The Stacks project

In some references, a perfect complex is defined as an object (of $D(R)$ or of the category of cochain complexes of $R$-modules) to which a bounded complex of finite projective $R$-modules is quasi-isomorphic. Is this definition equivalent to the above definition?

Hi! I think your question is a semantic one? Namely, the definition literally states what you say? So maybe you mean that it is often applied to, for example, complexes of $R$-modules without viewing them as objects of $D(R)$?

Or are you complaining that, given objects $K$ and $L$ of $D(R)$, we shouldn't say "$K$ and $L$ are quasi-isomorphic" but we should say "$K$ and $L$ are isomorphic"?

In both cases my answer to the question would be the same, namely, that objects of $D(R)$ are the same thing as complexes of $R$-modules (by our construction of the derived category). Hence in both cases the terminology makes sense, it is what we want to say, and we will use it later in the Stacks project. OK?

Probably I didn't state my question so well. In some references, a complex $X^\bullet$ is called perfect if there exists a bounded complex of finite projective modules $P^\bullet$ and a quasi-isomorphism $f: P^\bullet \to X^\bullet$. On the other hand, the definition in Stacks project says that $X^\bullet$ is perfect if there exists a bounded complex of finite projective modules $P^\bullet$ and a quasi-isomorphism $f:X^\bullet \to P^\bullet$ (here the arrow is reversed). However quasi-isomorphism is not a symmetric relation. But still are these two definitions of perfect complex equivalent?

Actually, no that is not what it says. The phrase "quasi-isomorphic" is supposed to mean that they are isomorphic as objects in $D(R)$. But since we never formally defined "quasi-isomorphic" there was no way for you to know that. Hmm... The solution is either to stop using the phrase "quasi-isomorphic" (it gets used 92 times), or to make a definition. For the moment I have just added a line to the chapter on Derived Categories saying what the terminology means, but I have stopped short of actually putting it in a definition, because now I am no longer so sure that it is standard terminology... Argh!

Anyway, thanks very much for pointing this out! You can find the change here. Note: By accident I placed committed this change to my development branch and so it will take some time because it gets merged into the actual project...

In fact, quasi-isomorphism is defined in tag 0115, 010Z for cochain and chain complexes. Probably the ''$P\xrightarrow{qis} X$ definition'' of perfect complexes is the right one as it makes $\mathbb Z/2 \mathbb Z$ a perfect complex. The ''$X\xrightarrow{qis} P$ definition'' does not seem to make $\mathbb Z/2\mathbb Z$ a perfect complex.

Yes, what you say is correct. To spell out what I said, if $K$ and $L$ are quasi-isomorphic this means there exists a third complex $M$ and maps $K \to M$ and $L \to M$ which are both quasi-isomorphisms. Equivalently there is a complex $N$ and quasi-isomorphisms $N \to K$ and $N \to L$. In any case the condition is symmetric in $K$ and $L$.

I think that adding this change is somewhat confusing because then it is not clear what does quasi-isomorphism mean in the definition of perfect complexes as there is already a definition of quasi-isomorphism in the book. If tag 0115 is left unchanged, then probably tag 0657 would require some modification or additional remark.

In the above comment, I guess that the quasi-isomorphisms $K\to M, L \to M, N\to K, N\to L$ are maps as given in this definition. If this is so, then is the existence of $N$ with quasi-isomorphisms (as here) $N\to K, N\to L$ (if I am not mistaken then this is precisely the statement $K\simeq L$ in $D(R)$) equivalent to the existence of $M$ with quasi-isomorphisms $K\to M, L\to M$? Could you add a word about it?

For now I am going to leave it like this.

The definition of arrows in the derived category $D(\mathcal{A})$ can be found in the chapter on derived categories. You start with the homotopy category $K(\mathcal{A})$ of complexes of $\mathcal{A}$ and then you invert the set of quasi-isomorphisms which is a multiplicative set compatible with the triangulated structure on $K(\mathcal{A})$. See 13.11.

Inverting the arrows is why we may not always have a quasi-isomorphism between isomorphic objects of the derived category.

In comment #617 on the last line I should have added "in a given direction" after quasi-isomorphism.

Thank you for your reply.