Singular homology of disjoint union of discs quotient boundaries

Question

I am trying to compute the singular homology of the space $X=D \coprod D\setminus \sim$ with $D$ the two dimensional disc and $\sim$ the equivalence relation that identifies the boundaries of the discs.

On one side I know that the homology of the disjoint union is the direct sum of the homologies and on the other side I know that $D\cup D\setminus \sim$ is the sphere $S^2$. I am not very used to disjoint unions so I don't know how I could link these two results, or if they are even useful to calculate the homology I want. How should I proceed?

I would suggest adding homological-algebra as a tag to this question — Morettin, May 01 '20 at 18:07

score 1 · Accepted Answer · answered May 01 '19 at 16:30

1

As you say, $X$ is homeomorphic to $S^2$ (intuitively, you can think of the two discs as two hemispheres, attached along the equator), so it has the same homology groups as $S^2$ (homology is a homeomorphism invariant, in fact a homotopy invariant).

answered May 01 '19 at 16:30

Jane Doé

388

Thank you I have some (stupid) confusions with disjoint unions – user404720 May 02 '19 at 15:31

Singular homology of disjoint union of discs quotient boundaries

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