Problem 20 part (b) of Chapter 1 asks us to show that the infinite-holed torus is homeomorphic to the "infinite jail cell window." His hint helped me to get started (I think).

(I apologize for not having a diagram, but one can be found by going to


and scrolling down to the text's page 24)

If one makes two additional cuts at the top (relative to the page) of the original outlined cylinder and perpendicular to it, another cylinder results: a cylinder perpindicular to the original, running along the top of the original.

This new cylinder can be stretched around the original cylinder until its ends almost meet, forming a hole. This is similar to the left end of the infinite-holed torus.

This is where I get stuck, because we've now distorted the cell windows adjacent to the one we started with, and the hole created in the paragraph above isn't attached to anything - and gluing isn't allowed.

Asking for online help for this kind of problem is a bit like asking for online help in learning to ride a bicycle. It is difficult to put into words, but I would be grateful for any help/hints.

This is not an assigned problem, but I thought the homework tag would be appropriate.

Thank you for your consideration.


  • 1
    $\begingroup$ Unfortunately, I don't understand the description of what you have done... I apologize if I am about to tell you to do the same thing. I believe Spivak is suggesting that you try to build a homeomorphism "inductively". To do this, think of (A) as obtained by starting with an annulus (cylinder) and then gluing infinitely many copies of a genus 0 surface with 4 boundary components (call this piece X). In the surface (C), he has given you the hint of how to find an annulus. You now want to find a way of exhausting (C) by a countable collection of X's, boundary glued to each other correctly. $\endgroup$
    – Sam Lisi
    Apr 10, 2012 at 15:54
  • $\begingroup$ Once you've figured this out, using these decompositions of (A) and (C) will allow you to construct the homeomorphism pretty easily. $\endgroup$
    – Sam Lisi
    Apr 10, 2012 at 15:55

1 Answer 1


For those not looking at the diagram, the "infinite jail cell" is the boundary of a tubular neighborhood of the integer grid (including horizontal a vertical lines) in the $z = 0$ plane of $xyz$-space.

If we take a tubular neighborhood of the line segments from $(0,0)$ to $(1, 0)$ to $(1,1)$ to $(0, 1)$ to $(0,0)$, we get something toroidal. The "cylinder" described by Spivak consists of half of each meridian of this torus -- the half that's closest to the point $(1/2, 1/2)$. Let's call that cylinder $C$; its boundary consists of two squared-off circles in the $z = \pm 1/2$ planes.

I believe that what Spivak is suggesting is that you next consider the part of the infinite jail cell that's between two larger squared-off circles, essentially ones that are parallel to the $z=0$-plane squares with corners $(-1, -1), (2, -1), (2, 2), (-1, 2)$, offset by $1/2$ in the positive and negative $z$ directions. Call this larger part $D$. It basically consists of 9 cells, the center one of which is $C$. The boundary of $D - C$ consists of four circles; the set $D-C$ is clearly a 2-manifold-with-boundary, hence homeomorphic to a $k$-holed torus with four disks removed for some $k$. (I'm pretty sure that $k$ is $8$, but I could easily be off by one or two.)

Now consider a $k+2$-holed torus, with the $k+2$ holes arranged in a line, like the infinite torus drawn by Spivak, and cut off the left and right half-tori. You end up with a $k$-holed-surface-with-boundary, having four circle-boundaries. This is clearly homeomorphic to D - C. And a single cylinder is homeomorphic to $C$, so the union, along common boundaries, gets you something that's homeomorphic to $D$.

You now proceed by induction: drawing a 5 x 5 pair of squared-off-circles, you see that the region $E$ between them and the 3x3 object $D$ is also homeomorphic to a $p$-holed chain with four circular ends, etc.

In other words, the region between adjacent circle-pairs is always homeomorphic to an $s$-holed-torus-with-boundary, having four boundary components. By the classification theorem for surface, this is the same as a truncated piece of a linear chain of tori. You thus build up a sequence of homeomorphisms

$h_i : U_i \rightarrow V_i$

where $U_i$ is a chain of $(2i + 1)^2 + 1$ tori with the last half-torus removed, and $V_i$ is the part of the infinite jail-cell enclosed by the two squared-off circles of edge-length $i$. The nice thing about this sequence is that $h_{i+1} = h_i$ on $U_i$, so the limit is well-defined and is a homeomorphism as well.


An alternative:

Let's just talk, from here on, about squares on the integer grid, OK? So what I did in the first solution was start with a 1x1 square, move to a concentric 3x3 square, and so on.

As an alternative, you can take, as your second square, a 2x2 square whose lower-left corner is the origin. The difference $D - C$ in this case clearly is homeomorphic to a 3-holed torus with some boundary, and the same sort of union argument applies. Now you take, as your third square, a 3x3 one whose lower-left is at $(-1, -1)$. Again, the difference $E - D$ is clearly homeomorphic to a 5-holed torus with some boundary. And in general, you can extend like this, alternating between adding a band to the northeast and adding a band to the southwest, gradually handling all the cells of the infinite jail-window.

The advantage of the second approach is the obvious homeomorphism between the added strip and a $2i+1$-holed torus with some disks removed. The disadvantage is that the gluing-map is more complex. I think it's about a wash.

  • 1
    $\begingroup$ This needs more upvotes. $\endgroup$
    – Rasmus
    Oct 28, 2013 at 16:19

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