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First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the "right" notion of Morita equivalence of topological groupoids.

Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.

Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $\mathcal G$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta_x = (s\pi(x),x)$.

Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continous section of $\pi$.

First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the "right" notion of Morita equivalence of topological groupoids.

Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.

Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $\mathcal G$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta_x = (s\pi(x),x)$.

Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continuous section of $\pi$.

First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the right notion of Morita equivalence of topological groupoids.

Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.

Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $\mathcal G$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta_x = (s\pi(x),x)$.

Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continous section of $\pi$.

First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the "right" notion of Morita equivalence of topological groupoids.

Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.

Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $\mathcal G$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta_x = (s\pi(x),x)$.

Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continous section of $\pi$.

First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the right notion of Morita equivalence of topological groupoids.

Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.

Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $X$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta_x = (s\pi(x),x)$.

Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continous section of $\pi$.

First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the right notion of Morita equivalence of topological groupoids.

Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.

Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $\mathcal G$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta_x = (s\pi(x),x)$.

Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continous section of $\pi$.