Suppose a manifold is covered as the union of $|A|$-many $n$-spheres, i.e. $M=\bigcup_{\alpha\in A}S^n$. We are allowed to move the spheres around so long as they cover $M$; that is, we can make the intersection as small as possible (e.g. a point). I would like to compute the homology groups of $M$; however, we have no information regarding the intersections of of the spheres (other than the fact that they cannot be disjoint).
Is there a way to "arrange" the spheres in such a way that they still cover $M$ but have homology that is computable? Could I contract the intersection to a point and then compute the homology of a wedge sum $H_k\left(\bigvee_{\alpha\in A} S^n\right)=\bigoplus_{\alpha\in A} H_k(S^n)$?
Note, $A$ can be made to be countably infinite or uncountably infinite.
Thanks in advance! Any help would be much appreciated.
For your new question, homology can be computed from the nerve of certain coverings by open balls, using the Nerve Theorem. You need all intersections of finitely many open balls to be either empty or contractible.
Edit: Here's a bit more information. You can define the nerve of any covering of a topological space by open sets. This is a simplicial complex that has a vertex for every open set in the cover and a simplex whenever a collection of finitely many open sets has a nonempty intersection. If these finite intersections are all contractible or empty, then we say the cover is a good cover. The Nerve Theorem states that the nerve of a good cover is homotopy-equivalent to the original space. Hence it will have the same homology. Admittedly, the nerve is often not any easier to compute than the original space. For my example of covering the circle by three intervals, the nerve is just a triangle, which is actually homeomorphic to the original circle.