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I want to assure me if I understand the part of the exactness axiom that "$\delta$ is natural" corretly and if not, then my question is: what does it mean?

The setting is the following definition:

A homology theory is a sequence $\{h_n\}_{n\in\mathbb{Z}}$ of covariant functors from an admissible category $\mathfrak{C}$ of $C^*$-algebras to abelian groups, which satisfies the following axioms:

  1. homotopy axiom

  2. Exactness axiom: Let $$0\to J \xrightarrow{\text{i}} A\xrightarrow{\text{j}} B\to 0$$ be a short exact sequence of $C^*$-algebras in $\mathfrak{C}$. Then there is a map $\delta_*:h_*(B)\to h_{*-1}(J)$ and a long exact sequence $$...\to h_n(J)\xrightarrow{i_n} h_n(A) \xrightarrow{j_n} h_n(B) \xrightarrow{\delta_n} h_{n-1}(J)\to ...$$ The map $\delta_*$ is natural with respect to morphisms of short exact sequences.

What does "The map $\delta_*$ is natural with respect to morphisms of short exact sequences" mean, is the following correct?:

Given a commutative diagram in $\mathfrak{C}$:

$$\require{AMScd} \begin{CD} 0 @> >> J @>i>> A @>j>> B @> >> 0 \\ @VV 0V @VV p_1 V @VV p_2 V @VV p_3 V @VV 0V \\ 0 @> >> J' @>i'>> A' @>j' >> B' @> >> 0 \\ \end{CD} $$

with exact rows. Then we have a commutative diagram with long exact rows

$$\require{AMScd} \begin{CD} ... h_n(J) @>h_n(i) >> h_n(A) @>h_n(j)>> h_n(B) @>\delta_n>> h_{n-1}(J) @> >>.. \\ @VV h_n(p_1) V @VV h_n(p_2) V @VV h_n(p_3) V @VV h_{n-1}(p_1) V \\ ... h_n(J') @>h_n(i') >> h_n(A') @>h_n(j')>> h_n(B') @>\delta_n>> h_{n-1}(J') @> >>.. \\ \end{CD} $$ Best.

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Yep, that's exactly what it means.

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