# How to show that a space has the homotopy type of wedge of spheres ?

Let me try and put the question in context. I am studying certain subsets of the tangent bundle of a sphere. I also have a regular CW complex which is a deformation retract of such a subset. Hence I have a description of the cells and the information that tells me which cell is in the boundary of which other cell. Fortunately this cell complex is a homotopy colimit of a diagram of spaces. As a result I can compute the cohomology groups (but not the product).

All my examples concerning $S^1$ and $S^2$ show that these subsets have the homotopy type of wedge of copies of $S^1$ and $S^2$ respectively. Hence I am trying to prove that this is the case in all dimensions. In this process the only thing I was able to prove that there is a retraction from these subsets to the underlying sphere.

So I would like to know about various methods to show that a space is a wedge of spheres.

I understand that this question might sound vague and the information too little.

• Your situation is pretty generic, so unless you have some further input data there's not a whole lot to say. Can you compute $\pi_1 X$ and show it's free? That would be a start. Commented Oct 26, 2010 at 19:15
• Computation of $\pi_1$ doesn't really help because in higher dimensions $S^1$ is absent. Commented Oct 26, 2010 at 22:26
• Of course it helps. If $\pi_1 X$ isn't a free group you know for certain $X$ doesn't have the homotopy-type of a wedge of spheres. Do you have anything to indicate $\pi_1 X$ is trivial? Commented Oct 26, 2010 at 22:56
• I can express $\pi_1$ as a colimit of fundamental groups of the spaces in a diagram, other than that I don't know anything in general. In case of the examples I could do by hand, I Was able to directly observe the wedge product hence I didn't bother to calculate the colimit and verify at the level of $\pi_1$. Commented Oct 27, 2010 at 0:58
• If you can't get your hands on $\pi_1$ I doubt there's much likelyhood you'll be able to prove this space is a wedge of spheres -- which is a much harder problem. Commented Oct 27, 2010 at 3:18

To follow up a bit on Mikael's answer, the notion of non-pure shellability is probably more relevant to your situation. Shellable simplicial complexes are wedges of spheres of equal dimension, but non-purity allows different dimensional spheres. You should look at papers by Michelle Wachs and Anders Bjorner if you're interested. However, this will require finding a simplicial decomposition of your space, which may be a challenge.

Added: Since this is now the accepted answer, I figure I should give the precise references. Both papers are on JSTOR (follow the links).

Björner, Anders; Wachs, Michelle L. Shellable nonpure complexes and posets. I.
Trans. Amer. Math. Soc. 348 (1996), no. 4, 1299–1327. http://www.jstor.org/stable/i311403

Björner, Anders; Wachs, Michelle L. Shellable nonpure complexes and posets. II. Trans. Amer. Math. Soc. 349 (1997), no. 10, 3945–3975. http://www.jstor.org/stable/i311413

(If I remember correctly) a shellable (simplicial) complex automatically has the homotopy type of a wedge of spheres: if you could find a shellable triangulation you should be done.

Of course, this'll only work if you're lucky enough that the structure you have admits nice combinatorial structures that happen to be shellable — but it's one way you can get to a wedge of spheres.

There is an algorithm that sometimes can be useful. You can reduce the number of cells of a CW-complex obtaing a new CW-complex homotopy equivalent to the first. This is done collapsing certain pairs of consecutive cells. The exact procedure is too long to explain. You can find it in "Combninatorial algebraic topolgy" by Kozlov chapter 11.The result you may need it's the main theorem of discrete Morse theory.

http://www.springer.com/mathematics/geometry/book/978-3-540-71961-8

• Thanks for the reply. I tried using acyclic matching but then I realized that I don't know whether the spaces under consideration admit the minimal complex (only cells are the homology cells). I have asked this question here: mathoverflow.net/questions/43799/… Commented Oct 27, 2010 at 14:16