I believe you circumvented the difficult part by claiming that R^2\J is path connected. If I knew this, I could take it from there.

I see that this works, thanks a lot for all your help!

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

I think I understand my lack of clarity. "Base Point" has many uses in topology, and so I apologize. The edit is above, but basically base point means the points we are identifying to form a wedge sum. Not the "Base point" of a loop, or anything like that. Sorry.

You claimed that my professor's assertion was false, so I was referring to this claim you made. My claim was that the wedge sum does not depend on where you connect the two spaces, up to homotopy equivalence. I do not understand why your response has 3 spaces in it. The question is if you wedge X and Y (path connected) together at x0 and y0, how is this homotopy equivalent to the same wedge, but done bringing together x1 and y1? You said this is not true in general. So, I am wondering if you have a counterexample? Thanks.

@Tom Goodwillie, meaning not even if X and Y are path connected? If so, do you have a counterexample? I was worried myself after so much effort that it would be false.

Yes, I know this... I have tried to use that in many ways. I have tried to show mutual homotopy equivalence to the disjoint union of X and Y (if these are the two spaces you are wedging.) Of course, that didn't work. It would be a cruel world in topology if it did, since it is blatantly false. I have tried to show homotopy to the dumb-bell like shape. (The wedge product except the point at which the spaces meet is extended to a line.) I believe it is harder than it first appears to think about this beyond the intuitive level, and to actually rigorously construct f and g.

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.

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.