Let $M$ be a topological space. Is it true that any map $H_k(M, \mathbb{Z}) \to \mathbb{Z}/2$ factors through the reduction of coefficients $H_k(M, \mathbb{Z}) \to H_k(M, \mathbb{Z}/2)$?
Factoring through reduction of coefficients in homology
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algebraic-topology
homology-cohomology
1 Answers
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When you talk about reduction of coefficients, do you mean in the sense of the universal coefficient theorem?
If so, $H_k (M, \mathbb Z/2) \cong H_k(M, \mathbb Z) \otimes \mathbb Z/2 \oplus {\rm Tor \ } (H_{k-1} (M, \mathbb Z),\mathbb Z/2)$, and your reduction of coefficients is the natural map from $H_k(M, \mathbb Z)$ into the $ H_k(M, \mathbb Z) \otimes \mathbb Z/2$ factor of $H_k (M, \mathbb Z/2)$ given by $x \mapsto x \otimes [1]$.
I believe that any map $H_k(M, \mathbb Z) \to \mathbb Z/2$ does factor through this natural map $H_k(M, \mathbb Z) \to H_k(M, \mathbb Z) \otimes \mathbb Z/2$ by the universal property for tensor products. (Or simply write $H_k(M, \mathbb Z)$ as a direct sum of copies of $\mathbb Z$ and $\mathbb Z_{p^n}$ and test this explicitly.)