ct.category theory - When do two objects become isomorphic in the stable category?

Suppose $X$ and $Y$ are stably isomorphic, so that there exist a morphism $f:Xto Y$ whose image $underline f:Xto Y$ in the stable category is an isomorphism. Then $underline f$ has an inverse: there exists $g:Yto X$ such that $underline gcircunderline f=1_X$ and $underline fcircunderline g=1_Y$, and this means in particular that there is a projective-injective $P$ and maps $r:Xto P$ and $s:Pto X$ such that $gcirc f-1_X=scirc r$.

This gives us maps $F=left(begin{smallmatrix}f\aend{smallmatrix}right):Xto Yoplus P$ and $G=left(begin{smallmatrix}g&-bend{smallmatrix}right):Yoplus Pto X$ such that $Gcirc F=1_X$. If now we assume that $mathcal A$ has all its idempotents split, then we can conclude that there is a $Q$ such that there is an isomorphism $H:Xoplus Qxrightarrow{cong} Yoplus P$ such that $F=Hcirciota:Xto Yoplus P$, with $iota:Xto Xoplus Q$ the canonical map.

Notice that $underline F$ and $underline H$ are isomorphisms in $underline{mathcal A}$, so that also $underlineiota$ is an isomorphism there. If $p:Xoplus Qto X$ is the projection, then $underline pcircunderlineiota=1_X$ in $underline{mathcal A}$, so in fact $(underlineiota)^{-1}=underline p$, and in consequence the composition $iotacirc p:Xoplus Qto Xoplus Q$ is the identity of $Xoplus Q$ in $underline{mathcal A}$. In other words, there exists a projective-injective $R$ and morphisms $u:Xoplus Qto R$ and $v:Rto Xoplus Q$ such that $pcirciota-1_{Xoplus Q}=vcirc u$.

If now $j:Qto Xoplus Q$ and $q:Xoplus Qto Q$ are the canonical maps, we have $qcirc vcirc ucirc j=-1_Q$, so that the morphism $underline{1_Q}:Qto Q$ is zero. This implies that in fact $Qcong 0$ in $underline{mathcal A}$. By what you showed in your question, this implies that $Q$ is a projective-injective in $mathcal A$.

All in all, we have shown that there exists projective-injectives $P$ and $Q$ such that $Xoplus Qcong Yoplus P$ in $mathcal A$, as you wanted.

(I do not think your question will have a positive answer when $mathcal A$ does not have all its idempotents split... I do not have a counterexample, though)

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