A question on the half-full functor I'm reading the book of Categories and Sheaves. 
book link
In the proof of the proposition 1.3.18. in the above link, I don't understand why $Hom_{C_0'} (F(X), F(Y))$ does not depend on the choice of $X, Y$. 
[My try]
Suppose that $F(X)=F(X_1), F(Y)=F(Y_1)$. 
I want to show that for any $F(f) \in Hom_{C_0'} (F(X), F(Y))$ where $f : X \to Y$, $F(f) \in Hom_{C_0'} (F(X_1), F(Y_1))$. 
Then by the sysmmetry, $Hom_{C_0'} (F(X), F(Y)) = Hom_{C_0'} (F(X_1), F(Y_1))$. 
From the hypothesis of half-fullness of $F$, there are isomorphisms $\alpha : X \to X_1 , \beta : Y \to Y_1$.   
So I tried to prove that $F(f) = F(\beta \circ f \circ \alpha^{-1})$ or somthing. 
But I can't. 
Please help me. Thanks in advance.
 A: I don't think that this is actually true. In particular, if $F\left(\operatorname{Hom}_\mathcal{C}(X,Y)\right)$ was really independent of the choice of $X$ and $Y$, then in particular, if $F(X)=F(Y)$, there should be an arrow $X\to Y$ whose image is the identity of $F(X)$, and while the hypothesis forces the existence of some isomorphism $X\to Y$, the definition explicitly says that the image by $F$ of such an isomorphism is not required to be equal to the identity.
With this in mind, we can build a counterexample as follows : let $\mathcal{C}$ be the "walking isomorphism" or codiscrete category with two objects. More explicitly, this means that $\mathcal{C}$ has two objects $X$ and $Y$, and four arrows : the identities of $X$ and $Y$, and two arrows $\alpha:X\to Y$ and $\beta:Y\to X$, with composition defined by $\alpha\circ \beta=id_Y$ and $\beta\circ \alpha=id_X$. Let $\mathcal{C}'=BG$ be any non-trivial group $G$, seen as a one-object category, and let $F:\mathcal{C}\to \mathcal{C}'$ be defined by $F(\alpha)=g$ and $F(\beta)=g^{-1}$ for some $g\neq 1_G$. Then $F$ is half-full since $X$ and $Y$ are isomorphic and faithful since its domain is a preorder; but even though $F(X)=F(Y)$, we have
$$F(\operatorname{Hom}_\mathcal{C}(X,Y))=F(\{\alpha\})=\{g\}\neq \{1_G\}=F(\operatorname{Hom}_{\mathcal{C}}(X,X)).$$
