I’ve decided to start collecting here some random exercises arose during discussions with my colleagues (usually during the weekly pizza-time) and I found interesting.

Let be a continuous map between topological spaces, and some coefficients. Let be the integers.

- If is an isomorphism, is one, too?
- If is an isomorphism, is one, too (for every )?

Assume that is a ring. can be defined as homology of , the chain complex which in degree is the free -module generated by singular -simplices of . This is a bounded below projective chain complex and thus if the map induced by is a quasi-isomorphism, it must be already a homotopy equivalence. Abstract no sense: in the standard model category structure for whose weak equiv are quasi-isomorphisms (See https://www.math.rochester.edu/people/faculty/doug/otherpapers/dwyer-spalinski.pdf Thm 7.2) and are both fibrant and cofibrat objects, since the inclusion is a monomorphism whose cokernel is degree-wise projective and the map is a degreewise epimorphism. Therefore by Whitehead theorem for Model Cats, a weak-equiv between fibrant-cofibrant objects is already an homotopy equivalence. Thus the map is still a quasi-isomorphism after applying which is one way to compute cohomology of a space.

For the second point, by replacing with the mapping cylinder of , we can always assume that is the inclusion. Thus it is enough to prove that when then also . **This is true under the added assumption that the homology of is finitely generated**, for which it is enough that both have finitely generated homology. In fact using this UCT for homology (See Davis & Kirk’s Lecture Notes in Algebraic Topology page 48 Theorem 2.36) we have

Therefore if both ends are zero, the middle one is zero as well, proving that .

Actually this last version of the UCT is the reason why I decided to write a little article about it, it’s a version which is always forgotten, and it’s a pity since it turns out it is very powerful.

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