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?


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

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


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

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.

  • 1
    $\begingroup$ Here's a relevant discussion on math.MO $\endgroup$
    – t.b.
    Commented Apr 18, 2011 at 19:11
  • $\begingroup$ 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.) $\endgroup$
    – user9402
    Commented Apr 18, 2011 at 19:17
  • $\begingroup$ 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 $X$. $\endgroup$
    – t.b.
    Commented Apr 18, 2011 at 19:26
  • $\begingroup$ The $\Sigma$ means the suspension, denoted $S$ on wikipedia. $\endgroup$
    – t.b.
    Commented Apr 18, 2011 at 19:28
  • 4
    $\begingroup$ 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 $X$. $\endgroup$
    – t.b.
    Commented Apr 18, 2011 at 19:53

3 Answers 3


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


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."

  • 1
    $\begingroup$ Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’. $\endgroup$
    – Zhen Lin
    Commented Mar 9, 2013 at 0:32
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Mar 9, 2013 at 0:56

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 $A \to A' \oplus A''$


You must log in to answer this question.