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Yes, there is such a universal coefficient theorem. $$0 \to Ext(H^{q+1}(X,R), G) \to H_q(X, G) \to Hom(H^q(X, R), G) \to 0$$ see Theorem 6.5.12 in Spanier's textbook "Algebraic Topology". It's on …

I suppose you could divide this problem into suitable sub-problems. 1) Does $M$ have sub-bundles of its tangent bundle? This can be reduced to a complicated cohomology problem. 2) Are any of thos …

A slight expansion on my comment, sort of complimentary to Tom's response. In complete generatlity $H_2X$ tells you nothing about $\pi_1 X$. If $X = A \times B$ with $A$ a $K(\pi,1)$ and $B$ a $K(\ …

As Fernando mentions, the desire to have an algorithm to determine if two maps are homotopic, generally-speaking is impossible to satisfy. This is because some such complexes have unsolvable word prob …

This isn't an application of the group-completion theorem, but the theorem provides context for a problem I'm quite interested in. Let $\mathcal K$ be the space of $C^\infty$-smooth embeddings of $ …

(3) Yes, this is smooth approximation theory. See Hirsch's "Differential Topology" textbook. I believe (1) and (2) were effectively answered by Larry Siebenmann in his ICM paper "Topological Manifol …

To expand on my comment, there's a very general tool to manipulate CW-complexes, due to Whitehead. It tells you when you can in effect remove a cell from a CW-decomposition via `elementary moves', us …

The super-classical example would be the use of the (?Serre's?) theorem that $H^n(X;G) = [X,K(G,n)]$ to deduce that co-dimension two knots have Seifert surfaces. This is written up in Kervaire and We …

There's a construction of $BG$ used quite often by geometric-oriented homotopy theorists that makes the homotopies quite explicit. In this model for $BG$ you start by "resolving" $G$ to the homotopy …

One not-very technical way to think of the Thom Isomorphism Theorem is that if you have a vector bundle, $p : E \to B$, if you remove the $0$-section $Z$ of the vector bundle from the Thom space $Th(p …

Given a disc bundle over a space $X$, there's the classifying map $$ X \to BDiff(D^n)$$ $Diff(D^n)$ has as a subgroup $O_n$, the orthogonal group. The disc bundle over $X$ is a vector bundle if an …

What you are talking about are called string links. String links have a stacking monoid operation. It's non-commutative (provided you have two or more strands). The invertible elements are precisel …

Expanded version of my comment: Please google "the steenrod realization problem". This is a foundational problem in algebraic topology. It has a nice answer -- Glen Bredon's book "Topology and Geomet …

I'll back away from your initial description a bit and use a little more notation for the handlebodies. Let $H_g$ be a genus $g$ handlebody, and $S_g$ the boundary surface. The 3-manifold $M$ is giv …

In a sense I think things are usually reversed. You might find it helpful to read the basics of obstruction theory, for example out of Whitehead's book (but Milnor and Stasheff is good, too). It in …