Let $X$ be a topological space, and let $\mathcal{A}$ be an open cover for $X$.

To say that $Open(X)$ is coherent with $\mathcal{A}$ means that $$B\in Open(X) \Leftrightarrow B\cap A\in Open(A)\forall A \in \mathcal{A}$$

I am trying to prove that $Open(X)$ is always coherent with $\mathcal{A}$, whenever $\mathcal{A}$ is an open cover of $X$.

(This is from Spanier's Algebraic Topology pp. 5)

However, I get stuck, because as far as I understand it, $Open(A)$ is not necessarily the usual subspace topology from $Open(X)$ (I thought so at first, but then the whole claim seems trivial and nonsensical), but some pre-specified topology, and we merely have to show $Open(X)$ is coherent with all these topologies.

So I start by assuming that $\mathcal{A}\subseteq Open(X)$ and $\bigcup_{A\in\mathcal{A}}A = X$ and, for the $\Longleftarrow$ direction, pick an arbitrary $B\subseteq X$ such that $(B\cap A)\in Open(A) \forall A\in\mathcal{A}$ and try to show that $B\in Open(X)$. But I don't know anything about $Open(A)$, so I don't know how to proceed.

  • $\begingroup$ This notion of coherence looks very interesting... $\endgroup$ – goblin Sep 9 '18 at 3:15

You do have to use the subspace topology ; otherwise it might not be true that the topology of $X$ is coherent with $\mathcal{A}$. For example, look at what happens when $\mathcal{A} = \{ X \}$.

  • $\begingroup$ Then perhaps I misunderstand the construction of the topology coinduced by a collection of maps $g_i:X_i\to X$. I thought it was given by $Open(X) := \{U\subseteq X\,|\,g_i^{-1}(U)\in Open(X_i) \forall i\}$, which, for the case of a collection of subsets of $X$, $\mathcal{A}$, with $g_i$ being the inclusion maps, reduces to $Open(X) = \{U\subseteq X\,|\,U\cap A\in Open(A)\forall A\in\mathcal{A}\}$. But if $Open(A) = \{U\cap A\,|\,U\in Open(X)\}$ is the subspace topology, then we get $Open(X) = \{U\subseteq X\,|\,U\in Open(X)\}$ which means that any collection of subsets of $X$ furnished with $\endgroup$ – PPR May 15 '15 at 13:42
  • $\begingroup$ the subspace topology makes $Open(X)$ coherent with it. So what's the sense of the definition?? $\endgroup$ – PPR May 15 '15 at 13:42
  • 1
    $\begingroup$ @PPR No, not any collection of subsets with the subspace topology makes $Open(X)$ coherent. If $U \in Open(X)$, then $U \cap A \in Open(A)$ by definition of the subspace topology. However, the converse is false : if $U$ is a subset of $X$ such that $U \cap A \in Open(A)$, you cannot conclude that $U \in Open(X)$ in general. All you can say is that $U \cap A = V \cap A$ for some $V \in Open(X)$. $\endgroup$ – Math536 May 15 '15 at 20:10
  • 1
    $\begingroup$ @PPR For example, take $\mathcal{A}$ to be the set of singletons of $X$. Then $Open(X)$ is coherent with $\mathcal{A}$ if and only if $X$ has the discrete topology. $\endgroup$ – Math536 May 15 '15 at 20:16
  • $\begingroup$ I understand my mistake. Thank you so much. $\endgroup$ – PPR May 15 '15 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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