Short exact sequence of functors which splits, but not naturally (revisited, motivated by the Universal Coefficient Theorem)

Question

I already took a look at this and this post, but I still don't really understand what the definition of a naturally split sequence is (which is mentioned in the Universal Coefficient Theorem). I tried to figure it out by myself, but it seems like I found something contradictory in my thoughts. Let me elaborate these:

Let $A,B,C: R\text{-MOD} \to R\text{-MOD}$ be covariant functors such that $ 0 \xrightarrow{} A \xrightarrow{\eta} B \xrightarrow{\psi} C \xrightarrow{} 0 $ forms

1. a natural short exact sequence and
2. a (not necessarily naturally) split exact sequence.

Furthermore, let $f:X\to Y$ be an $R$-module homomorphism. We can take a look at the diagram

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\ua}[1]{\bigg\uparrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\id}{\operatorname{id}} \begin{array}{c} 0 & \ra{} & A(X) & \ra{i_X} & A(X) \oplus C(X) & \ra{p_X} & C(X) & \ra{} & 0 \\ & & \ua{\id_{A(X)}} & & \ua{\theta_X} & & \ua{\id_{C(X)}} \\ 0 & \ra{} & A(X) & \ra{\eta_X} & B(X) & \ra{\psi_X} & C(X) & \ra{} & 0 \\ & & \da{A(f)} & & \da{B(f)} & & \da{C(f)} & \\ 0 & \ras{} & A(Y) & \ras{\eta_Y} & B(Y) & \ras{\psi_Y} & C(Y) & \ras{} & 0 \\ & & \da{\id_{A(Y)}} & & \da{\theta_Y} & & \da{\id_{C(Y)}} \\ 0 & \ra{} & A(Y) & \ra{i_Y} & A(Y) \oplus C(Y) & \ra{p_Y} & C(Y) & \ra{} & 0 \\ \end{array} $$

where $i$ denotes the natural embedding and $p$ denotes the natural projection. The upper and lower part of this diagram commutes because of 2. and the middle part is commutative because of 1. Hence, the whole diagram commutes.

Since this diagram commutes, we get a commutative diagram

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\ua}[1]{\bigg\uparrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\id}{\operatorname{id}} \begin{array}{c} A(X) & \ra{i_X} & A(X) \oplus C(X) & \ra{p_X} & C(X) \\ \da{A(f)} & & \da{\theta_Y \circ B(f) \circ \theta_X^{-1}} & & \da{C(f)} \\ A(Y) & \ra{i_Y} & A(Y) \oplus C(Y) & \ra{p_Y} & C(Y) \\ \end{array}. $$

One can show that such a commutative diagram must lead to $\theta_Y \circ B(f) \circ \theta_X^{-1} = A(f) \oplus C(f)$, so $\theta_Y \circ B(f) = \left(A(f) \oplus C(f)\right) \circ \theta_X$. But this shows that $\theta: B \to A \oplus C$ is a natural isomorphism which is the definition of naturally split, isn't it?

But this seems to be wrong since the Universal Coefficient Theorem states that there is a split exact sequence which does not split naturally.

Could you explain me what's wrong with my reasoning? Thank you.

Answer 1

Expanding on Roland's comments, just because you have two split exact sequences, doesn't mean that morphisms between these two exact sequences must preserve the splitting. In other words, it cannot be shown that $\mu = A(f) \oplus C(f)$. This is something even more basic than category theory, it's mere linear algebra at this point. Roland gave the following counterexample: $$\require{AMScd} \begin{CD} 0 @>>>\mathbb{Z} @>>> \mathbb{Z} \oplus \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\ @. @| @VfVV@|@. \\ 0 @>>>\mathbb{Z} @>>> \mathbb{Z} \oplus \mathbb{Z} @>>>\mathbb{Z}@>>>0 \end{CD}$$ where $f$ is the linear map given by $f(x,y) = (x,x+y)$. The problem with the proof you wrote in the comments is that you are implicitly assuming that $\mu(x,y) = (\mu_1(x), \mu_2(y))$ for some $\mu_1$, $\mu_2$, basically. Knowing the projection of $\mu(x,0)$ on the first space and the projection of $\mu(0,y)$ on the second space isn't enough to determine $\mu(x,y)$. So the argument fails.

Now why do we say that the universal coefficients isomorphism is not split naturally? From Exercise 11 in Chapter 3.1 of Hatcher's book: there exists a space (specifically, a Moore space $X = M(\mathbb{Z}/m\mathbb{Z}, n)$) such that the quotient map $p : X \to X/S^n = S^{n+1}$ induces a trivial map on $\tilde{H}_i(-;\mathbb{Z})$. If the splitting were natural, i.e. if it were functorial wrt $p$, then by the five lemma this would imply that $H^i(p) = 0$ too. However, one can show that this isn't the case.