MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm afraight this might be obviously true or false, but anyway: Let $({\mathcal A},{\mathcal E})$ be a Frobenius category and $X,Y\in{\mathcal A}$. If there exist projective-injective $P,Q\in{\mathcal A}$ such that $X\oplus P\cong Y\oplus Q$ in ${\mathcal A}$, then $X\simeq Y$ in $\underline{\mathcal A}$. Is there converse true?

Considering the case $Y=0$ the condition $X\simeq 0$ in $\underline{\mathcal A}$ is equivalent to the existence of a projective-injective object $P$ and maps $p: P\to X$, $\sigma: X\to P$ such that $p\circ\sigma=\text{id}_X$. Thus if ${\mathcal A}$ is idempotent complete, then this implies that $X$ is a summand of $P$ and thus itself projective-injective. How can one proceed in the general case ?

share|cite|improve this question
up vote 5 down vote accepted

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

This gives us maps $F=\left(\begin{smallmatrix}f\\\a\end{smallmatrix}\right):X\to Y\oplus P$ and $G=\left(\begin{smallmatrix}g&-b\end{smallmatrix}\right):Y\oplus P\to X$ such that $G\circ F=1\_X$. 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 $H:X\oplus Q\xrightarrow{\cong} Y\oplus P$ such that $F=H\circ\iota:X\to Y\oplus P$, with $\iota:X\to X\oplus Q$ the canonical map.

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

If now $j:Q\to X\oplus Q$ and $q:X\oplus Q\to Q$ are the canonical maps, we have $q\circ v\circ u\circ j=-1\_Q$, so that the morphism $\underline{1\_Q}:Q\to Q$ is zero. This implies that in fact $Q\cong 0$ 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 $X\oplus Q\cong Y\oplus P$ 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)

share|cite|improve this answer
Great! Thank you, Mariano! By the way: Did you see my question concerning your paper in…? – Hanno Becker Jan 28 '10 at 6:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.