I'm working out the notion of spectrum of a quasi-coherent algebra over a scheme.
$\require{AMScd}$ Let $(X,\mathcal O_X)$ be a scheme and $\mathcal A$ a quasi-coherent $\mathcal O_X$-algebra, i.e. a sheaf of ring $\mathcal A$ together with a morphism $\mathcal O_X \to \mathcal A$ making it a quasi-coherent $\mathcal O_X$-module. Then from the previous morphism, for every affine open subset $U$ of $X$, we have $$ \mathrm{Spec}\,(\mathcal A(U)) \to U.$$ Taking another affine open subset $V$ of $X$ such that $V \subseteq U$, the diagram $$\begin{CD} @. V \\ @. @VVV \\ \mathrm{Spec}\,(\mathcal A(U)) @>>> U \end{CD}$$ admit the limit $\mathrm{Spec}\,(\mathcal A(U)) \times_U V \simeq \mathrm{Spec}\,(\mathcal A (U) \otimes_{\mathcal O_X (U)} \mathcal O_X(V)) = \mathrm{Spec}\,(\mathcal A(V))$, the last equality because of the quasi-coherence of $\mathcal A$. So the arrow $\mathrm{Spec}\,(\mathcal A(V)) \to \mathrm{Spec}\,(\mathcal A(U))$ is an open immersion and we can define the relative spectrum as the glueing $$ \mathrm{Spec}\,\mathcal A := \mathrm{colim}_{\mathcal I}\,\mathrm{Spec}\,(\mathcal A(-))$$ where $\mathcal I$ is the opposite category of affine open subset of $X$.
Then, it is said in my notes that for all affine open subset $U$ of $X$, one have $$ \boxed{\mathrm{Spec}\,(\mathcal A(U)) = \mathrm{Spec}\,\mathcal A \times_X U,}$$ which I can't see. Obviously, the map $\mathrm{Spec}\,\mathcal A \to X$ being defined by the universal property of colimits on the maps $\mathrm{Spec}\,(\mathcal A(U)) \to U \to X$, we have a commutative square $$\begin{CD} \mathrm{Spec}\,(\mathcal A(U)) @>>> U \\ @VVV @VVV \\ \mathrm{Spec}\,\mathcal A @>>> X. \end{CD}$$ However, I don't see why it is cartesian...