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I was working on a few proofs about topological equivalence, and I realized that things would be a lot easier if I could form some shortcut that exempted me from proving the continuity of both a function mapping two topological spaces and its inverse. So I conjectured the following, and attempted to prove it:

If a function $f: A\to B$ is bijective and continuous, then its inverse $f^{-1}:B\to A$ is also continuous.

However, that was false, which surprised me. I found a counterexample at this previously asked math SE question. In the given counterexample, the set $A$ from which $f$ was mapping was split into two separate intervals.

My question is this: under what conditions is my conjecture true? What added property must $f$, $A$, and $B$ have for continuity to imply bicontinuity? My intuition tells me that most functions have this property, but I don't know how to tell when.

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  • $\begingroup$ Suggestion: $f$ is a bijection, so you may as well use it to identify $B$ with $A$. So now $f$ is just the identify function on $A$ viewed under two topologies. If every open set in the codomain topology is open in the domain topology, then $f$ is continuous. If every open set in the domain topology is open in the codomain topology, then $f$ is open (i.e., maps open sets to open sets). Both of these conditions have to hold for $f$ to be bicontinuous. $\endgroup$ – Rob Arthan Aug 8 '17 at 0:19
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One useful sufficient condition is that if the domain is compact and the range is Hausdorff then $f$ is bicontinuous.

Another sufficient condition with a weaker constraint on the domain says that if the domain is locally compact, and the range is Hausdorff, and $f$ is a proper map, then $f$ is bicontinuous.

You can find the proofs of these in most topology books, for example Munkres "Topology".

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You add that $f$ is an open map.

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  • $\begingroup$ How would I prove that that works? $\endgroup$ – Franklin Pezzuti Dyer Aug 7 '17 at 23:59
  • $\begingroup$ Let $g=f^{-1}$. If $f(U)$ is open for every open set $U\subset A$, then $g^{-1}(U)$ is open, so $g$ is continuous. $\endgroup$ – ajotatxe Aug 8 '17 at 0:02
  • $\begingroup$ $f$ is open means that the image of an open subset is open. Let $U$ be an open subset of $A$, $({f^{-1}})^{-1}(U)=f(U)$ is open. $\endgroup$ – Tsemo Aristide Aug 8 '17 at 0:03
  • $\begingroup$ @TsemoAristide Would it suffice to say that $A$ is connected? This is an entirely different restriction, but it resolves the counterexample that I posed. $\endgroup$ – Franklin Pezzuti Dyer Aug 8 '17 at 23:05
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A map $f:A\to B$ is continuous if $f^{-1}(V)$ is open for every open set $V\subset B$. To ensure that $f^{-1}$ is continuous you need the condition that only the open sets of $B$ meet this property. That is, if there is a subset $V'\subset B$ that is not open, but $f^{-1}(V')$ is open in $A$, then $f^{-1}$ is not continuous.

Intuitively, this means that $B$ has a coarser topology than $A$. That is, there are open sets in $A$ that correspond to non-open sets of $B$.

For an "extreme" example, take a set $X$ with more than one point and consider in it the discrete topology $\tau$ (every subset is open) and the coarse topolgy $\tau'$ (only $\emptyset$ and $X$ are open). Let $f$ be the identity map. Then $f:(X,\tau)\to (X,\tau')$ is continuous, but its inverse is not.

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