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.
 A: $\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$
Note:  The splitting is actually natural in $G$ contrary to what is written in many texts.  What she is referring to is the following:$\newcommand{\z}{\mathbb{Z}}$. Lets fix $G=\z/2$. 
Thm:  There is no possibility of a natural transformation from the functor $H_*(-,\z/2)$ to $(H_{*}(-,\z))^* \oplus Ext(H_{*-1}(-,\z),\z/2)$. 
Proof:  Suppose if possible that for all spaces $X,Y$ and every continuous map $X \to Y$, there is an induced a commuting square 
$ \begin{CD}
H_*(X,\z/2) @>\cong>> Tor(H_{*-1}(X),\z/2) \oplus H_*(X) \otimes \z/2 \\
@VVV @VVV\\
H_*(Y,\z/2) @>\cong>> Tor(H_{*-1}(Y),\z/2) \oplus H_*(Y) \otimes \z/2 \\
\end{CD}$
.  
Then there would have to be a commuting square induced by $\RP^2 \to \RP^2/\RP^1 \to S^2$:
$ \begin{CD}
H_2(\RP^2,\z/2) @>\cong>> Tor(H_{1}(\RP^2),\z/2) \oplus H_2(\RP^2) \otimes \z/2 \\
@| @V0VV\\
H_2(S^2,\z/2) @>\cong>> Tor(H_{2-1}(S^2),\z/2) \oplus H_2(S^2) \otimes \z/2 \\
\end{CD}$.  Contradiction since the 0 map would need to be an isomorphism.
Note: One could not have expected there to be such a natural transformation because $H_*(X)$ splits into objects of different homogenous degree, while the induced maps of topological maps map elements of the same degree onto elements of the same degree.
Note 2: The diagram $$ \begin{CD}
H^2(\RP^2,\z/2) @>\cong>> Ext(H_{1}(\RP^2),\z/2) \oplus H_2(\RP^2)^*  \\
@| @V0VV\\
H^2(S^2,\z/2) @>\cong>> Ext(H_{1}(S^2),\z/2) \oplus H_2(S^2)^*  \\
\end{CD}$$ shows that the exact sequence of the cohomological universal coefficient theorem does not split naturally with respect to the topological space.
A: 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."
A: Here is the definition for a natural splitting.
Given an abelian category C, for each element A in C, there is a short exact splitting sequence $$0 \to A' \to A \cong A' \oplus A'' \to A'' \to 0$$, such that $j_A \circ i_A = id_{A'}$, $\pi_A \circ \gamma_{A} =id_{A''} $
Given $f: A \to B$, define a functor T from C to C, such that
T(A) is $A' \oplus A''$, $T(f): A' \oplus A'' \to B' \oplus B''$, such that $$T(f) = f' \oplus f''$$
$$f' : A' \to B', f' = j_B \circ f \circ ia$$
$$f'':A'' \to B'', f'' = \pi_B \circ f \circ \gamma_{A} $$
Go check that T is indeed a functor.
Then natural splitting means that there is a natural isomorphism from identity functor to this T. The components of the natural isomorphism are given by the isomorphism from $A \to A' \oplus A''$
