Before going into the main problem, let me just write down the relevant definitions for the sake of completeness.
Definition 1. Let $\mathbf{X}$ be a category. A concrete category over $\mathbf{X}$ is a pair $(\mathbf{A},U)$, where $\mathbf{A}$ is a category and $U : \mathbf{A} \to \mathbf{X}$ is a faithful functor.
Definition 2. If $(\mathbf{A},U)$ and $(\mathbf{B}, V)$ are concrete categories over $\mathbf{X}$ , then a concrete functor from $(\mathbf{A},U)$ to $(\mathbf{B}, V)$ is a functor $F : \mathbf{A}\to \mathbf{B}$ with $U = V \circ F$. We denote such a functor by $F : (\mathbf{A},U)\to (\mathbf{B},V)$.
Definition 3. A concrete functor $F$ between two concrete categories $(\mathbf{A},U)$ and $(\mathbf{B}, V)$ over $\mathbf{X}$ is said to be a concrete equivalence if $F:\mathbf{A}\to\mathbf{B}$ is an equivalence.
Definition 4. A concrete category $(\mathbf{A},U)$ over $\mathbf{X}$ is said to be transportable provided that for every $\mathbf{A}$-object $A$ and every $\mathbf{X}$-isomorphism $UA \overset{k}{\to}X$ there exists an $\mathbf{A}$-object $B$ with $UB = X$ such that $A \overset{k}{\to}B$ is an $\mathbf{A}$-isomorphism
In Joy of Cats at page 73 the authors provide the proof of the following proposition (number unchanged),
LEMMA 5.35. For every concrete category $(\mathbf{A},U)$ over $\mathbf{X}$, there exists a transportable concrete category $(\mathbf{B}, V)$ over X and a concrete equivalence $E : (\mathbf{A},U)\to (\mathbf{B}, V)$.
In the proof of the theorem they define the category $\mathbf{B}$ as follows:
Each $\mathbf{B}$-object is a triple $(A, a, X)$, with $A \in Ob(\mathbf{A}), X \in Ob(\mathbf{X})$, and $a : UA \to X$ an $\mathbf{X}$-isomorphism. For each $(A,a,X),(\hat{A},\hat{a},\hat{X})\in Ob(\mathbf{B})$ the $\text{hom}$-sets are defined as follows, $$\text{hom}_{\mathbf{B}}((A, a,X), (\hat{A},\hat{a},\hat{X})) = \text{hom}_{\mathbf{A}}(A,\hat{A})$$ Identities and composition are as in $\mathbf{A}$.
However, strictly speaking, by the definition of a category of the book (see Def. 3.1, page 21), $\mathbf{B}$ is not a category since the $\text{hom}$-sets are not disjoint. Although by Remarks 3.2(3) we do not need to show that $\text{hom}$-sets are not disjoint.
But I was wondering, is it possible to modify the definition of the category $\mathbf{B}$ in such a way that the proof goes through and the $\text{hom}$-sets become disjoint?