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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](http://mathoverflow.net/questions/59938/examples-for-non-naturality-of-universal-coefficients-theorem) on math.MO2011-04-18
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    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.)2011-04-18
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    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 $X$.2011-04-18
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    The $\Sigma$ means the suspension, denoted $S$ on [wikipedia](http://en.wikipedia.org/wiki/Suspension_(topology)).2011-04-18
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    I edited my question. Am I correct? If so, the rest of it makes sense.2011-04-18
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    **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 $X$.2011-04-18

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