Is Quotient Map Closed? Proof and Explanation | Math Homework

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SUMMARY

The quotient map \( f: X \to Y \) is established as an open map but not necessarily a closed map. The discussion clarifies that while there is a one-to-one correspondence between open sets in \( X \) and open sets in \( Y \), this does not extend to closed sets. A counterexample provided is the projection map \( \pi_1: \mathbb{R}^2 \to \mathbb{R} \), which demonstrates that an open quotient map can fail to be closed. Thus, the assertion that all open quotient maps are closed is definitively false.

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  • Understanding of quotient maps in topology
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  • Knowledge of projection maps, specifically \( \pi_1: \mathbb{R}^2 \to \mathbb{R} \)
  • Basic concepts of set theory and mappings
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Homework Statement


THe quotient map f is open but is it also closed?

The Attempt at a Solution


I think it is. Consider f: X->Y

FOr every open set V in Y there exists by definition an open set f^-1(V) in X. There is a one to one correspondence between open sets in X and open sets in Y by definition.

So for every closed set V complement in Y there exists a closed set f^-1(V) complement in X. So f is both closed and open.
 
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Are you asking whether or not an open quotient map is closed? If so, then what you did does not prove that. There isn't a 1-1 correspondence between open sets in X and open sets in Y. Look carefully at the definition. If f:X->Y is a quotient map, then U is open in Y iff f^-1(U) is open in X. This does not exhaust all the open sets in X. So you haven't proved that f takes closed sets to closed sets.

Anyway, this is false. For a counterexample, let \pi_1 : \mathbb{R}^2 \to \mathbb{R} be the projection map onto the first coordinate. Then \pi_1 is an open surjection, so it's a quotient map. However it's not closed. (I'll let you find a closed set that doesn't get mapped to a closed set.)
 
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