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I'm reading Abstract and Concrete Categories and am currently attempting to solve the very first exercise, which asks to show that a thin category is determined by its graph up to isomorphism.

Here, a category $mathbf{A}$ is a quadruple $(mathcal{O},mathrm{hom},mathrm{id},circ)$, such that

• 
  $mathcal{O}$ is a class of objects, alternatively denoted $mathrm{Ob}(mathbf{A})$,


• 
  $mathrm{hom}colonmathcal{O}^2rightarrowmathcal{U}$ is a mapping, where each element of its image is called a hom-set (here, $mathcal{U}$ is the universe, i.e. the class of all sets),


• 
  $mathrm{id}colonmathcal{O}rightarrowmathrm{Mor}(mathbf{A})colon=bigcup_{Xinmathrm{hom}(mathcal{O}^2)}X$ is a mapping that sends an $mathbf{A}$-object $A$ to an identity morphism $mathrm{id}_Ainmathrm{hom}(A,A)$ and


• 
  $circcolonmathrm{Mor}(mathbf{A})timesmathrm{Mor}(mathbf{A})leadstomathrm{Mor}(mathbf{A})$ is a partial mapping that is defined on $(g,f)$ iff $finmathrm{hom}(A,B)$ and $ginhom(B,C)$ for $A,B,Cinmathrm{Ob}(mathbf{A})$; in that case, $gcirc fcolon=circ(g,f)inmathrm{hom}(A,C)$.

These are subject to the conditions that

• composition is associative, i.e. $(hcirc g)circ f=hcirc(gcirc f)$ whenever defined,


• for any morphism $finmathrm{hom}(A,B)$, we have $mathrm{id}_Bcirc f=f=fcircmathrm{id}_A$ and


• hom-sets are disjoint.

A morphism $finmathrm{Mor}(mathbf{A})$ is by definition a member of a hom-set $mathrm{hom}(A,B)$ with $A,Binmathrm{Ob}(mathbf{A})$ and, by disjointedness of hom-sets, it is the member of exactly one hom-set. Thus, setting $mathrm{dom}(f)=A$ and $mathrm{cod}(f)=B$ produces two well-defined mappings $mathrm{dom},mathrm{cod}colonmathrm{Mor}(mathbf{A})rightarrowmathrm{Ob}(mathbf{A})$, giving what is called domain and codomain of a morphism respectively.

Furthermore, a large graph is defined as a quadruple $(V,E,d,c)$, where $V$ and $E$ are classes (called the class of vertices and edges respectively), and $dcolon Erightarrow C$ and $ccolon Erightarrow C$ are mappings giving what is called domain or codomain of each edge respectively. The graph $G(mathbf{A})$ of a category $mathbf{A}$ is the graph with $V=mathrm{Ob}(mathbf{A})$, $E=mathrm{Mor}(mathbf{A})$, $d=mathrm{dom}$ and $c=mathrm{cod}$.

A thin category is a category where any hom-set contains at most one element. Let $mathbf{A}=(mathcal{O},mathrm{hom},mathrm{id},circ)$ and $mathbf{A}^prime=(mathcal{O}^prime,mathrm{hom}^prime,mathrm{id}^prime,circ^prime)$ be two thin categories such that $G(mathbf{A})=G(mathbf{A}^prime)$. I.e.
$$(mathrm{Ob}(mathbf{A}),mathrm{Mor}(mathbf{A}),mathrm{dom},mathrm{cod})=G(mathbf{A})=G(mathbf{A}^prime)=(mathrm{Ob}(mathbf{A}^prime),mathrm{Mor}^prime(mathbf{A}),mathrm{dom}^prime,mathrm {cod}^prime),$$
where the $^prime$ denotes being relative to the category $mathbf{A}^prime$. From this, it follows immediately by definition of a tuple that $mathcal{O}=mathrm{Ob}(mathbf{A})=mathrm{Ob}(mathbf{A}^prime)=mathcal{O}^prime$, $mathrm{Mor}(mathbf{A})=mathrm{Mor}(mathbf{A}^prime)$, $mathrm{dom}=mathrm{dom}^prime$ and $mathrm{cod}=mathrm{cod}^prime$. Let $A,Binmathcal{O}=mathcal{O}^prime$. We then have by definition that
$$mathrm{hom}(A,B)=mathrm{dom}^{-1}(left{Aright})capmathrm{cod}^{-1}(left{Bright})=(mathrm{dom}^prime)^{-1}(left{Aright})cap(mathrm{cod}^prime)^{-1}(left{Bright})=mathrm{hom}^prime(A,B).$$
Thus, $mathrm{hom}=mathrm{hom}^prime$. Now let $A,B,Cinmathcal{O}$, $finmathrm{hom}(A,B)$ and $ginmathrm{hom}(B,C)$ be arbitrary. Then, $gcirc finmathrm{hom}(A,C)$ exists, i.e. the hom-set $mathrm{hom}(A,C)$ has at least on element, but, since $mathbf{A}$ is thin, it contains at most one element, hence it has exactly one element. Since $mathrm{Ob}(mathbf{A})=mathrm{Ob}(mathbf{A}^prime)$ and $mathrm{Mor}(mathbf{A})=mathrm{Mor}(mathbf{A}^prime)$, we may regard $f,g$ as morphisms in $mathbf{A}^prime$ and thus $gcirc^prime f$ exists. By nature of composition, we have $gcirc^prime finmathrm{hom}(A,C)$, but since this set has exactly one element, it follows that $gcirc f=gcirc^prime f$ and, since $f,g$ were arbitrary, it follows that $circ=circ^prime$.

Finally, let $A,Binmathrm{Ob}(mathbf{A})$ and $finmathrm{hom}(A,B)$ be arbitrary. By the previous result, it holds that $mathrm{id}_Bcirc^prime f=mathrm{id}_Bcirc f=f$ and $fcirc^primemathrm{id}_A=fcircmathrm{id}_A=f$. Hence, the identity morphisms in $mathbf{A}$ are also identity morphisms in $mathbf{A}^prime$ and, by uniqueness of identities, it follows that $mathrm{id}=mathrm{id}^prime$.

Finally, by the definition of a tuple, it follows that $mathbf{A}=mathbf{A}^prime$.

Now, this doesn't only imply that a graph determines a thin category up to isomorphism (as was required to be shown), but it shows that a thin category is uniquely determined by its graph. This claim is clearly stronger than the initial one and since the proposed proof is pretty trivial, I suspect there must be an error as otherwise the way the exercise was stated seems off. My question now is where any error or potential misunderstanding lies.