I'm studying algebraic geometry from the book by Wedhorn/Görtz (Algebraic Geometry I: Schemes).
There is an isomorphism which I quite do not understand. In Chapter 11 (p. 289), on vector bundles, the authors say the following:
I don't understand equation (11.2.5). His explanation is not clear to me. Why do we have $\text{Hom}_{X^{'}} (T^{'}, \text{Spec} \beta \times_X X^{'}) = \text{Hom}_X (T^{'}, \text{Spec} \beta)$. Also, does he mean $ (\text{Spec} \beta) \times_X X^{'}$ or $\text{Spec} ( \beta \times_X X^{'})$ ?
We have $g \circ f^{'}: T^{'} \rightarrow X$. Therefore, if I apply the bijection (11.2.1) (see below) to this, by definition, I have $$ \text{Hom}_{X} (T^{'}, \text{Spec} \beta) \cong \text{Hom}_{\mathcal{O}_X\text{-alg}} (\beta, (g \circ f^{'})_{*} \mathcal{O}_{T^{'}}) \cong \text{Hom}_{\mathcal{O}_X^{'}\text{-alg}} (g^{*} \beta, f_{*}^{'} \mathcal{O}_T^{'}). $$ But I don't understand how he obtains the first line after (11.2.5). I think this is crucial in understanding the argument.
This whole explanation follows the theorem on relative schemes (prop. 11.1):
$\require{AMScd}$The isomorphism $\operatorname{Hom}_{X'}(T',\operatorname{Spec}\mathscr B\times_X X')\cong\operatorname{Hom}_X(T',\operatorname{Spec}\mathscr B)$ follows from the universal property of the fiber product. Here, $\operatorname{Spec}\mathscr B\times_X X'$ means $(\operatorname{Spec}\mathscr B)\times_X X',$ the fiber product of schemes.
We fix an $X'$ scheme $f' : T'\to X'.$ Let us first suppose that we have a morphism of $X$-schemes $T'\to\operatorname{Spec}\mathscr B.$ (This makes sense, because we may view $T'$ as a scheme over $X$ via $T'\to X'\to X.$) Because we assume the the morphism $T'\to \operatorname{Spec}\mathscr B$ is a morphism of $X$-schemes, it follows that the diagram \begin{CD} T' @>>> \operatorname{Spec}\mathscr B \\ @VVV @VVV\\ X' @>>> X \end{CD} commutes, and so we obtain a unique morphism $T'\to\operatorname{Spec}\mathscr B\times_X X'$ such that \begin{CD} T'\\ @VVV\\ \operatorname{Spec}\mathscr B\times_X X' @>>> \operatorname{Spec}\mathscr B \\ @VVV @VVV\\ X' @>>> X \end{CD} commutes. Thus, we have a map of $X'$-schemes $T'\to \operatorname{Spec}\mathscr B\times_X X'.$
Conversely, suppose we have a map of $X'$-schemes $T'\to \operatorname{Spec}\mathscr B\times_X X'.$ Projection onto the first factor gives a map $T'\to\operatorname{Spec}\mathscr B,$ and the structure map $\operatorname{Spec}\mathscr B\to X$ lets us view $T'$ as an $X$-scheme. (We could also view $T'$ as an $X$-scheme via the composition $T'\to X'\to X$; commutativity of the diagram \begin{CD} \operatorname{Spec}\mathscr B\times_X X' @>>> \operatorname{Spec}\mathscr B \\ @VVV @VVV\\ X' @>>> X \end{CD} implies that these two $X$-scheme structures on $T$ are the same.)
This gives you maps $$\operatorname{Hom}_{X'}(T',\operatorname{Spec}\mathscr B\times_X X')\to\operatorname{Hom}_X(T',\operatorname{Spec}\mathscr B)$$ and $$\operatorname{Hom}_X(T',\operatorname{Spec}\mathscr B)\to\operatorname{Hom}_{X'}(T',\operatorname{Spec}\mathscr B\times_X X'),$$ which I leave you to check are mutually inverse. This will again be an application of the universal property of the fiber product.