# Bijection abstract simplicial complex

Given two compact Hausdorff spaces $X$ and $Y$ and $h \colon X \to Y$ a homeomorphism, how can I prove that $h_{\mathfrak{A}} : N(\mathfrak{A}) \to N(h(\mathfrak{A}))$ is a bijection where $N(\mathfrak{A})$ and $N(h(\mathfrak{A}))$ are the nerves of the coverings $\mathfrak{A}$ and $h(\mathfrak{A})$?

[Edit: copied over from duplicate question simplicial complex bijection the following clarification]

ps: I use this definition: If $\mathfrak{A} = \{A_{i}\}_{i \in I}$ is an open covering of X, $N(\mathfrak{A})$ is an abstract simplicial complex such that: the vertices are the open sets of $\mathfrak{A}$ and a collection $\{A_{0}, A_{1}, \dots, A_{p}\}$ of such vertices constitutes a p-simplex if and only if $\bigcap_{i = 0}^p A_{i}$ is not empty.

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Suppose $\mathfrak{A}=\{U_\lambda\subset X\:|\:\lambda\in\Lambda\}$ is an open covering of $X$ and let $I\subset\Lambda$ be in the nerve $N(\mathfrak{A})$. This implies that $\bigcap_{\lambda\in I}U_\lambda\neq\emptyset$. We'll use the same indexing set $\Lambda$ for $N(h(\mathfrak{A}))$ so that we just need to show that the nerves $N(\mathfrak{A})$ and $N(h(\mathfrak{A}))$ are equal (and so the map $h_{\mathfrak{A}}$ is the identity map which is clearly a bijection).
In order to show that $I\in N(h(\mathfrak{A}))$ we only need to show that $\bigcap_{\lambda\in I}h(U_\lambda)\neq\emptyset$. This is clear though because if $x\in\bigcap_{\lambda\in I}U_\lambda$ then for all $\lambda\in I$, $x\in U_\lambda$ and so for all $\lambda\in I$ we also have $h(x)\in h(U_{\lambda})$ hence $h(x)\in\bigcap_{\lambda\in I}h(U_\lambda)$ and so $\bigcap_{\lambda\in I}h(U_\lambda)\neq\emptyset$. It follows that $I\in N(h(\mathfrak{A}))$ and so $N(\mathfrak{A})\subseteq N(h(\mathfrak{A}))$.
We can follow this argument again but using $h^{-1}$ (which exists because $h$ is a homeomorphism) instead of $h$ to get $N(\mathfrak{A})\supseteq N(h(\mathfrak{A}))$ and so $N(\mathfrak{A})= N(h(\mathfrak{A}))$.