A false reasoning for a better intuition behind the universal coefficient theorem

Question

Let $X$ be a topological space. I am trying to understand the natural homomorphism $$H^*(M,\mathbb{Z}) \to H^*(M,\mathbb{R}),$$ induced by the inclusion $\mathbb{Z} \hookrightarrow \mathbb{R}$.

The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.

Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by: $$C^k(M,\mathbb{Z}) := Hom(C_k(M),\mathbb{Z}), \quad C^k(M,\mathbb{R}) := Hom(C_k(M),\mathbb{R}).$$ Then the boundary operator $\partial$ on $C_k(M)$ dualizes to boundary operators $\delta_{\mathbb{Z}}$ and $\delta_{\mathbb{R}}$ on the two above groups respectively.

Let

$$Z^k_{\mathbb{Z}} := ker(\delta_{\mathbb{Z}} : C^k(M,\mathbb{Z}) \to C^{k+1}(M,\mathbb{Z})), \quad Z^k_{\mathbb{R}} := ker(\delta_{\mathbb{R}} : C^k(M,\mathbb{R}) \to C^{k+1}(M,\mathbb{R})),$$ and $$B^k_{\mathbb{Z}} := im(\delta_{\mathbb{Z}} : C^{k-1}(M,\mathbb{Z}) \to C^k(M,\mathbb{Z})), \quad B^k_{\mathbb{R}} := im(\delta_{\mathbb{R}} : C^{k-1}(M,\mathbb{R}) \to C^k(M,\mathbb{R})).$$

1. Since $Hom(M,-)$ is left exact, the inclusion $\mathbb{Z} \hookrightarrow \mathbb{R}$ yields an injection $C^k(M,\mathbb{Z}) \hookrightarrow C^k(M,\mathbb{R})$.
2. Since $\delta_{\mathbb{R} | C^k(M,\mathbb{Z})} = \delta_{\mathbb{Z}}$, we have inclusions $$B^k_{\mathbb{Z}} \subset B^k_{\mathbb{R}}, \quad Z^k_{\mathbb{Z}} \subset Z^k_{\mathbb{R}},$$ which yield a natural homomorphism $\rho : H^k(M,\mathbb{Z}) \to H^k(M,\mathbb{R})$ in cohomology.

Suppose that $\alpha \in Z^k_{\mathbb{Z}}(M)$ is such that $\rho([\alpha]) = 0$. This means that $\alpha \in B^k_{\mathbb{R}}$. But $\alpha \in C^k(M,\mathbb{Z})$, so $\alpha \in B^k_{\mathbb{R}} \cap C^k(M,\mathbb{Z}) = B^k_{\mathbb{Z}}$, and therefore $[\alpha] = 0$.

Could someone tell me where this reasoning fails ? Thanks a lot !

Answer 1

The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.

For instance if you consider a chain complex $\mathbb Z \overset{2}\to \mathbb Z$ and change coefficients to $\mathbb R$ then in degree 0 degree 1 one has $B^1_{\mathbb R} \cap C^1_{\mathbb Z} = \mathbb Z$ but $B^1_{\mathbb Z} = 2\mathbb Z$.