What is the kernel of the map $H_i(X;\mathbb Z)\to H_i(X;\mathbb Z_2)$?

Question

Let $X$ be a topological space. By Universal Coefficient Theorem for Homology we have the exact sequence $$0\to H_i(X;\mathbb Z)\otimes\mathbb Z_2\to H_i(X;\mathbb Z_2)\to \text{Tor}_1(H_{i-1}(X;\mathbb Z),\mathbb Z_2)\to0$$

Also there is a natural map from $H_i(X;\mathbb Z)\to H_i(X;\mathbb Z)\otimes\mathbb Z_2$ given by $g\mapsto g\otimes1$. This gives a map $\phi_i:H_i(X;\mathbb Z)\to H_i(X;\mathbb Z_2)$. I am interested in the object $\ker\phi_i$.

In particular for the case $i=1$, $\text{Tor}_1(H_0(X;\mathbb Z),\mathbb Z_2)=0$ so that $H_1(X;\mathbb Z)\otimes\mathbb Z_2\cong H_1(X;\mathbb Z_2)$. What is $\ker\phi_1$? Does it have some nice structure, for example is it free or trivial?

Thank you.

Answer 1

The short exact sequence $0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to \mathbb{Z}_2 \to 0$ gives rise to a Bockstein exact sequence

$$\cdots \to H_{i+1}(X, \mathbb{Z}_2) \to H_i(X, \mathbb{Z}) \xrightarrow{2} H_i(X, \mathbb{Z}) \to H_i(X, \mathbb{Z}_2) \to H_{i-1}(X, \mathbb{Z}) \to \cdots$$

from which it follows by exactness that the kernel of the map $H_i(X, \mathbb{Z}) \to H_i(X, \mathbb{Z}_2)$ is the image of the multiplication by $2$ map $H_i(X, \mathbb{Z}) \xrightarrow{2} H_i(X, \mathbb{Z})$. This requires no assumptions about anything being finitely generated.

The interesting question is about the image of this map; it is not necessarily surjective, so the intuition that all it does is "reduce $\bmod 2$" is a bit misleading. Instead its image is the kernel of a connecting homomorphism $H_i(X, \mathbb{Z}_2) \to H_{i-1}(X, \mathbb{Z})$, so e.g. is zero if $H_{i-1}(X, \mathbb{Z})$ has no $2$-torsion, but not in general.

Answer 2

If $H_1(X;\mathbb Z)$ is finitely generated, you can calculate pretty explicitly what the map $H_i(X;\mathbb Z) \to H_i(X;\mathbb Z) \otimes \mathbb Z_2$ does--it takes everything to its value modulo 2. This is because $H_1(X;\mathbb Z) \simeq \mathbb Z^k \oplus \bigoplus \mathbb Z_{p^i}^{d_{p^i}}$ by the structure theorem for finitely generated abelian groups. The tensor product of this with $\mathbb Z_2$ turns components either into $\mathbb Z_2$ or 0. Therefore the natural map from this group to its tensor product with $\mathbb Z_2$ takes everything in a $\mathbb Z$ component modulo 2, preserves everything in a $\mathbb Z_2$ component, and sends everything with torsion not equal to 2 to zero. The map $H_i(X;\mathbb Z)\otimes \mathbb Z_2 \to H_i(X;\mathbb Z_2)$ is injective, so composing with it does not change the kernel. So the kernel of the combined map is $(2\mathbb Z)^k \oplus \bigoplus_{p \neq 2} \mathbb Z_{p^i}^{d_{p^i}}\oplus \bigoplus_{i > 1}2\mathbb Z_{2^i}^{d_{2^i}}$.