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January 14

An elementary homotopy equivalence

Continuing with my attempts to grasp the basics of quotient spaces and homotopy, I find myself quickly running into trouble when trying to show that two simple spaces are homotopically equivalent. Consider, for example, the problem of showing that the two spaces "S² with a straight line joining north and south poles" and "wedge sum of S² with a circle" are homotopically equivalent.

I can intuitively see some vague ways of deforming these spaces into each other; for example, one can take the second space and imagine the circle standing inside S² and joined at the north pole, and then quotient half the circle into the side of the sphere; the result should by rights be homeomorphic to the first space. But when it comes to actually writing down explicit maps or homotopies, I'm lost; the text (linked above) is rather light on examples, so it's not even clear what a correct solution would look like. How do topologists approach problems like this? — merge 12:30, 14 January 2010 (UTC)[reply]

This is example 0.8 in Hatcher's Algebraic Topology (relevant chapter here). He does it by collapsing a curve in the sphere from N to S to a point, using some general technical lemmas to make this precise. Algebraist 13:17, 14 January 2010 (UTC)[reply]

Hmm. Thanks for the reference, but it doesn't seem of much use for this problem; in Bredon's book, it's posed in Chapter 1 just after introducing the most basic definitions and properties of homotopy. There are certainly no CW-complexes in sight, and no propositions about homotopy equivalence of quotient spaces (at least that I can identify as such). — merge 16:23, 14 January 2010 (UTC)[reply]

Here's an approach that might help. Imagine the sphere as a (hollow) cube and line as a semicircular "handle" above two points near the centre of one face of the cube. Now apply a homotopy that moves the two points together while preserving the boundary of the cube - it should be reasonably easy to define an explicit formula to do this. Extend to the whole cube using the identity map on the other faces, and you're done (messy details left as an exercise..) (Disclaimer - it's been a long time since I did this sort of thing) AndrewWTaylor (talk) 17:16, 14 January 2010 (UTC)[reply]

Just because the approach of using CW complexes and a few technical lemmas isn't the intended approach of your text doesn't mean it can't be the best approach to such problems. Hatcher clearly thinks it is. Algebraist 17:28, 14 January 2010 (UTC)[reply]

I have no opinion on what approach is best; just trying to solve the problem with the tools I've been given. — merge 19:31, 14 January 2010 (UTC)[reply]

To AndrewWTaylor: Your homotopy takes place outside of both spaces, so with the same reasoning the OP's first space is homotopic to the sphere, which is impossible because one has a non-trivial fundamental group and the other doesn't. Correct me if I'm wrong I'm a novice. Money is tight (talk) 17:34, 14 January 2010 (UTC)[reply]

I recall being messily taught this example in an introductory algebraic topology course - the lecturer used a (non-obvious) deformation retraction of both spaces onto a common space...but I can't fully recall what that was. Not that that matters really, since Brenon seems to want it done using basic definitions. Icthyos (talk) 19:19, 14 January 2010 (UTC)[reply]

Heh. This section does contain definitions for deformation retracts and some homotopy properties of mapping cylinders, for what it's worth. :) If this problem is bad, the next two problems on the page must be much worse... — merge 19:31, 14 January 2010 (UTC)[reply]

How about this: If X is the one-point union of the sphere and the circle at point x₀ and Y is the other space, let f:Y → X map the points on the closed left side of the Y sphere to x₀ and the rest of the Y sphere to the X sphere minus x₀, and then the line gets mapped to the circle minus x₀ in the way you'd expect. Then let g:X → Y map the X sphere to the Y sphere identically with the point x₀ at the north pole and then map half of the circle to the line connecting the poles and the other half to a path along the left side of the sphere connecting the poles. I'm pretty sure fg and gf are homotopic to 1 but it might be annoying to show. Rckrone (talk) 22:10, 14 January 2010 (UTC)[reply]

I've come up with a couple of ideas along these lines, but it always ends up seeming far from obvious how to even show the maps in question are continuous. — merge 23:35, 14 January 2010 (UTC)[reply]

Yeah, that's fairly similar to the attempt at a solution I provided - I was told it wasn't clear that the compositions of f and g were homotopic to the identity maps, then we moved on to some hand-waving. Icthyos (talk) 00:21, 15 January 2010 (UTC)[reply]

Consecutive integers with same sum of proper divisors

At one point, I was having trouble remembering the definition of Ruth-Aaron pairs, and that got me into some other interesting questions.

Are there pairs of consecutive integers with the same sum of all their divisors? Yes. (14,15) and (206,207) are the first two such pairs; this is (sequence A002961 in the OEIS).
What if we look only at proper divisors? (2,3) works and there are no others under 5,000. Note that this sum is always the same parity as the number itself, unless the number is a square or twice a square, so if we want two consecutive ones to be the same, then one of them must be of that type. Before I spend a lot of time programming this to at least start things off, it would be useful to know if there is already a result in the literature.

Ma^t_chups 14:48, 14 January 2010 (UTC)[reply]

A few minutes search with PARI/GP says (2,3) is the only case where n² or 2n² with n < 10⁶ is part of such a pair. I don't know whether others have studied the problem. PrimeHunter (talk) 02:11, 16 January 2010 (UTC)[reply]