Let $\mathcal{A}$ be an abelian category with enough projectives and let $\underline{\mathcal{A}}$ be its stable category with loop functor $\Omega: \underline{\mathcal{A}} \to \underline{\mathcal{A}}$ (for definitions see remark 3 below). Define $\Omega^0 := id_{\underline{\mathcal{A}}}$ and $\Omega^{n+1} := \Omega \circ \Omega^n$ and denote the hom-groups of $\underline{\mathcal{A}}$ by $[-,-]$. Then, if

$$ 0 \to B^' \to B \to B^{''} \to 0$$

is a short exact sequence in $\mathcal{A}$, there is a sequence (starting with $n=0$ from the left):

$$ ... \to [\Omega^n(A),B^'] \to > [\Omega^n(A),B] \to > [\Omega^n(A),B^{''}] \to > [\Omega^{n+1}(A),B^'] \to ...$$

and the composition of two consecutive maps is zero.

Is this sequence exact, or equivalently, is $[\Omega^n(A),-]_{n > \ge 0}$ a delta-functor ?

**Remark 1:** I know that the following is a long exact sequence (ending with $n=0$ at the right):

$$ ... \to [A, \Omega^n(B^')] \to [A, \Omega^n(B)] \to [A, \Omega^n(B^{''})], \to [A, \Omega^{n-1}(B^')] \to ...$$ Therefore is guess that the sequence above is also exact.

**Remark 2:** There is a natural epimorphism
$$Ext_\mathcal{A}^n(A,B) \to [\Omega^n(A),B].$$
If $\mathcal{A}$ satisfies $Ext_\mathcal{A}^n(-,P)=0$ for all projectives $P$ and all $n > 0$ then the epimorphism is actually an isomorphism (for $n >0$) and the
exactness of the sequence follows from the long exact $Ext$-sequence.

**Remark 3:** The stable category $\underline{\mathcal{A}}$ is defined as follows: It has the same objects as $\mathcal{A}$ and the hom's are given by
$$[A,B] := Hom_{\underline{\mathcal{A}}}(A,B) := Hom_\mathcal{A}(A,B) / P(A,B)$$
where $P(A,B)$ is the subgroup of homomorphisms that factor through a projective. The endo-functor $\Omega$ is obtained by taking fixed projective presentations in $\mathcal{A}$:
$$0 \to \Omega(A) \to P \to A \to 0.$$