How to prove that the map $H^n(M, \mathbb{Z})\otimes \mathbb{R}/\mathbb{Z} \to H^n(M, \mathbb{R})/i(H^n(M, \mathbb{Z})) $ is an isomorphism?

Question

I am trying to get the middle isomorphism in eq 3.3 of 'Quadratic functions in geometry, topology, and M-theory by Hopkins and Singer (here) from the mentioned Mayer-Vietoris sequence.

It suffices to prove that $H^n(M, \mathbb{Z})\otimes \mathbb{R}/\mathbb{Z} $ is isomorphic to $ H^n(M, \mathbb{R})/i(H^n(M, \mathbb{Z})) $. (Here $M$ is a manifold.)I am able construct a map $H^n(M, \mathbb{Z})\otimes \mathbb{R}/\mathbb{Z} \to H^n(M, \mathbb{R})/i(H^n(M, \mathbb{Z})) $ in the obvious way, and check that it is indeed well defined. However I am unable to show that this map is an isomorphism. I would appreciate any clues. Thanks

Answer 1

From the short exact sequence $0\rightarrow\mathbb{Z}\xrightarrow{i}\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}\rightarrow 0$ you get an exact sequence

$$\dots\rightarrow H^n(M)\otimes\mathbb{Z}\xrightarrow{1\otimes i} H^n(M)\otimes\mathbb{R}\rightarrow H^n(M)\otimes \mathbb{R}/\mathbb{Z}\rightarrow 0$$

by tensoring with $H^n(M)$ (integral coefficients are assumed unless otherwise stated). The left-hand map is not in general injective, but this is of no consequence for what we want to prove.

Now there are isomorphisms $H^n(M)\otimes\mathbb{Z}\cong H^n(M;\mathbb{Z})$ and $H^n(M)\otimes\mathbb{R}\cong H^n(M;\mathbb{R})$, which are natural in the second variable, meaning that the map $1\otimes i$ in the exact sequence above becomes the change of coefficients map $i:H^n(M;\mathbb{Z})\rightarrow H^n(M;\mathbb{R})$. Thus we have exactness of

$$\dots\rightarrow H^n(M;\mathbb{Z})\xrightarrow{i} H^n(M;\mathbb{R})\rightarrow H^n(M)\otimes \mathbb{R}/\mathbb{Z}\rightarrow 0,$$

and we may safely identify

$$H^n(M)\otimes\mathbb{R}/\mathbb{Z}\cong H^n(M;\mathbb{R})/i(H^n(M;\mathbb{Z})).$$

Edit: Fixed an embarrassing error pointed out by the op.