# What does “splitting naturally” mean in the Universal Coefficients Theorem

The Universal Coefficients Theorem states that

$0\rightarrow H_n(X)\otimes G\rightarrow H_n(X;G)\rightarrow\operatorname{Tor}(H_{n-1}(X),G)\rightarrow 0$

splits, but not naturally. In all the algebraic topology contexts I've come across, "natural" implies commutativity, but I don't see what a "natural split" means. Can someone give me an explicit definition?

Edit:

From what I've read... if I understand this correctly, splitting naturally implies that if

$0\rightarrow A\rightarrow A\oplus C\rightarrow C\rightarrow 0$

and

$0\rightarrow A'\rightarrow A'\oplus C'\rightarrow C'\rightarrow 0$

and given maps $a:A\rightarrow A'$ and $c:C\rightarrow C'$, the map $A\oplus C\rightarrow A'\oplus C'$ has to be the map $(a,c)$ in order for the diagram to commute.

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Here's a relevant discussion on math.MO – t.b. Apr 18 '11 at 19:11
I already did see that discussion, but I still don't understand how the OP's question relates to the Universal Coefficients Theorem. I don't see how those induced maps on homology and homology with coefficients relate to the exact sequence in the UCT. (Also, I don't know what $\Sigma$ means, so that may help me.) – user9402 Apr 18 '11 at 19:17
Ok, so here's the thing. The sequence splits, that is to say that the middle term is the direct sum of the two outer terms. If this splitting were natural in $X$, the map in the middle would have to be the direct sum of the outer maps for all maps $f: X \to Y$. Dylan asks about a map that induces the identity on the two outer terms in the UCT, but fails to be the identity in the middle, so this shows that the splitting isn't natural in $X$. – t.b. Apr 18 '11 at 19:26
The $\Sigma$ means the suspension, denoted $S$ on wikipedia. – t.b. Apr 18 '11 at 19:28
Yes, what you say is correct. However, it becomes clearer if you write $0 \to A(X) \to B(X) \to C(X) \to 0$ and $0 \to A'(X) \to B'(X) \to C'(X) \to 0$ (the sequences depend naturally on $X$). Now they split. That is to say, $B(X) \cong A(X) \oplus C(X)$, for some isomorphism, but this isomorphism does not depend naturally on $X$. – t.b. Apr 18 '11 at 19:53

However, we cannot. The reason is that there are endomorphisms of spaces that induce the identity map on $H_*(X)$ but not on $H_*(X;G)$. This means, in particular, that the functor would give an endomorphism of exact sequences that is the identity on the ends but not in the middle. It is easy to check that no such endomorphism of a split exact sequence exists, and so the functor does not lift.
@ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_{n-1}(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation. – Dylan Wilson Mar 9 '13 at 0:56