Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

3 Answers

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.

Here is part of the general picture


and here is a picture of the special case you asked about:


share|improve this answer
add comment

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 $Y$.

share|improve this answer
No, the map is trivial on $H^2$. –  Ryan Budney Aug 7 '12 at 19:01
add comment

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).

share|improve this answer
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
show 3 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.