely because $0$ is not in the image of $f$. Note that $\phi \circ f$ is an extension of $f|_{S^{n-1}}$ to all of $D^n$.

The following proposition finishes off the problem:

Suppose $g:S^{n-1}\rightarrow S^{n-1}$ extends to a map $G:D^n\rightarrow S^{n-1}$. Then $g$ has degree $0$.

(In fact, the converse is also true.)

Proof: Define $\sim$ on $S^{n-1}\times [0,1]$ to collapse $S^{n-1}\times \{0\}$ to a point. Call the collapsing map $\pi$. It is easy to see using polar coordinates that $S^{n-1}\times [0,1]/\sim$ is homeomorphic to $D^n$. Call such a homeomorphism $\psi$ and note that $\psi$ can be chosen to have the property that $\psi(\omega,1) = \omega \in S^{n-1}\subseteq D^n$.

We will now define a homotopy $F:S^{n-1}\times [0,1]\rightarrow S^{n-1}$ between $g$ and a constant map.

We let $F(\omega, t) = G(\psi(\pi(\omega,t)))$. This is a composition of continuous functions, so is continuous.

Now, note that \begin{align*} F(\omega,1) &= G(\psi(\pi(\omega,1)))\\ &= G(\psi(\omega,1))\\ &= G(\omega)\\ &= g(\omega)\end{align*} because $G|_{S^{n-1}}= g$.

Also, \begin{align*} F(\omega, 0) &= G(\psi(\pi(\omega,0))) \\ &= G(\psi([S^{n-1}\times \{0\}]))\\ &= G(0),\end{align*} so $F(\omega,0)$ is constant.

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