Recently I encountered an alleged fact about restrictions of quotient maps, and I tried proving it. Arriving, with some help, at a proof sketch. But today I was told that the fact wasn't true, so I wonder what is wrong in this proof. (Edit: the error was found and is explained below.)

False proposition: For a top. space $X$ with a projection $\pi$ to a quotient space $X/_\sim$, the restriction of the map to an open subset $\mathcal U\subseteq X$ with subspace topology is a quotient map.

Corollary: If $p$ is bijective, it is an homeomorphism.

False Proof: Consider $\pi : X \to X/_\sim$ with the restriction $\pi|_u = p : \mathcal U \to \pi(\mathcal U)$ as follows.

$$\begin{array} AX & \stackrel{\pi}{\longrightarrow} & X/_\sim \\ \uparrow{i} & & \uparrow{j} \\ \mathcal U & \stackrel{p}{\longrightarrow} & \pi(\mathcal U) \end{array} $$

Let $W \subseteq \pi(\mathcal U)$ be open.

Subspace top: $W$ is open iff $W=V\cap\pi\mathcal (U) = j^{-1}(V)$ for a open $V\subseteq X/_\sim$.

Quotient top: $V$ is open iff $\pi^{-1}(V)\subseteq X$ is open.

Since $\mathcal U$ is open, the inclusion $i$ is an open map. Thus, $\pi^{-1}(V)$ is open iff $i^{-1}(\pi^{-1}(V))$ is open. ($\star$)

And $i^{-1}(\pi^{-1}(V))=\pi^{-1}(V)\cap\mathcal U = p^{-1}(j^{-1}(V))=p^{-1}(W)$.

The whole chain of implications gives: $p^{-1}(W)$ open iff $W$ open, i.e. $p$ is a quotient map.

Edit: The step marked ($\star$) is the wrong one. $i$ is open and continuous, but that $i^{-1}(\pi^{-1}(V))$ is open does not imply that $\pi^{-1}(V)$ is open. Thanks to user martini for pointing this out.

Also note that this proposition holds if $p^{-1}(p(W))=W$, which indeed is the case in the corollary, according to A question about the restriction of quotient maps to subsets of domain.

  • $\begingroup$ Why is $\pi^{-1}(V)$ open iff $i^{-1}(\pi^{-1}(V))$? I can see $\pi^{-1}(V)$ open implies $i^{-1}\pi^{-1}(V)$ is open. Why is the other way true? $\endgroup$ – R_D Oct 23 '15 at 13:54

Note that one in your chain of equivalences is just an implication, namely: You state that

"[...] the inclusion $i$ is an open map. Thus $\pi^{-1}(V)$ is open iff $i^{-1}\pi^{-1}(V)$ is open."

That's wrong. Any surjective open map is a quotient (that is what you state, that $i$ is a quotient), but if $U \ne X$, $i$ will not be onto. Here only the implication $\pi^{-1}(V)$ open $\Rightarrow$ $i^{-1}\pi^{-1}(V)$ open is true, but you need both. For example consider $U = (0,1) \subseteq \mathbf R = X$ then $\pi^{-1}(V) = [0,1]$ is not open in $X$, allthough $i^{-1}\pi^{-1}(V) = (0,1)$ is open in $U$.

  • $\begingroup$ Thank you, that was a stupid oversight by me. $\endgroup$ – neptun Oct 23 '15 at 14:05

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