Conditions to extend functions Continuously into the Boundary (D^1/S^1)

[WWGD]

TL;DR Summary

Other than for null-homotopic maps, which continuous maps defined on D^1 --> D^1 extend continuously to maps B^1--> B^1 , which maps can be extended in opposite direction?

Other than for null-homotopic maps, which continuous maps defined on $D^1 \rightarrow D^1$ (Open disk)extend continuously to maps $B^1 \rightarrow B^1$ ,($B^1$ the closed disk) which maps can be extended in opposite direction, i.e., continuous maps $f: S^1 \rightarrow S^1$ that extend to the interior ? I assume if we have the needed homotopy/(homology?) class being trivial, this is sufficient. Is this also necessary? I think @lavinia and/or @Infrared may know?
EDIT: I know this is obstruction theory but trying to see if someone has a clearer/different explanation.

 

[Infrared]

My reading of your question is the following: For which continuous functions $f:S^1\to S^1$ do there exist a continuous extension $F:B^2\to B^2$ such that $F(S^1)\subset S^1$ and this restriction agrees with $f$?

If so, the answer is all of them! One way to extend is just $F(re^{i\theta})=rf(e^{i\theta})$.

On the other hand, if you want an extension $F: B^2\to S^1$, then this is possible if and only if $f$ is nulhomotopic. Definitely this is a necessary condition. On the other hand, if $f$ is nulhomotopic, then there is a homotopy $H:S^1\times [0,1]\to S^1$ such that $H(x,0)=f(x)$ and $H(S^1,1)=\{\text{pt}\}$. So, $H$ induces a map on the quotient $\tilde{H}:\left(S^1\times [0,1]\right)/(S^1\times\{1\})\to S^1$. The domain is a disk, and we choose this map to be our $F$. It extends $f$ since $H(x,0)=f(x)$.

Please clarify if neither of these were your intended problems.

 

[WWGD]

My reading of your question is the following: For which continuous functions $f:S^1\to S^1$ do there exist a continuous extension $F:B^2\to B^2$ such that $F(S^1)\subset S^1$ and this restriction agrees with $f$?

If so, the answer is all of them! One way to extend is just $F(re^{i\theta})=rf(e^{i\theta})$.

On the other hand, if you want an extension $F: B^2\to S^1$, then this is possible if and only if $f$ is nulhomotopic. Definitely this is a necessary condition. On the other hand, if $f$ is nulhomotopic, then there is a homotopy $H:S^1\times [0,1]\to S^1$ such that $H(x,0)=f(x)$ and $H(S^1,1)=\{\text{pt}\}$. So, $H$ induces a map on the quotient $\tilde{H}:\left(S^1\times [0,1]\right)/(S^1\times\{1\})\to S^1$. The domain is a disk, and we choose this map to be our $F$. It extends $f$ since $H(x,0)=f(x)$.

Please clarify if neither of these were your intended problems.

I don't understand, if a point is in S^1, isn't it of the form $e^{i\theta}=1e^{1\theta}$?

 

[Infrared]

Yes, and if a point $z$ is in $B^2$, then it is of the form $z=re^{i\theta}$ for $0\leq r\leq 1$. Since $f:S^1\to S^1$ is already defined, we know that $f(e^{i\theta})\in S^1$ is well-defined. So our extension is just $F(z)=F(re^{i\theta})=r f(e^{i\theta})\in B^2$. This is well-defined near $0$ since we included the factor of $r$.

 

[Infrared]

I think you also asked which continuous maps $f:\text{int}(B^2)\to\text{int}(B^2)$ extend to $B^2\to B^2$. This isn't a topology question (any such map is nulhomotopic). It is possible to extend such an $f$ if and only if $f$ is uniformly continuous. I can write out a proof if you'd like, but I think it's a doable exercise (there's a similar exercise in one dimension in Rudin iirc).

Sorry, it wasn't clear to me exactly which extension problem(s) you are trying to solve.

 

[WWGD]

Edit: Sorry, I keep forgetting the closed ball is ocntractible. Let me rethink.

