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.

$\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$
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.
