I apologize for the title, but it was difficult to find a more appropriate one, feel free to edit if you think it is necessary.
My questions comes as part of the proof that, for a scheme $X$, and a quasi-coherent $\mathcal{O}_X$-algebra $\mathcal{B}$, the functor $$Sch_X^{op}\rightarrow Sets$$ $$(f:T\rightarrow X)\mapsto\text{Hom}_{(\mathcal{O}_X-\text{Alg.})}(\mathcal{B},f_{*}\mathcal{O}_T)$$ is representable. In this context we reduce ourselves to prove it in the affine case: $X=\text{Spec} (A)$. This will mean that $\mathcal{B}\simeq \tilde{B}$ for some $A$-module $B$.
Then I would like to prove that for every $X$-scheme $f:T\rightarrow X$ we have a natural bijection $$\text{Hom}_{(\mathcal{O}_X-\text{Alg.})}(\mathcal{B},f_{*}\mathcal{O}_T)\simeq \text{ Hom}_{(A\text{-Alg.})}(B,\Gamma(T,\mathcal{O}_T)).$$
Let me remark that I cannot use $\text{Spec}(\mathcal{B})$ since my question refers to part of the proof of its existence.
In the case in which $T$ is affine this is true since $f_{*}\mathcal{O}_T$ is a quasi-coherent algebra. Should we use gluing then? Or is there a way to prove it directly? In both cases I would like to have some details. Thank you.
We cannot use the construction of $\operatorname{Spec}(\mathcal{B})$ but we are allowed to use $\operatorname{Spec}(B)$. By glueing you can prove $\operatorname{Hom}_{A-\textit{alg}}(B,\Gamma(T,\mathcal{O}_T))=\operatorname{Hom}_{\operatorname{Spec}(A)}(T,\operatorname{Spec}(B))$ (see for example Proposition 2.2.4 in EGA I, where the proof it is done in the case $A=\mathbb{Z}$ but for general $A$ the proof is the same). Now the equality $\operatorname{Hom}_{\operatorname{Spec}(A)}(T,\operatorname{Spec}(B))=\operatorname{Hom}_{\mathcal{O}_X-\textit{alg}}(\mathcal{B},f_*(\mathcal{O}_T))$ is exactly the content of the first paragraph of the proof of Proposition 1.2.7 in EGA II.