Correspondence between Ext group and extensions (from Weibel's book)

I am trying to understand the proof of Theorem 3.4.3 from Weibel's book Introduction to homological algebra.

The statement is the following.

Let $R$ be a ring. Given $R$-modules $A$ and $B$, an extension $\xi$ of $A$ by $B$ is an exact sequence $0\to B\to X\to A\to0$. Then the connecting homomorphism $$\delta:\text{Hom}(A,A)\to\text{Ext}^1(A,B) \tag{1}$$ induces the 1-1 correspondence $\theta$ between equivalence classes of extensions of $A$ by $B$ and $\text{Ext}^1(A,B)$, which is given by $\theta:\xi\mapsto\delta(\text{id}_A)$.

In the beginning of the proof it is written the following

Fix an exact sequence $0\stackrel{j}\to M\to P\to A\to0$ with $P$ projective. Applying $\text{Hom}(-,B)$ yields an exact sequence $$\text{Hom}(P,B)\to\text{Hom}(M,B)\stackrel{\delta}\to\text{Ext}^1(A, B)\to0.\tag{2}$$

I don't understand why the connecting homomorphism in the sequence (2) is denoted by the same letter as in (1). Are they really the same?

$\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Ext}{Ext}$I must have a different edition than you do, here's the statement of Theorem 3.4.3 that I read:

Theorem 3.4.3. Given two $R$-modules $A$ and $B$, the mapping $\Theta : \xi \mapsto \partial(\operatorname{id}_A)$ establishes a 1-1 correspondence $$\{\text{equivalence classes of extensions of } A \text{ by } B \} \leftrightarrow \Ext^1(A,B)$$ in which the split extension corresponds to the element $0 \in \Ext^1(A,B)$.

Proof Fix an exact sequence $0 \to M \xrightarrow{j} P \to A \to 0$ with $P$ projective. Applying $\Hom(-, B)$ yields an exact sequence $$\Hom(P, B) \to \Hom(M,B) \xrightarrow{\partial} \Ext^1(A,B) \to 0.$$ [...]

Now the mapping $\Theta$ is actually explained above the statement of the theorem. $\xi$ is an exact sequence, of the form $0 \to B \to X \to A \to 0$. If you apply the theory of derived functors to compute $\mathbb{R}\Hom(A,-)$, you get a long exact sequence: $$0 \to \Hom(A,B) \to \Hom(A,X) \to \Hom(A,A) \xrightarrow{\partial} \Ext^1(A,B) \to \Ext^1(A,X) \to \dots$$ where $\partial$ is the connecting morphism, for this particular exact sequence $\xi$. And the image of $\operatorname{id}_A$ under this $\partial$ is $\Theta(\xi)$.

But in the proof, Weibel takes also another exact sequence, $0 \to M \xrightarrow{j} P \to A \to 0$, and applies again derived functors, but on the other side of the $\Hom$! (So you're computing $\mathbb{R}\Hom(-, B)$.) You get a long exact sequence: $$0 \to \Hom(A,B) \to \Hom(P,B) \to \Hom(M,B) \xrightarrow{\partial} \Ext^1(A,B) \to \Ext^1(P,B) \to \dots$$ there is also a connecting morphism. We can call it $\partial$ again, but it's not the same $\partial$ as before, it's just that connecting homomorphisms are called $\partial$ in general (or $\delta$, or some other variant on the letter "d"). In general the context is enough to determine what $\partial$ we are talking about.

It's like how the differential of a chain complex is usually simply called "$d$", even if there are multiple chain complexes that we are considering... You could call the first connecting morphism $\partial_\xi^A$ or something like that, and find an appropriate notation for the second one, but carrying all these subscripts and superscripts around becomes annoying very quickly. (And don't forget that this is only the first connecting morphism, there are higher Ext groups! So it should be something like $\partial_{\xi, A}^1$...)

Bonus: In the proof, $P$ is projective, which is why $\Ext^1(P,B) = 0$ and why Weibel omitted it in the long exact sequence.