 

[lavinia]

Edit: Sorry, I keep forgetting the closed ball is ocntractible. Let me rethink.

@WWGD

I thought maybe there is a way to look at the closed ball which you might find helpful.

For any topological space $X$ there is another space called the cone on $X$. It is the quotient space of the Cartesian product $X×[0,1]$ of $X$ with the closed unit interval obtained by identifying $X×1$ to a point. It generalizes the idea of the usual geometric cone since topologically the usual cone that is just the cone on the circle. Like the usual cone in which $X$ is a circle, there is a closed line segment starting at each point in $X×0$ and all of these segments meet at a single "vertex" point $X×1$ crushed to a point.

For the cone on any sphere in Euclidean space one can imagine projecting it into the hyperplane of the sphere and this gives the ball that the sphere surrounds. So the ball is like a flattened teepee over the sphere. In the case of the cone over the circle, stand the cone on a plane and project it vertically onto the plane. The picture one gets of the ball is that its center is the vertex of a teepee and the radius lines are the poles.

If a mapping of a space into itself is null homotopic, this means by definition that the map can be extended to $X×[0,1]$ with $X×1$ mapped to a single point in $X$. But this is the same as saying that the map can be extended to the cone on $X$. So a map of a sphere into itself is null homotopic if and only if it can be extended to the cone on the sphere: that is if an only if it can be extended to the ball.

Notes:

- The cone on any space is contractible since the space can be shrunk to the vertex.

- The map from the ball into the disk that you ask about,you probably meant it to map the entire ball into the circle. This makes a big difference. If you only ask the map to be extended to the disk then as was pointed out, this can always be done. But if you require the image of the map to land in the circle then the map on the boundary must be null homotopic.

- For me, the classical proof of Brower's Fixed Point Theorem helped in understanding these ideas. To recall, this theorem says that any continuous map of the disk into itself has a fixed point.

The proof goes by contradiction by showing that if $f(x)≠x$ for all points in the disk then there is a continuous map from the disk to the circle $H:B^1→S^1$ defined by following the line segment from $f(x)$ through $x$ until it reaches the boundary of the disk. By its construction this map is the identity on $S^1$. So one has extended the identity map on the circle to the disk and this says that the identity map on the circle is null homotopic.

One then shows that the identity map is not null homotopic for instance using that the circle's first integer homology group is non-trivial .

Note that this theorem applies for any closed manifold since the identity is again not null homotopic. That is: there is no map from the cone on a closed manifold into the manifold that restricts to the identity on the manifold. This is often stated as 'no closed manifold is the retract of the cone over it'

 

[Infrared]

If a mapping of a space into itself is null homotopic, this means by definition that the map can be extended to $X×[0,1]$ with $X×1$ mapped to a single point in $X$. But this is the same as saying that the map can be extended to the cone on $X$. So a map of a sphere into itself is null homotopic if and only if it can be extended to the cone on the sphere: that is if an only if it can be extended to the ball.

I just wanted to second this, as it's exactly the same point of view as I take in post 2.

Note that this theorem applies for any closed manifold since the identity is again not null homotopic. That is: there is no map from the cone on a closed manifold into the manifold that restricts to the identity on the manifold. This is often stated as 'no closed manifold is the retract of the cone over it'

This is one generalization of the (lemma used to prove the) Brouwer fixed point theorem. I think a more natural one is: "No compact manifold with boundary retracts onto its boundary". This can also be proven directly by homology arguments, but in the smooth case, there's a nice argument by Milnor: Let $r:M\to\partial M$ be a smooth retraction. By Sard, there is a regular value $p\in\partial M$. Then $r^{-1}(p)$ is a compact 1-dimensional submanifold of $M$. Since $r$ is a retraction, $r^{-1}(p)\cap\partial M=\{p\}$. On the other hand, $\partial r^{-1}(p)=r^{-1}(p)\cap\partial M$. This is a contradiction because the boundary of a compact 1-manifold cannot be a singleton.