An alternative definition of pseudo-coherent complex

Question

Let $(X,\mathcal{O}_X)$ be a scheme or a general ringed space. First recall that a complex of $\mathcal{O}_X)$-modules $\mathcal{E}^{\bullet}$ is called strictly perfect if $\mathcal{E}^{\bullet}$ is a two-side bounded complex of finitely generated locally free $\mathcal{O}_X)$-modules.

Then we have the following definition of pseudo-coherent complex of $\mathcal{O}_X)$-modules:

Let $(X, \mathcal{O}_X)$ be a ringed space. Let $\mathcal{E}^{\bullet}$ be a complex of $\mathcal{O}_X$-modules. Let $m \in \mathbf{Z}$.

1. We say $\mathcal{E}^\bullet$ is $m\textit{-pseudo-coherent}$ if there exists an open covering $X = \bigcup U_i$ and for each $i$ a morphism of complexes $\alpha_i : \mathcal{E}_i^\bullet \to \mathcal{E}^\bullet|_{U_i}$ where $\mathcal{E}_i$ is strictly perfect on $U_i$ and $H^j(\alpha_i)$ is an isomorphism for $j > m$ and $H^m(\alpha_i)$ is surjective.

2. We say $\mathcal{E}^\bullet$ is $\textit{pseudo-coherent}$ if it is $m$-pseudo-coherent for all $m$.

See http://stacks.math.columbia.edu/tag/08CA

If $\mathcal{E}^\bullet$ is pseudo-coherent, then locally the cohomology $H^{\bullet}(\mathcal{E})$ is bounded above but not bounded below.

It seems that even if $\mathcal{E}^\bullet$ is pseudo-coherent, for different $m$ we may choose different cover $\{U_i\}$ and different $\mathcal{E}_i^\bullet$.

$\textbf{My question}$ is: is the definition of pseudo-coherent complex equivalent to the following:

We say $\mathcal{E}^\bullet$ is $\textit{pseudo-coherent}$ if there exists an open covering $X = \bigcup U_i$ and for each $i$ a morphism of complexes $\alpha_i : \mathcal{E}_i^\bullet \to \mathcal{E}^\bullet|_{U_i}$ where $\mathcal{E}_i$ is a $\textit{bounded above}$ complex of finitely generated locally free sheaves on $U_i$?

Answer 1

No, this is true under noetherian hypothesis. See the relevant exposé by Illusie in SGA 6, It is related to the phenomenon that not every finitely presented module is coherent whenever the ring is not coherent itself.

Answer 2

Yes, provided that $X$ is a scheme (or, for instance, an algebraic stack). In particular, we need not assume that $X$ is locally noetherian.

Let $(X, \mathcal{O}_X)$ be a locally ringed space (or a locally ringed topos). Let $\mathcal{E}^\bullet$ be a complex of $\mathcal{O}_X$-modules and let $U_\alpha$ be a covering of $X$. Then it is clear that $\mathcal{E}^\bullet$ is pseudo-coherent if and only if $\mathcal{E}^\bullet|_{U_\alpha}$ is pseudo-coherent for all $\alpha$. Moreover, it is clear that a bounded above complex of finite locally free $\mathcal{O}_X$-modules is pseudo-coherent (the map from the brutal truncation at $m$ gives a perfect complex witnessing that it is $m$-pseudo-coherent). In particular, your alternative definition (stated at the end of the question) is stronger than the usual definition.

Now assume that $X$ is an affine scheme. Then a complex of $\mathcal{O}_X$-modules is pseudo-coherent if and only if its is quasi-isomorphic to a bounded above complex of finite free $\mathcal{O}_X$-modules. Indeed, if $\mathcal{E}^\bullet$ is pseudo-coherent, then it has quasi-coherent cohomology, so it is quasi-isomorphic to a complex of quasi-coherent modules. This reduces the problem to the corresponding statement for modules over a ring, which is Stacks Project, Tag: 064U.

This shows that your alternative definition is equivalent to the usual definition when $X$ is a scheme (or an algebraic stack).

I don't believe that your alternative definition is equivalent to the usual one for arbitrary locally ringed spaces, but I don't have a counterexample.

Note: Your definition of strictly perfect complex is not the correct one for arbitrary ringed spaces. A complex on a ringed space is strictly perfect if it is a bounded complex of direct summands of locally free complexes.