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A definition of wedge sum can be found here:

http://en.wikipedia.org/wiki/Wedge_sum

My professor has claimed that wedge sums of path connected spaces X and Y are well-defined up to homotopy equivalence, independently of choice of base points x0 and y0. Base point here means the points that are identified under the equivalence relation forming the wedge product out of the disjoint union topology of X and Y.

Recall homotopy equivalence of X and Y means that there is f:X->Y and g:Y->X continuous with gf and fg homotopic to the identity.

With these definitions, please prove my professor's claim, which I have failed to do for a week. (It is left as an exercise in his lecture.)

Thanks.

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1. Mathoverflow isn't for homework. 2. This is a fun homework assignment and you should think about it more. –  Richard Kent Jul 3 '10 at 20:57
    
This is not homework. It is an exercise left by my professor, which was not assigned for homework. If you don't believe me, you can look at exercise 2.35 in his set of lecture notes: math.caltech.edu/~ma109a/109anotes.pdf You may notice the class is over. I want this fact proven for research I'm doing this summer. –  Jeff Jul 3 '10 at 20:59
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One week's worth of effort, as well as the failure to find anything useful on wikipedia, google-book's preview, and in Caltech's own library says otherwise. Also, none of my peers who have taken Math 109a can handle it either. I have asked, trust me. Do you have a hint? I first tried elementary things like inclusion maps and projections (remember the wedge product is a quotient space.) I then thought about examples (I can do well-behaved cases in R^n) and I thought about showing a homotopy with a dumb-bell like object where the wedged point becomes a line. –  Jeff Jul 3 '10 at 21:08
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I mean, what's the question? True: If $X$, $Y$, and $Z$ are based spaces and $X$ and $Y$ are homotopy equivalent in the based sense then $X\vee Z$ and $Y\vee Z$ are homotopy equivalent in the same sense. False: If $X$, $Y$, and $Z$ are path connected spaces and $X$ and $Y$ are homotopy equivalent then $X\vee Z$ and $Y\vee Z$ are homotopy equivalent no matter what base points you use to stick things together. –  Tom Goodwillie Jul 3 '10 at 21:30
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Tom, I think he's not varying the homeomorphism type of the two factors, only the placement of the basepoint. –  Richard Kent Jul 3 '10 at 21:35
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1 Answer 1

up vote 16 down vote accepted

A counterexample is shown on the cover of the paperback edition of the classic textbook Homology Theory by Hilton and Wylie. This can be viewed on the amazon webpage for the book. The example consists of the wedge of two copies of a cone, the cone on the sequence 1/2, 1/3, 1/4, ... together with its limit point 0. With one choice of basepoints the wedge is not contractible, but with other choices it is.

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Hi, I am assuming you mean that the special point at which one can attach these two cones to get a noncontractible space is the limit point, 0. But it seems to me that any which way you attach these two spaces, you will get a contractible space. I believe this to be true because both cones are themselves contractible, and so I can contract one first, and then the other. It is because these cones are a subset of R^2 that I can contract one first, and then the other. For instance, this process works to contract the space shown on the cover of the book you suggested. Thanks for your response –  Jeff Jul 3 '10 at 23:06
    
I see that this works, thanks a lot for all your help! –  Jeff Jul 4 '10 at 1:39
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