Lemma 99.16.5. In Situation 99.16.3. Denote $\mathcal{X} = \mathcal{C}\! \mathit{omplexes}_{X/B}$. Then $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces.
Proof. Consider two objects $x = (T, g, E)$ and $y = (T, g', E')$ of $\mathcal{X}$ over a scheme $T$. We have to show that $\mathit{Isom}_\mathcal {X}(x, y)$ is an algebraic space over $T$, see Algebraic Stacks, Lemma 94.10.11. If for $h : T' \to T$ the restrictions $x|_{T'}$ and $y|_{T'}$ are isomorphic in the fibre category $\mathcal{X}_{T'}$, then $g \circ h = g' \circ h$. Hence there is a transformation of presheaves
\[ \mathit{Isom}_\mathcal {X}(x, y) \longrightarrow \text{Equalizer}(g, g') \]
Since the diagonal of $B$ is representable (by schemes) this equalizer is a scheme. Thus we may replace $T$ by this equalizer and $E$ and $E'$ by their pullbacks. Thus we may assume $g = g'$.
Assume $g = g'$. After replacing $B$ by $T$ and $X$ by $X_ T$ we arrive at the following problem. Given $E, E' \in D(\mathcal{O}_ X)$ satisfying conditions (1), (2) of Lemma 99.16.2 we have to show that $\mathit{Isom}(E, E')$ is an algebraic space. Here $\mathit{Isom}(E, E')$ is the functor
\[ (\mathit{Sch}/B)^{opp} \to \textit{Sets},\quad T \mapsto \{ \varphi : E_ T \to E'_ T \text{ isomorphism in }D(\mathcal{O}_{X_ T})\} \]
where $E_ T$ and $E'_ T$ are the derived pullbacks of $E$ and $E'$ to $X_ T$. Now, let $W \subset B$, resp. $W' \subset B$ be the open subspace of $B$ associated to $E, E'$, resp. to $E', E$ by Lemma 99.16.1. Clearly, if there exists an isomorphism $E_ T \to E'_ T$ as in the definition of $\mathit{Isom}(E, E')$, then we see that $T \to B$ factors into both $W$ and $W'$ (because we have condition (1) for $E$ and $E'$ and we'll obviously have $E_ t \cong E'_ t$ so no nonzero maps $E_ t[i] \to E_ t$ or $E'_ t[i] \to E_ t$ over the fibre $X_ t$ for $i > 0$. Thus we may replace $B$ by the open $W \cap W'$. In this case the functor $H = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, E')$
\[ (\mathit{Sch}/B)^{opp} \to \textit{Sets},\quad T \mapsto \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X_ T}}(E_ T, E'_ T) \]
is an algebraic space affine and of finite presentation over $B$ by Lemma 99.16.1. The same is true for $H' = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (E', E)$, $I = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, E)$, and $I' = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (E', E')$. Therefore we can repeat the argument of the proof of Proposition 99.4.3 to see that
\[ \mathit{Isom}(E, E') = (H' \times _ B H) \times _{c, I \times _ B I', \sigma } B \]
for some morphisms $c$ and $\sigma $. Thus $\mathit{Isom}(E, E')$ is an algebraic space. $\square$
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