homology theory for $C^*$-algebras, map is natural wrt morphisms of short ecact sequences 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:


*

*homotopy axiom

*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.

Yep, that's exactly what it means.
