MHB Inverse image of a set under the restriction of a function

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The discussion focuses on proving the equality in Theorem 18.2 Part (f) from Munkres' "Topology," specifically regarding the inverse image of a set under the restriction of a function. The theorem states that the inverse image of an open set V in Y, intersected with Uα, is equal to the inverse image of V under the restricted function f|Uα. Participants clarify the definitions of the sets involved, showing that both sides of the equation represent the same collection of elements. The explanation emphasizes understanding the definitions to see why the equality holds. This discussion highlights the importance of grasping set definitions in topology to prove statements about functions and their restrictions.
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I am reading Munkres book, "Topology" (Second Edition).

I need help with an aspect of Theorem 18.2 Part (f) concerning the inverse image of a set under the restriction of a function ...

Theorem 18.2 Part f reads as follows:View attachment 4194
View attachment 4195
View attachment 4196In the above text we read:

" ... ... Let $$V$$ be an open set in $$Y$$.

Then

$$f^{-1} (V) \cap U_{ \alpha } = {(f | U_{ \alpha }) }^{-1} (V)$$ ... ...

... ... "I would like to prove that:

$$f^{-1} (V) \cap U_{ \alpha } = { (f | U_{ \alpha }) }^{-1} (V)
$$... BUT ... cannot see how to do this ...Can someone please help ...

Peter
 
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Hi Peter,

I will just write down what this sets are.

$f^{-1}(V)=\{x\in X \ : \ f(x)\in V \}$

$f^{-1}(V)\cap U_{\alpha}=\{x\in U_{\alpha} \ : \ f(x)\in V \}$

Did you see now why the equality holds?
 
Fallen Angel said:
Hi Peter,

I will just write down what this sets are.

$f^{-1}(V)=\{x\in X \ : \ f(x)\in V \}$

$f^{-1}(V)\cap U_{\alpha}=\{x\in U_{\alpha} \ : \ f(x)\in V \}$

Did you see now why the equality holds?
Thanks for the help, Fallen Angel ...
 
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