# Operation on spaces and impact on algebraic invariants

I'm trying to get a feel for why different operations on spaces are useful. I realize this question is very long if someone wants to give a response to all the cases. With ''operations on spaces'' I mean:

product, wedge product, cone, suspension, smash product, loop space etc.

My question is essentially what is their relation to the following.

1. Homology & cohomology including cup product structure.
2. Homotopy groups

Some of these have answers I already know of. Like the product commutes with homotopy and the Kunneth formula takes care of products of homology and cohomology. However, I don't know what, if anything, can be said about the resulting cup product structure if it's known for each individual space. The cone is also trivial, because it's contractible, but other things like the smash product I'm completely clueless about.

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For reduced cohomology: The wedge product induces a ring isomorphism from the direct sum of the cohomology rings of the summands, the suspension induces group isomorphisms with a shift of degree +1 (and kills all cup products!), the smash product can be determined from the lexseq of a pair. I don't know if there's an easy description of what happens with the loopspace, but I know that it makes your homology into a ring under the "Pontrjagin product". As for homotopy, the wedge product does not give you any easy information. Some people say that this is fundamentally what makes homotopy... –  Aaron Mazel-Gee Apr 29 '11 at 18:05
difficult, is that the homotopy excision theorem is pretty weak and the result depends on how connected the initial spaces are (a space is n-connected if $\pi_{\leq n-1}=0$). However, taking loops shifts all homotopy groups down by 1, so that's nice. Smash product certainly gives you a map $\pi_n(X)\times \pi_m(Y)\rightarrow \pi_{n+m}(X\wedge Y)$, but I don't know what there is to say about it. This should be closely related to the question of excision, though, because of how the smash product is defined. –  Aaron Mazel-Gee Apr 29 '11 at 18:10

In K-theory you have Bott periodicity, relating $\mathrm K(X)$ to $\mathrm K(S^2X)$, where $S^2X$ denotes the twice suspended space.
In fact one often defines $K^0(X)=K(X)$ and $K^{-n}(X)=K(S^nX)$, for the following reason: in singular cohomology, there is the suspension isomorphism $\tilde{H}^*(X)\cong \tilde{H}^{*+1}(SX)$. (If you think about cellular cohomology, this should be pretty clear.) So the statement you make reduces to saying that $K^0(X)\cong K^{-2}(X)\cong K^{-4}(X)\cong...$. And as a bonus, this also answers a small bit of the original question... –  Aaron Mazel-Gee Apr 29 '11 at 17:06