[증명] If each set p⁻¹({y}) is connected, and if Y is connected, then X is connected.

Proposition

Let p : X → Y be a quotient map.

If each set p⁻¹({y}) is connected, and if Y is connected, then X is connected.

 

Proof.

I will use reduction of absurdity. Start with that X is not connected,

Then there exists seperation U and V of X.

p⁻¹({y}) is connected, so p⁻¹({y}) ⊂ U or p⁻¹({y}) ⊂ V for all y ∈ Y.

 

Let W = { y | p⁻¹({y}) ⊂ U } and Z = { y | p⁻¹({y}) ⊂ V }. 

W ∪ Z = Y and W ∩ Z=∅.

Then, p⁻¹(W) = U, p⁻¹(Z) = V, both of which are open.

Because p is a quotient map, W and Z should be open, making them a seperation of Y.

This is a contradiction to the fact that Y is connected.

 

Therefore, X is connected.

 

 

[comment]

I myself am a bit unsure about the conclusion that p⁻¹(W) = U, p⁻¹(Z) = V derived from W = { y | p⁻¹({y}) ⊂ U } and Z = { y | p⁻¹({y}) ⊂ V }. 

So, if there are any errors, please let me know in the comments.

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