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Let's have an algebraic topology thread.
Yesterday I learned about Eilenberg-MacLane spaces. That is for a group G and a natural number n you can construct a topological space K(G,n) such that G is the n-th homotopy group and all other homotopy groups are trivial.
Can I get a similar result if I want more homotopy groups to be non-trivial and to equal other previously specified groups? For example if I have a set of groups (Gi) with i in some finite or countably infinite index set, can I find a space where the i-th homotopy group is Gi for all i? Or is there some obstruction making such a space impossible?
Maybe there is a glueing of the K(Gi,nj)'s which does this, I'd ask on SE
>>7767928
I know category theory and need to learn some basic algebraic topology for a homotopy type theory course. What's a good introductory resource?
>>7768076
OP here. I like Tom Dieck's "Algebraic Topology" and Davis & Kirk's "Lecture Notes in Algebraic Topology". Bredon's "Topology and Geometry" also contains many chapters on algebraic topology, but I only briefly looked into that book.
>>7767928
What you're looking for is called a product.
>>7768076
Hatcher is the standard reference. His book is certainly the most well written and provides a lot of geometric intuition. On the down side he avoids using category theory as much as he can, but if you'll be taking a homotopy theory class then it's no loss.
>>7770028
K(G,n) ^ K(G,m) is m+n-1-connected (first nontrivial homotopy group is the n+m th), where ^ is the smash product.
>>7770095
Anyway OP, you can construct your space by playing with bouquets of spheres, similar to the standard CW complex construction of the K(G,n).
If you take a sphere for each generator of G (m of them, say), you can form K(G,n) by taking the wedge sum of the spheres, giving you [math]\bigvee^{m} S^{n}[/math], using attaching maps to impose relations, then adding cells in higher dimension to kill the higher homotopy.
Also, look up the Borel construction.
>>7770095
Wrong product, just use the cartesian one.
>>7770263
Yes, they do. Just apply the universal property for products to the relevant hom space and then look at path components on both sides of the homeomorphism.
