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 $U\_i$?



An alternative definition of pseudo-coherent complex
**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.


