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I'm trying to show that $S^2 \times S^3$ is not homotopy equivalent to $S^2 \vee S^3 \vee S^5$. My argument is that removing any point from the first space leaves a space which is connected (since it is a 5-manifold, so locally looks like $\mathbb{R}^5$, and removing a point from $\mathbb{R}^5$ leaves a connected space), while removing the point where the second space is wedged together does not. Does this argument work? Also, could I have appealed to cup product structure?

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No. $\mathbb{R}^2$ is homotopy equivalent to $\mathbb{R}$, and now you see that your point-method doesn't work. –  Chris Gerig Aug 1 '12 at 1:56
    
@ChrisGerig I see we think alike. –  Alex Becker Aug 1 '12 at 1:59
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Yes, you could apply the cup product structure. In the first space, the generators in degrees 2 and 3 multiply to give something in degree 5, while in the second space they multiply to give 0. –  Aaron Aug 1 '12 at 2:13
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@Matt, I believe the reason the point-method doesn't work (besides providing that obvious example) is that homotopy-equivalence allows a "continuous deletion of points", with points on the space collapsing down to other points; whereas a homeomorphism forbids that by injectivity (it only "squishes" or "spreads" the points). –  Chris Gerig Aug 1 '12 at 4:23
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There's no well-defined operation of "deleting a point" from a homotopy type -- even a connected one. (Consider the unit interval.) –  Aaron Mazel-Gee Aug 2 '12 at 0:41
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Just posting my comment as an answer:

Yes, you could apply the cup product structure. In the first space, the generators in degrees 2 and 3 multiply to give something in degree 5, while in the second space they multiply to give 0.

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No, the argument does not work. For example, the real line is homotopy equivalent to the plane, yet the line has a cut-point while the plane does not.

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