I. If X is compact then Y is compact
II. If X is Hausdorff space then Y is Hausdorff space
II. If X is compact and Y is Hausdorff space then f^(1) exist
Which are correct?
I am not good at topology, hope someone could help! Thx!
f: X>Y is continous bijection

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 Joined: Sun Sep 23, 2012 11:38 am
Re: f: X>Y is continous bijection
I. This is true. The continuous image of a compact set is compact. Take any open cover $$\{ V_\alpha \}$$ of Y. Then $$X \subseteq \bigcup f^{1} (V_\alpha)$$ so $$\{ f^{1} (V_\alpha) \}$$ is an open cover of X. Since X is compact, you can take a finite subcover $$\{ f^{1} (V_i) \}_{i=1}^N$$. Then $$\{ V _i \}_{1}^N$$ covers Y. Thus any open cover of Y has a finite subcover.
II. This is false. Take the set {1,2,3}. Give it the discrete topology on X, and give it some nonHausdorff topology on Y, e.g. with open sets {1,2,3}, {}, {1}, {2,3}. Then $$f: X \rightarrow Y$$ defined by f(1)=1, f(2)=2, f(3)=3 is a continuous bijection, since by the discrete topology every subset of X is open so $$f^{1}(U)$$ is open for all U open in Y.
III. This is true. To prove that the inverse of f is continuous, we can show that the inverse of the inverse of f maps closed sets to closed sets. So we need to show that f(F) is closed for every closed subset F in X. Since F is a closed subset in a compact space X, then F is compact. Therefore by continuity, f(F) is compact, since the image of a compact set under a continuous map is compact. Since compact subsets of a Hausdorff space are closed, we have that f(F) is closed.
II. This is false. Take the set {1,2,3}. Give it the discrete topology on X, and give it some nonHausdorff topology on Y, e.g. with open sets {1,2,3}, {}, {1}, {2,3}. Then $$f: X \rightarrow Y$$ defined by f(1)=1, f(2)=2, f(3)=3 is a continuous bijection, since by the discrete topology every subset of X is open so $$f^{1}(U)$$ is open for all U open in Y.
III. This is true. To prove that the inverse of f is continuous, we can show that the inverse of the inverse of f maps closed sets to closed sets. So we need to show that f(F) is closed for every closed subset F in X. Since F is a closed subset in a compact space X, then F is compact. Therefore by continuity, f(F) is compact, since the image of a compact set under a continuous map is compact. Since compact subsets of a Hausdorff space are closed, we have that f(F) is closed.
Re: f: X>Y is continous bijection
Another counterexample for the 2nd one is that if the range has an indiscreet topology (which is nonHausdorff), then the function is automatically continuous.
One thing that may help with intuition is that if you have a bijection from one set to another, then they have the same number of elements. If it's a continuous bijection, then the topology of the range has fewer open sets than the domain. If a set is compact, then it has relatively few open sets (in the sense that if you remove open sets it'll still be compact, but if you add open sets then it might not be compact). If a set is Hausdorff then it has a relatively large number of open sets (if you add open sets, it'll still be Hausdorff; take some away and it might not). So if the domain is compact and the range has fewer open sets, the range will be compact. If the domain is Hausdorff and the range has fewer open sets, then it won't necessarily be Hausdorff.
One thing that may help with intuition is that if you have a bijection from one set to another, then they have the same number of elements. If it's a continuous bijection, then the topology of the range has fewer open sets than the domain. If a set is compact, then it has relatively few open sets (in the sense that if you remove open sets it'll still be compact, but if you add open sets then it might not be compact). If a set is Hausdorff then it has a relatively large number of open sets (if you add open sets, it'll still be Hausdorff; take some away and it might not). So if the domain is compact and the range has fewer open sets, the range will be compact. If the domain is Hausdorff and the range has fewer open sets, then it won't necessarily be Hausdorff.