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.

It doesn't seem like anyone has said it the following way, yet, which might help clear things up:
What does it mean for an exact sequence to split? It means, among other things, that you have given a section of the last map in the sequence. Now, we have a functor from spaces to short exact sequences and we might ask whether or not we can lift this to a functor from spaces to short exact sequences together with a section (i.e. split short exact sequences.)
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.
So perhaps a better thing to say is that the sequence does not split "functorially."
