In their book "Sheaves in Geometry and Logic" (paragraph III.9: continuous group action) Saunders MacLane and Ieke Moerdijk define a complete subcategory $\mathsf{S}(G)$ of a category $\mathsf{B}(G)$ of "topological" $G$-sets for a topological group $G$. Objects of this category are sets of the right cosets $G / U$, where $U$ is any open subgroup of $G$. Then, it is shown, that the morphisms $\varphi : G \setminus U \to G \setminus V$ in this category are in the direct correspondence with group elements $g$, such that $U \subset g^{-1}V g.$
The point of this theorem (III.9.1) is to show that $\mathsf{B}(G)$ is a Grothendieck topos by constructing the equivalence to the category of sheaves over the small category $\mathsf{S}(G)$. This equivalence is constructed explicitly as a pair of functors. And one of the functors is defined as a colimit over open subgroups of $U$:
$$\Psi(P) = {\lim_\to}_U P(G / U).$$
Inclusions, which correspond to the group unit $e_G$, are the morphisms of this diagramm. Then, the elements of $\Psi(P)$ for a presheaf $P$ are equivalence classes of form $[U,x]$ with $x \in P(G / U)$. And $[U,x] = [V,y]$, if there is an open group $W \subset U \cap V$ with $P_{W,U}(e)(x) = P_{W,V}(e)(y)$. Then the group action is defined by $$[U,x]g = \Big[g^{-1}Ug,P_{g^{-1}Ug,U}(g)(x)\Big].$$
The continuity of this action is nontrivial, as the colimit is computed in the category $\mathsf{SET}$ and not $\mathsf{B}(G)$. Note, that MacLane and Moerdijk considers all continuous $G$-sets ( that is objects of $\mathsf{B}(G)$) to be equipped with a discrete topology. So, it was shown in the exercise that the group action over $X$ is continuous if an isotropy subgroup $I_x$ is open for every $x \in X$. So, it is just claimed that $ U \subset I_{[U,x]}$ in $\Psi(P)$ without any proof. But I can't see why this is true?
Assume $g \in U$, then $[U,x]g = [U,P_{U,U}(g)(x)]$. Thus, I need to find an open subgroup $V \subset U$, such that $ P_{V,U}(g)(x) = P_{V,U}(e)P_{U,U}(g)(x) = P_{V,U}(e)(x)$. But this condition means that $e = g$, And this is not the case in general! So, is this a wong argument? And at what moment?
Overall, I feel that the content of this paragraph is beautiful. And I wish to know if I'm missing something here.
The key point here is how the correspondence between morphisms of $\mathsf{S}(X)$ and elements of $G$ is defined. The morphism $\varphi : G / U \to G / V$ corresponds to $g \in G$ exactly then $\varphi(Ux) = Vgx$ for any $x \in G$. So, if $g \in U$ and $\varphi \in \mathrm{End}_{\mathsf{S}(G)}(G/U)$, then $\varphi(Ux) = Ux$ and hence $\varphi = \mathrm{id}_{G/U}$. And in general case $g = f$ as a morphisms to $G / U$ if $g,f \in U$.
So the equation $e = g$ above holds for morphisms and not for group elements. In fact, $P_{U,U}(g)(x) = P_{U,U}(\mathrm{id})(x) = x$. Hence, the statement $[U,x]g = [U,x]$ for $g \in U$ is actually trivial.