First of all:

The Schonflies Theorem: Let $f : S^1 \rightarrow \mathbb R^2$ be an embedding. Then there exists a homeomorphism $F : \mathbb R^2 \rightarrow \mathbb R^2$ such that $F_{|S^1} = f$.

And also according to the same book: enter image description here

But there are some things that I don't understand:

1- How possible is it that the figure(s) are embedding of $S^2$ in $\mathbb R^3$? The object which I colored it blue has two holes and the object named $A$ has four holes, and so on increasing the number of holes as 'the process' goes on; however, $S^2$ has no holes so must be embedding of $S^2$ in $\mathbb R^3$

2- A question based on intuition: the loop $L$ is still outside, in finitely many repetition of the process (i.e. turning the blue object to the object labeled $A$ and so on for finitely many times); will it remain outside when $n \rightarrow \infty$? (sometimes infinity has strange characteristics, for example $\bigcap_1^k (-\dfrac{1}{i}, \dfrac{1}{i})$ is always an open interval for finitely many $k=n$ but becomes one point (closed set in Standard Topology) for infinity!).

3- I don't understand the explanation of the book why the loop $L$ cannot be deformed to a point outside the Alex Horned Sphere, at all! Why disc $D$? I mean, how the disc appeared here, since it's irrelevant to the subject? Why less than $\epsilon$ makes the problem solved ($\epsilon>0$ so still possible to move the loop to the outside)? Would someone please give some simple detailed explanation?

PS all the texts are from the book 'C Adams and R Franzosa - Introduction to Topology. Pure and Applied'.

EDIT - Regarding questions 2 and 3: Alex Horned Sphere example considering some assumptions which I believe are not correct. No matter how close the two horns come to each other by the process, they can't touch together since it wouldn't be embedding of a sphere anymore. And since the 'thickness' of a disk or its embedded is zero so it can pass through. Contradiction! (to illustrate: in real lines $[-a,b_n)$ and $(c_n,d]$ can have zero distant in limit but never touch when limit of $b_n$ and $c_n$ are same and $c_n>b_n$ for all $n$; but this is not same the case of the two intervals be compact.)

  • $\begingroup$ If the appearance of the disk is confusing you, you can replace it with $S^1\times [0,1]$ (the domain of a homotopy) and the argument works exactly the same. The disk just comes from taking $S^1\times[0,1]$ and collapsing $S^1\times\{1\}$ to a point, since at time $1$ of the homotopy you are supposed to get a constant map. $\endgroup$ Aug 14, 2015 at 8:16
  • $\begingroup$ @EricWofsey: thanks a lot for your comment. Actually any of the problems has been solved as I commented below Dorebell's answer. The main problem now again is what I have written in EDIT. :) $\endgroup$
    – user200918
    Aug 14, 2015 at 9:03
  • $\begingroup$ I'm not sure exactly what your confusion is about "touching", but it might be helpful to think about a shape with two linked toroidal hooks coming out of it (i.e., what you get from "connecting the ends" in the middle of the right half of Figure 11.8). In that case, the two arms still don't touch each other, but it's still impossible to pass a loop through it. You can't see this sort of behavior in your example in $\mathbb{R}$ with intervals, because you need at least 3 dimensions to have such a "link". $\endgroup$ Aug 14, 2015 at 9:37
  • $\begingroup$ @EricWofsey: "the two arms still don't touch each other, but it's still impossible to pass a loop through it" and "...you need at least 3 dimensions to have such a link": these are exactly the claims that I don't understand! Would you please guide me a proof for the mentioned two sentences? Thank you $\endgroup$
    – user200918
    Aug 14, 2015 at 9:45
  • 1
    $\begingroup$ To follow up on the idea of a "link" or "chain": you should think of the limiting process of the Alexander sphere not as two points getting closer, but rather a strange, knotted, chain thing which gets progressively more complex. This kind of complexity is not possible in lower dimensions, and that's why the Shonflies theorem is true for them! $\endgroup$
    – Dorebell
    Aug 14, 2015 at 16:45

1 Answer 1


Note: in case you, like me, were confused by the notation, the author appears to use $A$ to refer to the entire sphere, not just the part you colored blue.

Question 1: The blue image, and all subsequent images generated at each finite stage of the process, have no holes. The blue image is the embedding of $S^2$ into $\mathbb{R}^3$ given by taking the standard one, drawing two circles on it, and "pulling" them out to create the "handles" in the image. Nothing is ripped in the process, and it can be undone by shortening the edges of the handle until the circles are again flush with the surface. (Note that this process involves "corners" as drawn, but it's easy to round them out and thus get something smooth). The process is animated here.

Question 2: You're right to be wary of the role of approaching a limit here. The Alexander Horned Sphere itself is an illustration of what can go wrong by taking a process to infinity. Note that at every finite stage of its construction process, this embedding is actually equivalent to the normal one! (Imagine smushing the handles back into the sphere to go backwards a step, and keep doing this until you get back to where you started). In particular, at each finite stage, the exterior of the sphere is simply connected. However, it's not too hard to see that the loop $L$ remains outside the sphere even in the limit. This is because at each finite stage, the amount of space in $\mathbb{R}^3$ where things are changing is smaller. You can draw a little $3$-dimensional ball $B$ around the loop $L$ such that at every stage past the first, nothing changes inside this ball. In symbols, if $A_n$ is the sphere obtained at the $n^{th}$ stage of the process, then $B \cap A_n = B \cap A_1$ for all $n \geq 1$. This means that, since $L \subset B$, $$L \cap A_n = (L \cap B) \cap A_n = L \cap (B \cap A_n) = L \cap (B \cap A_1) = (L \cap B) \cap A_1 = L \cap A_1 = \emptyset$$.

