## How to find homology groups of a suspension of a simplicial complex? [closed]

The task is to express homology groups of SX through homology groups of X. (X is a simplicial complex)

I don't really understand how to solve the problem.

Every book, where i tried to read the explanation, operates with "reduced homology groups". And the solution is simple - $H'_k(SX) = H'_{k-1}(X)$

But i don't understand, how to move from reduced homology groups to ordinary homology groups.

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The relationship between reduced and unreduced homology is given by the homology long exact sequence of a pair and is also covered in most intro algebraic topology texts. IMO your question is more appropriate for math.stackexchange. I've voted to close. – Ryan Budney May 10 2011 at 18:25
The difference between reduced and unreduced homology groups is only in H_0, where the latter has one more dimension. I also vote to close. – André Henriques May 10 2011 at 19:17
Dear unknown(google), your question seems more appropriate for the nice people at math.stackexchange.com – S. Carnahan May 10 2011 at 20:05