# 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. –  Ryan Budney Oct 26 '10 at 19:15
Computation of $\pi_1$ doesn't really help because in higher dimensions $S^1$ is absent. –  Priyavrat Deshpande Oct 26 '10 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? –  Ryan Budney Oct 26 '10 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 $\pi_1$. –  Priyavrat Deshpande Oct 27 '10 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. –  Ryan Budney Oct 27 '10 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.

-