What does "is natural in $A$" mean in this context? While reading Bredon's Topology and Geometry, I've come across the following claim:

Naturality in $A$ of the sequence defining $\text{Ext}(A,G)$ shows that $\text{Ext}(A,G)$ is a contravariant functor of $A$.

I am confused. I know what is a natural transformation, and I understand, for example, that the map induced by the boundary map in singular homology is a natural transformation, because I know the category involved and the respective functors. However, I can't transfer my understanding of "natural transformation" to the above context. "Naturality in $A$" is a phrase which does not have a meaning to me, and the text doesn't make it clear. 
So, my question is, what is Bredon meaning by this?
"The sequence defining $\text{Ext}(A,G)$" is the exact sequence below:
$$0 \to \text{Hom}(A,G) \to \text{Hom}(A,I) \to \text{Hom}(A,J) \to \text{Ext}(A,G) \to 0,$$
where $0 \to G \to I \to J \to 0$ is a injective resolution of $G$.
UPDATED: I have started a bounty on this question, and not accepted any of its answers, because I am not sure that "natural" is only being used as an intuitive meaning. Bredon uses the same term later when talking about the splitting on the Universal Coefficients Theorem and goes on to say why the split with respect to something cannot be natural.
 A: The claim of naturality is usually done in the context of a morphism: one says that some morphism from an object constucted from $A$ to some other object constructed from $A$ is natural, and then it means precisely that both the domain and the codomain of the morphism depend functorially on $A$ —that is,they are the result of evaluating functors at $A$— and that the morphism is in fact the component at $A$ of a natural transformation.
In your particular case, you have the sequence $$0 \to \text{Hom}(A,G) \to \text{Hom}(A,I) \to \text{Hom}(A,J) \to \text{Ext}(A,G) \to 0$$ Each of the six abelian groups appearing there (incuding the zeroes) is the result of evaluating a certain functor at $A$, and the claim is that each of the five morphisms appearing in the sequence is natural in the sense I explained above.
Even more precisely, since $\operatorname{Ext}$ is not, at that point, define to be a functor (because you are still trying to define its action on maps), the claim is really that the sequence $$0 \to \text{Hom}(A,G) \to \text{Hom}(A,I) \to \text{Hom}(A,J)$$ is natural in $A$, so that this implies that the cokernel of the last arrow, which is defined to be $Ext$, is also a functor.
A: What the author might be saying is the following. He has defined the Ext functor as a cokernel of a map, and it is a general fact that if you have a commutative diagram of exact rows 
A' ----> A ---> A'' ---> 0 
|        |      
B' ----> B ---> B'' ---> 0

There is a unique arrow $A''\longrightarrow B''$ making the diagram commutative so that "taking cokernels" is, in some way, functorial. If you now have two abelian groups $A_0,B_0$ (and a universal choice of injective resolutions), you can construct both exact sequences and use that hom is a functor to obtain a partial diagram as above. You then obtain a unique map that yields Ext a functor, too. 
A: In order to define a functor $F : \textbf{A} \to \textbf{B}$, as Bredon does, without defining $F(f)$ for arrows $f$ of $\textbf{A}$, there had better be an obvious choice. That the sequence is “natural” means that there is an obvious choice for the arrows in the new exact sequence that you form with the $\text{Hom}$’s, and then the arrows between the $\text{Ext}$’s are predetermined, since they are obtained by taking homology. Thus, saying that the sequence is natural in $A$ means, perversely, that “the implicit maps that Bredon is not specifying should be taken to be the natural ones.”
Then what does “in $A$” mean? Bredon intends to specify that, specifically, the map which sends $A$ to that given exact sequence is natural for each injective resolution of $G$. This is not a mathematical statement—it is merely for the sake of precision of language. It is the same usage as when you say that the map $f : \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y) = \lfloor x\rfloor y$ is continuous in $y$ but not in $x$.
