Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

I'll use the definition of a nerve of an open covering which wikipedia uses.

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 $N(\mathfrak{A})= N(h(\mathfrak{A}))$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.