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@@ -2163,8 +2163,7 @@ We recall that in the above diagram $\sigma_3$ is a bisimulation, and the rest a
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% We have $(Fp_1)^\dagger\comp\sigma$.
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%\end{proof}
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\begin{example}
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And another counter-example!!!
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Assuming that the functor is the powerset endofunctor over the category of sets and injective maps. We show the functor with $\powfi$. For every $X$, we define the order on $\powfi X$ as $A\appr B$ whenever $|A|\leq |B|$, where $|A|$ and $|B|$ are just cardinalities of $|A|$ and $|B|$ respectively. This is a preorder. Then we define the order over every $\Hom(X,\powfi Y)$ pointwise. We have chosen the category of sets with injective maps because we could not define a functor from $\Set$ to $\preord$ with the mentioned ordering.\\
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Assuming that the functor is the powerset endofunctor over the category of sets and injective maps. Let us call this functor $\powfi$. For every $X$, we define the order on $\powfi X$ as $A\appr B$ whenever $|A|\leq |B|$, where $|A|$ and $|B|$ are just cardinalities of $|A|$ and $|B|$ respectively. This is a preorder. Then we define the order over every $\Hom(X,\powfi Y)$ pointwise. We have chosen the category of sets with injective maps because we could not define a functor from $\Set$ to $\preord$ with the mentioned ordering.\\
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We take $R=\{(1,1),(1,2),(2,1),(2,2)\}$, and $X=\{1,2,3\}$. $\alpha$ is defined as below:
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\begin{gather*}
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\alpha(x)=
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