Let $M$ be a smooth $n$-manifold, and $\iota: \mathbb{Z}\hookrightarrow \mathbb{R}$ be the natural inclusion homomorphism of abelian groups. Then, it is easy to check that we obtain a well defined induced map $\iota_*:H_i(M,\mathbb{Z})\rightarrow H_i(M,\mathbb{R})$ for each $0\leq i\leq n$. Need this map be injective? I suspect that it might not be given that $H_i(M,\mathbb{Z})$ could have a torsion subgroup which would necessarily have to be mapped identically to zero given that $H_i(M,\mathbb{R})$ must be be torsion free.
If we instead consider cohomology, then I am unsure of how to obtain an induced $H^i(M,\mathbb{Z})\rightarrow H^i(M,\mathbb{R})$ since it seems that the natural map on the cochain complexes should go $S^i(M,\mathbb{R})\rightarrow S^i(M,\mathbb{Z})$ and be given by the pullback of $\iota$. But then the induced cohomology map goes $H^i(M,\mathbb{R})\rightarrow H^i(M,\mathbb{Z})$ so I must be doing some thing wrong here.
Any help on ironing out these ideas without appealing to the universal coefficient theorem would be lovely.
Best Answer
Your first paragraph is correct: integral homology may have torsion, and therefore the map $H_n(M; \mathbb{Z}) \to H_n(M; \mathbb{R})$ cannot always be an inclusion.
The definition of the $n$th cochain group with coefficients in an abelian group $A$ is $$ C^n(M; A) = \textrm{Hom}(C_n(M; \mathbb{Z}), A). $$ (Note the coefficients in the chain group.) Therefore given a group homomorphism $\phi: A \to B$, there is a map $$ C^n(M; A) \to C^n(M; B) $$ defined by $f \mapsto \phi \circ f$. This will commute with the differential and so it will induce a map on cohomology. In particular, there is a map $$ H^n(M; \mathbb{Z}) \to H^n(M; \mathbb{R}), $$ which cannot always be an inclusion for the same reason as for homology.