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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.

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    $\begingroup$ How do you show that the last commutative diagram leads to $\theta_Y\circ B(f)\circ\theta^{-1}_X=A(f)\oplus C(f)$ ? Especially, if you take $c\in C(X)$, then why $\theta_Y\circ B(f)\circ \theta_X^{-1}(c)=(0,c')$ ? $\endgroup$
    – Roland
    Apr 3, 2018 at 13:07
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    $\begingroup$ Think about $$\require{AMScd} \begin{CD} 0@>>>\mathbb{Z}@>>>\mathbb{Z\oplus Z}@>>>\mathbb{Z}@>>>0\\ @.@|@V\begin{pmatrix}1&1\\0&1\end{pmatrix}VV@|@.\\ 0@>>>\mathbb{Z}@>>>\mathbb{Z\oplus Z}@>>>\mathbb{Z}@>>>0 \end{CD}$$ where the horizontal exact sequences are the injection into the first component followed by the projection onto the second component. $\endgroup$
    – Roland
    Apr 3, 2018 at 13:12
  • $\begingroup$ Regarding your first comment: Let's set $\mu:=\theta_Y \circ B(f) \circ \theta_X^{-1}$. Since the diagramm commutes, we have $\mu \circ i_X = i_Y \circ A(f)$ and $p_Y \circ \mu = C(f) \circ p_X$. Let $(a,c) \in A \oplus C$ and define $(x,y):=\mu(a,c)$. Now I would like to show $x = A(f)(a)$ and $y=C(f)(c)$. - First, $ y = p_Y(x,y) = p_2 \circ \mu(a,c) = C(f) \circ p_X(a,c) = C(f)(c).$ Especially $c=0 \Rightarrow y=0$. - Second, $(x,0) = \mu(a,0) = \mu \circ i_X(a) = i_Y \circ A(f)(a) = (A(f)(a),0)$, so $x = A(f)(a)$. If there is something wrong with my solution, please let me know. $\endgroup$
    – Diglett
    Apr 3, 2018 at 13:23
  • $\begingroup$ Regarding your second comment: Could you explain me, what your example has to do with functors? Could it be that it has something to do with the homology of the torus? I'm not sure but I think I saw something similar like this before... $\endgroup$
    – Diglett
    Apr 3, 2018 at 13:29
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    $\begingroup$ You only have shown that $y=C(f)(c)$ and $(x,0)=(A(f)(a),0)$. You didn't prove that $(x,y)=(A(f)(a),C(f)(c)$. Yes $x=A(f)(a)$ if $c=0$, but what if $c\neq 0$ as I suggested ? Take the diagram I wrote above : the middle vertical map is $(a,c)\mapsto (a+c,c)$. Clearly it commutes, but the map is not of the form $(A(a),C(c))$. $\endgroup$
    – Roland
    Apr 3, 2018 at 13:30

1 Answer 1

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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.

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  • $\begingroup$ Knowing $\mu(x,0)$ and $\mu(0,y)$ is enough to know $\mu(x,y)$ by linearity, of course-you probably meant something like knowing the first projection of $\mu(x,0)$ and the second projection of $\mu(0,y)$ isn't enough to determine $\mu(x,y)$, no? $\endgroup$ Apr 3, 2018 at 15:28
  • $\begingroup$ @KevinCarlson You're right, thanks! $\endgroup$ Apr 3, 2018 at 15:44
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    $\begingroup$ @NajibIdrissi isn't it important to keep $f(x,y)=(x+y,y)$ as suggested by Roland and not $f(x,y)=(x,x+y)$ in order for the diagram to commute? $\endgroup$
    – roi_saumon
    Jun 3, 2019 at 13:44

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