1
$\begingroup$

$\require{AMScd}$ I'm reading through the proof of the Universal Coefficient Theorem (for homology) given in Massey's Singular Homology Theory, in which he claims that the split SES $$0 \to H_n(K) \otimes G \overset{\alpha}{\to} H_n(K \otimes G) \overset{\beta}{\to} \operatorname{Tor}(H_{n-1}(K),G) \to 0$$ is natural with respect to coefficient homomorphisms but not with respect to chain maps. In the proof the author writes (after showing that the sequence splits), "Using this procedure, it is easy to prove that the direct sum decomposition is natural vis-a-vis coefficient homomorphisms." The procedure he refers to is the construction of a left inverse $\pi$ for the map $\alpha$ above, from which it follows that the sequence splits.

What does it mean for a direct sum decomposition to be natural? I've only seen the term used in reference to maps. Can I interpret this to mean that the map $\pi$ is natural in the ordinary sense, that is, the following diagram commutes for any homomorphism $h : G_1 \to G_2$? $$\begin{CD} H_n(K \otimes G_1) @>{\pi_1}>> H_n(K) \otimes G_1 \\ @V{h_\sharp}VV @V{1 \otimes h}VV \\ H_n(K \otimes G_2) @>{\pi_2}>> H_n(K) \otimes G_2 \end{CD}$$

My other guess for the meaning of the term is that if we write $H_n(K \otimes G_1) = B_1 \oplus B_2$ and $H_n(K \otimes G_2) = C_1 \oplus C_2$ as determined canonically by the map $\pi$, then $h_\sharp(B_1) \subset C_1$ and $h_\sharp(B_2) \subset C_2$. Are either of these guesses correct? Are they equivalent? What if we proved the sequence split by finding a right inverse $s$ for $\beta$ instead of a left inverse for $\alpha$? Would naturality then take the form of a commutative diagram with $s$?

EDIT: It might be relevant that the maps $\alpha$ and $\beta$ have already been shown to be natural with respect to coefficient homomorphisms.

$\endgroup$
2
  • $\begingroup$ Do you know the concept of a natural transformation between two functors? $\endgroup$
    – Paul Frost
    Nov 15, 2020 at 18:42
  • $\begingroup$ @PaulFrost A little bit, but I don't want to get too abstract here at the moment. $\endgroup$
    – Nick A.
    Nov 15, 2020 at 18:51

1 Answer 1

2
$\begingroup$

$\require{AMScd}$

Your interpretation (based on the diagram in your question) is correct. Also your other guess is correct.

For a short exact sequence of abelian groups $$0 \to A \overset{\alpha}{\to} B \overset{\beta}{\to} C\to 0$$ we have canonical bijections between the following three sets:

  1. The set of "left splitting homomorphisms" $\pi : B \to A$ (characterized by $\pi \circ \alpha = id_A$).

  2. The set of isomorphisms $\phi: B \to A \oplus C$.

  3. The set of "right splitting homomorphisms" $\iota : C \to B$ (characterized by $\beta \circ \iota = id_C$).

Note that these sets are non-empty precisely when the sequence splits. Let us give a sketch without going into details of proofs.

(1) Given $\pi$, define $\phi_\pi : B \to A \oplus C$ by $\phi_\pi(b) = (\pi(b),\beta(b))$ and show that is an isomorphism.

(2) Given $\iota$, define $\psi_\iota : A \oplus C \to B$ by $\psi_\iota(a,c) = \alpha(a) + \iota(c)$ and show that it is an isomorphism. Then define $\phi_\iota = \psi_\iota^{-1}$.

(3) Given $\phi$, define $\pi_\phi = p_A \circ \phi$, where $p_A : A \oplus C \to A$ denotes projection.

(4) Given $\phi$, define $\iota_\phi = \phi^{-1}\circ i_C$, where $i_C : C \to A \oplus C$ is given by $i_C(c) = (0,c)$.

Let $h_\sharp : H_n(K \otimes G_1) \to H_n(K \otimes G_2)$ and $h_* : \operatorname{Tor}(H_{n-1}(K),G_1) \to \operatorname{Tor}(H_{n-1}(K),G_2)$ be the induced homomorphisms. The commutativity of your diagram is equivalent to the commutativity of $$\begin{CD} H_n(K \otimes G_1) @>{\phi_1}>> H_n(K) \otimes G_1 \oplus \operatorname{Tor}(H_{n-1}(K),G_1)\\ @V{h_\sharp}VV @V{1 \otimes h \oplus h_*}VV \\ H_n(K \otimes G_2) @>{\phi_2}>> H_n(K) \otimes G_2 \oplus \operatorname{Tor}(H_{n-1}(K),G_2) \end{CD}$$ In fact, if your diagram commutes, then $$((1 \otimes h) \oplus h_*)(\phi_1(b)) = ((1 \otimes h) \oplus h_*)(\pi_1(b),\beta_1(b)) = ((1 \otimes h)(\pi_1(b)), h_*(\beta_1(b))) \\ = (\pi_2( h_\sharp(b)),\beta_2(h_\sharp(b))) = \phi_2(h_\sharp(b)) .$$ Thus the above diagram commutes. Conversely, if the above diagram commutes, then $$(1 \otimes h)(\pi_1(b)) = (1 \otimes h)((p_1 \circ \phi_1)(b)) = p_2((1 \otimes h) \oplus h_*)(\phi_1(b)) = p_2 \phi_2(h_\sharp(b)) \\ = \pi_2(h_\sharp(b)) .$$ Thus your diagram commutes.

For the diagram with the right splitting homomorphisms it can be shown similarly that its commutavitity is equivalent to the commutativity of the above diagram.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .