>>12617428

Next, we ask what happens if we tensor the complex with some abelian group [math]A[/math] (over the integers). We obtain the complex [math]C_*(X; A) = \cdots \to C_{n+1}\otimes A \xrightarrow{\partial_{n+1} \otimes 1_A} C_n\otimes A \xrightarrow{\partial_n \otimes 1_A} C_{n-1}\otimes A \to \cdots[/math], the homology of which gives us [math]H_*(X; A)[/math], the singular homology of [math]X[/math] with coefficients in [math]A[/math]. This is where many who see this the first time would assume "oh but [math]H_n(X; A) \cong H_n(X; \mathbb{Z}) \otimes A[/math]", which is not true in general. Instead, there is an exact sequence [math]0 \to H_n(X; \mathbb{Z})\otimes A \to H(X; A) \to \text{Tor}(H_{n-1}(X; \mathbb{Z}), A) \to 0[/math] for every [math]n\ge 0[/math] ([math]H_{-1}(X; \mathbb{Z}) = 0[/math]). Even better, yet not completely perfect: this sequence is split exact: [math]H_n(X; A) \cong H_n(X; \mathbb{Z})\otimes A \oplus \text{Tor}(H_{n-1}(X; \mathbb{Z}), A)[/math], but not naturally (for example Dylan Wilson's answer in https://math.stackexchange.com/questions/33685/what-does-splitting-naturally-mean-in-the-universal-coefficients-theorem).

Some immediate things that follow from the UCT:

(1) If the group [math]A[/math] is torsion-free, then [math]Tor(-, A)=0[/math], and so [math]H_n(X; A) \cong H(X; \mathbb{Z}) \otimes A[/math].

(2) If [math]H_n(X; \mathbb{Z}) \otimes A = 0[/math], then [math]H_n(X; A) \cong \text{Tor}(H_{n-1}(X; \mathbb{Z}, A)[/math].

(3) [math]H_0(X; A) \cong H_0(X; \mathbb{Z}) \otimes A[/math].

If now [math]H_n(X; \mathbb{Z})[/math] is all torsion, or even a finite group, then it vanishes when tensored with [math]\mathbb{Q}[/math], and (2) applies. Furthermore, the rationals are torsion-free, so (1) applies and we get [math]H_n(X; \mathbb{Q}) = 0[/math]. Many such cases!

Reminder to get WASTED!