# How to see short exact sequence is split

In Hatcher's Algebraic Topology, chapter 3.1 (Universal Coefficient Theorem), on page 193 equation (v), he says that the following short exact sequence splits:

$$0 \rightarrow \text{Coker}(i^*_{n-1}) \rightarrow H^n(C; G) \xrightarrow[]{h} \text{Hom}(H_n(C), G) \rightarrow 0$$

$$C$$ is a chain complex of free abelian groups, $$G$$ is any abelian group, $$H_n$$ is the nth homology group of the chain complex, $$H^n$$ is the nth cohomology group with coefficients in G, $$i^*_{n-1}$$ is the map which restricts homomorphisms on the subgroup of cycles ($$Z_{n-1}$$) in $$C_{n-1}$$ to homomorphisms on the subgroup of boundaries ($$B_{n-1}$$), and h is the map which restricts maps from $$C_n$$ which vanish on $$B_n$$ to maps on $$Z_n$$ which vanish on $$B_n$$, then takes the induced map from $$Z_n / B_n \rightarrow G$$ given by the universal property of the quotient.

My question is just how we can prove that this sequence splits. In the absence of obvious maps that could be used to satisfy the splitting lemma, I would want to show that either the 2nd group is an injective $$\mathbb{Z}$$ module, or that the 4th is projective, but I can't see a way to do that.

• Neither of those (the injectivity or the projectivity) is true in general. Aug 11, 2020 at 4:20

"In the absence of obvious maps that could be used to satisfy the splitting lemma...."

But we are not absent these. (And as mentioned in comment your approach does not work.)

Suppose $$\psi \in \text{Hom}(H_n(C), G)$$. We want to construct a cohomology class $$E(\psi)$$ so that $$(hE)(\psi) = \psi$$. How can we do this? We need $$\psi$$ to now take as input chains instead of homology classes. The picture will be: we will need a way to take an arbitrary chain $$c$$ and turn it into a cycle $$c'$$, which we then feed $$\psi$$. That is, we need to somehow "kill off" the part of $$c$$ with boundary. How do we make sense of this?

Note that $$C_n/Z_n \to B_{n-1}$$ is an isomorphism. The latter, being a subgroup of a a free abelian group, is free abelian; we may thus choose a lift $$F: B_{n-1} \to C_n$$, so that $$(\partial F)(c) = c$$. We have constructed a (noncanonical!) homomorphism taking every boundary to a chain which gives it as boundary, and thus a splitting $$C_n \cong Z_n \oplus F(B_{n-1})$$. Write $$p: C_n \to Z_n$$ for the map $$p(c) = c - F(\partial b)$$. The map $$p$$ is a group homomorphism, which sends every chain to a cycle, and is the identity on cycles.

This is how you construct $$E(\psi)$$: by formula, using $$p$$. Set $$E(\psi)(c) = \psi([pc]),$$ where $$[pc]$$ is the homology class of the cycle $$pc$$.

Then $$\delta(E(\psi))(c) = E(\psi)(\partial c) = \psi([p \partial c]) = \psi([\partial c]) = \psi([0]) = 0.$$ So $$E(\psi)$$ is a cocycle. This provides a construction of a homomorphism $$E: \text{Hom}(H_n(C), G) \to Z^n(C;G);$$ passing to cohomology I claim that the resulting map (still called $$E$$ by abuse of notation) is your section.

For $$(hE)(\psi)([c]) = E(\psi)([c]) = \psi([pc]) = \psi([c]),$$ which is precisely the claim that $$hE(\psi) = \psi$$.