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Let $X$ be the $2$-sphere with two pairs of points identified, say $(1,0,0) \sim (-1,0,0)$ and $(0,1,0) \sim (0,-1,0)$. Write $Y$ for the wedge sum of two circles with a $2$-sphere: if it matters, the sphere is in the "middle," so the circles are attached at two distinct points on the sphere.

Now I think one can show, using Mayer-Vietoris and van Kampen, that these spaces have the same homology (that of a torus) and fundamental group (free on two generators). But are they homotopy equivalent?

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the cup product structure on the cohomology ring of $Y$ is trivial; I doubt this is true on $X$ - you could check with a simplicial decomposition; I'll try to think of a less awful way. – user29743 Aug 7 '12 at 18:34
Hint: $X$ and $Y$ have the same cohomology ring, the same homotopy groups, homotopy algebras, $K$-theory, etc, etc... – Ryan Budney Aug 7 '12 at 18:59

Yes. Taking the wedge sum with a circle is the same as identifying two points (up to homotopy, with a nice space like the sphere which is homogeneous).

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Nice. How does one prove such a thing? This reminds me of a quote I heard recently: "Something is obvious if it is easy to write down the proof. Note that this does not apply in topology." – Justin Campbell Aug 7 '12 at 21:02
Clearly identifying two points is homotopy equivalent to gluing the endpoints of an interval to the space. To finish the argument, hopefully we can contract a path (this time in the space) to a point without affecting homotopy type, but this is not clear to me. This looks like the question I asked the other day about wedge sums... – Justin Campbell Aug 7 '12 at 21:30
Think of it this way (my friend explained it to me): given two points on a space $M$, attach an arc connecting them. We can "slide those two points together" on $M$ to get something that looks like $M$ wedge a circle. Or we can contract the arc to get something that looks like $M$ with two points identified. – user29743 Aug 7 '12 at 22:05
But why is sliding these two points together a homotopy equivalence? Surely this depends on $M$ being suitably nice. – Justin Campbell Aug 7 '12 at 22:11
When attaching cells to a space, if you homotope the attaching map you get homotopy-equivalent spaces. This is an exercise in Hatcher's text. – Ryan Budney Aug 7 '12 at 22:44

This is actually a special case of the result that if $A \to X$ is a closed cofibration, and $f:A \to B$ is a map, then the natural map $M(f) \cup X \to B \cup_f X$ is a homotopy equivalence: here $M(f)$ is the mapping cylinder of $f$, and the result is 7.5.4 of my book Topology and groupoids.

For your case, you have to take $A$ to consist of $4$ points, and $B$ to consist of $2$ points.

Here is part of the general picture


and here is a picture of the special case you asked about but with just one pair of points identified:


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Is this right? EDIT: no. $X$ admits a surjection to $\mathbb{RP}^2$ which induces an inclusion on the $\mathbb{Z}/2$ valued cohomology rings. Therefore the cup product structure on $X$ is non-trivial and so $X$ cannot be homotopy equivalent to $Y$.

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No, the map is trivial on $H^2$. – Ryan Budney Aug 7 '12 at 19:01

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