Question 3: First, we'll see why the disc $D$ matters. A loop $L$ in a space $X$ should be thought of as a continuous function $\gamma: S^1 \rightarrow X$. (Point of clarification here: we don't require $\gamma$ to be an embedding and allow self-intersections). Then, a deformation of $L$ to a point $x \in X$ is exactly a map $\Gamma: S^1 \times [0,1] \rightarrow X$ such that $\Gamma(s, 0) = \gamma(s)$ and $\Gamma(s, 1) = x$ for all $s$. You're supposed to think of it as a continuously varying family of loops $\gamma_t(s) = \Gamma(s,t)$. For each fixed $t$, changing the $s$ variable describes a loop. For $t = 1$, this is the constant loop at the point $x$. Now, since $\gamma_1(s)$ is a constant function in $s$, we can think of $\Gamma$ as actually being defined on $D$, as follows:

Using polar coordinates on $D$, we can write $d \in D$ as $(r, \theta)$ with $r \in [0,1]$ and $\theta \in S^1$. We can define $\tilde{\Gamma}: D \rightarrow X$ by $\tilde{\Gamma}(d) = \Gamma(1-r, \theta)$. Since $\Gamma(1, \theta)$ is constant in $\theta$, this is well-defined and continuous.

Thus, we may assume by way of contradiction that there is some function $F: D \rightarrow \mathbb{R}^3 \setminus A$ such that, in polar coordinates, the map $\gamma: S^1 \rightarrow X$ described by $\gamma(\theta) = F(1, \theta)$ coincides with $L$. Now, as stated in the text, there is some definite number $\epsilon > 0$ such that for every point $y \in F(D)$, the distance between $y$ and every point $a \in A$ is at least $\epsilon$. (This is because both $A$ and $F(D)$ are compact and they don't intersect by assumption). This number $\epsilon$ depends on $F$, but not on $y$ and $a$. However, we are assuming there is a certain fixed $F$ which satisfied our hypotheses. Now, for each sphere $A_n$, you can see by staring at the picture that the only way to shrink the loop $L$ to a point is to move it through the little gaps in the picture. In particular, the image of the disc given by this deformation will include some points within the little gaps. (As a hand-wavy explanation, think about the fact that $L$ has some points "under" the biggest handle and some points "over" it. In order to shrink it to a point, we have to bring the "under" parts up or the "over" parts down). Now at each finite stage, this is no problem: we can always choose a disc small enough to get through the littlest gap. However, for the full limiting construction $A$, there is no littlest gap; the handles are interwoven so tightly that any gap would be smaller than $\epsilon$ for any $\epsilon$ whatsoever. Of course, this is impossible to really visualize, so it's easier to convince yourself with a limiting argument. By drawing each $A_n$ carefully, you can arrange it such that the distance between the two sides of the little gap in $A_n$ is less than half of the distance in the previous step. Thus, the limiting size of the little gaps is less than $\frac{1}{2^n}A_1$ for all $n$, and thus it is $0$! In particular, there is some finite stage $A_n$ such that the gap in $A_n$ is less than the $\epsilon$ we fixed earlier. Since the size of the gap in the limit is smaller than the size of the gap at every finite stage, this too is less than $\epsilon$. But since $F(D)$ has to contain some points inside the little gap, it has to contain a point $y$ such that $y$ is within distance $\epsilon$ of the wall, a contradiction.

This is a strange and somewhat disturbing construction. It's a little hard to believe that this limiting object really exists, and that it is still an embedding of $S^2$. It takes a bit more argument to show that this is true; see here. (Does anyone know of a "canonical" reference for this fact?)

  • $\begingroup$ Thank you very much for your answer. First part was brilliant. I still can't understand the second and third answers. In one or two days if I still can't understand it I will write back. :) $\endgroup$
    – user200918
    Aug 13, 2015 at 8:09
  • $\begingroup$ Please let me know if you have any specific questions! This construction and the proofs involved belong to the large class of things in topology that are easy to explain with pictures and a blackboard but much harder to state rigorously. $\endgroup$
    – Dorebell
    Aug 13, 2015 at 9:10
  • $\begingroup$ In q.2, I believe that the ball which you have considered includes L and a part of one of the handles but doesn't include any thing of 'the gap'; if L is going to out then it 'passes' the gap. Now I understand everything in text (not incl. the link in q.3) but one: the A.H.S is intrinsically inconsistent! : we construct some configuration in the gap of $A_1$ which shouldn't matter what is it, rather what matters is that in final we admit that L can't go outside anymore (since 'the gate's size' is zero) which logically is equivalent that L is not outside of the A.H.S ? thank you. $\endgroup$
    – user200918
    Aug 14, 2015 at 3:12
  • $\begingroup$ Unless we say some illustrative comparison: a line can pass through $b$ in $A=[a,b) \cup (b,c]$ but distance between the line and $A$ will be zero. Does it like what A.H.S tries to say? Thank you $\endgroup$
    – user200918
    Aug 14, 2015 at 3:16
  • $\begingroup$ Example A=[a,b)∪(b,c] cannot be true since it is not compact. Would you please explain why loop can't go out while it is outside, since the A.H.S has it's boundary and in infinity distance between two gaps to be zero means that finally A.H.S intersects with itself. ? Thank you $\endgroup$
    – user200918
    Aug 14, 2015 at 6:05

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