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Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimesnion 2, and $\bar X^n$ is the underlying topological space. We can assume moreover that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$.

Is it true that for simplicial homologies of $\bar X^n$ we have $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$?

If not, what is a simplest contretemps, and what is the correct statement? If yes, what would be a reference?

PS. It seems to me that this should be true in the case when $X^n$ is a global quotient of a manifold by a finite group, because I guess in this case the simplicial homology of $X^n$ should be equal to the invariants of the action of $G$ on $H_k(X'^n)$. At the same time actions on $H_k(X'^n)$ and $H_{n-k}(X'^n)$ are dual. Is this reasoning correct?

Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimension 2, and $\bar X^n$ is the underlying topological space. We can assume, moreover, that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$.

Is it true that, for simplicial homologies of $\bar X^n$, we have $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$?

If not, what is a simplest counterexample, and what is the correct statement? If yes, what would be a reference?

PS. It seems to me that this should be true in the case when $X^n$ is a global quotient of a manifold by a finite group, because I guess in this case the simplicial homology of $X^n$ should be equal to the invariants of the action of $G$ on $H_k(X'^n)$. At the same time, actions on $H_k(X'^n)$ and $H_{n-k}(X'^n)$ are dual. Is this reasoning correct?

Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimesnion 2, and $\bar X^n$ is the underlying topological. We can assume moreover that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$.

Is it true that for simplicial homologies of $\bar X^n$ we have $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$?

If not, what is a simplest contrexample? And what is the correct statement. If yes, what would be a reference?

PS. It seems to me that this should be true in the case when $X^n$ is a global quotient of a manifold by a finite group, because I guess in this case the simplicial homology of $X^n$ should be equal to the invariants of the action of $G$ on $H_k(X'^n)$. At the same time actions on $H_k(X'^n)$ and $H_{n-k}(X'^n)$ are dual. Is this reasoning correct?

Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimesnion 2, and $\bar X^n$ is the underlying topological space. We can assume moreover that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$.

Is it true that for simplicial homologies of $\bar X^n$ we have $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$?

If not, what is a simplest contretemps, and what is the correct statement? If yes, what would be a reference?

PS. It seems to me that this should be true in the case when $X^n$ is a global quotient of a manifold by a finite group, because I guess in this case the simplicial homology of $X^n$ should be equal to the invariants of the action of $G$ on $H_k(X'^n)$. At the same time actions on $H_k(X'^n)$ and $H_{n-k}(X'^n)$ are dual. Is this reasoning correct?

Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimesnion 2, and $\bar X^n$ is the underlying topological. We can assume moreover that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$.

Is it true that $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$?

If not, what is a simplest contrexample? And what is the correct statement. If yes, what would be a reference?

Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimesnion 2, and $\bar X^n$ is the underlying topological. We can assume moreover that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$.

Is it true that for simplicial homologies of $\bar X^n$ we have $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$?

If not, what is a simplest contrexample? And what is the correct statement. If yes, what would be a reference?

PS. It seems to me that this should be true in the case when $X^n$ is a global quotient of a manifold by a finite group, because I guess in this case the simplicial homology of $X^n$ should be equal to the invariants of the action of $G$ on $H_k(X'^n)$. At the same time actions on $H_k(X'^n)$ and $H_{n-k}(X'^n)$ are dual. Is this reasoning correct?