# Composition of Two Continuous Functions ... Browder, Proposition 3.12

Gold Member

## Summary:

I need help with an aspect of Andrew Browder's proof that the composition of two continuous functions is continuous ...

## Main Question or Discussion Point

I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help in understanding the proof of Proposition 3.12 ...

Proposition 3.12 and its proof read as follows:

In the above proof by Browder we read the following:

" ... ... Since $f(I) \subset J$, $f^{ -1 } ( g^{ -1 }(V) ) = f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)$ ... ... "

My question is as follows:

Can someone please explain exactly why/how $f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)$ ... ...

Help will be much appreciated ...

Peter

## Answers and Replies

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fresh_42
Mentor
You have an equation of the type $A \cap B = A$. If you draw a Venn diagram, you will see that this is equivalent to $A \subseteq B$. Now $I \subseteq f^{-1}(J)$ but $I$ is all we have, so $I=f^{-1}(J)$.

More interesting is why $f^{-1}(U\cap J) =f^{-1}(U) \cap f^{-1}(J)$. Can you tell why this is true?

Gold Member
You have an equation of the type $A \cap B = A$. If you draw a Venn diagram, you will see that this is equivalent to $A \subseteq B$. Now $I \subseteq f^{-1}(J)$ but $I$ is all we have, so $I=f^{-1}(J)$.

More interesting is why $f^{-1}(U\cap J) =f^{-1}(U) \cap f^{-1}(J)$. Can you tell why this is true?

Thanks fresh_42 .... can now see why $f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)$ ... ...

As for $f^{-1}(U\cap J) =f^{-1}(U) \cap f^{-1}(J)$ ... I know this is true for any sets U and J (See Topology by Munkres, Exercise 2(c) page 20 ...) ... but I think you are asking me to give a sense of why it is true ... but I have not yet got any ideas regarding this ...

Thanks again for your help ...

Peter

fresh_42
Mentor
Such equations are often shown elementwise: choose an arbitrary element $x\in f^{-1}(A\cap B)$ and show that it is in $f^{-1}(A)$ and in $f^{-1}(B)$, too. For the other direction, choose an element $x\in f^{-1}(A)\cap f^{-1}(B)$ and show that it is an element of $f^{-1}(A\cap B)$. As long as it is true for an arbitrary element, as long is it true for all elements.

Gold Member
Such equations are often shown elementwise: choose an arbitrary element $x\in f^{-1}(A\cap B)$ and show that it is in $f^{-1}(A)$ and in $f^{-1}(B)$, too. For the other direction, choose an element $x\in f^{-1}(A)\cap f^{-1}(B)$ and show that it is an element of $f^{-1}(A\cap B)$. As long as it is true for an arbitrary element, as long is it true for all elements.

Oh ... so you were asking for a formal proof ...

That, I think would proceed as follows:

To prove ... $f^{-1} (A \cap B) =f^{-1} (A) \cap f^{-1} (B)$

Now ... $x \in f^{-1}(A \cap B)$

$\Longrightarrow f(x) \in A \cap B$

$\Longrightarrow f(x) \in A$ and $f(x) \in B$

$\Longrightarrow x \in f^{-1} (A)$ and $x \in f^{-1} (B)$

$\Longrightarrow x \in f^{-1} (A) \cap f^{-1} (B)$

So ... we have $f^{-1} (A \cap B) \subset f^{-1} (A) \cap f^{-1} (B)$ ... ... ... ... ... (1)

Now ... $x \in f^{-1} (A) \cap f^{-1} (B)$

$\Longrightarrow x \in f^{-1} (A)$ and $x \in f^{-1} (B)$

$\Longrightarrow f(x) \in A$ and $f(x) \in B$

$\Longrightarrow f(x) \in A \cap B$

$\Longrightarrow x \in f^{-1}(A \cap B)$

So ... we have $f^{-1} (A) \cap f^{-1} (B) \subset f^{-1} (A \cap B)$ ... ... ... ... ... (2)

Now ... (1) and (2) imply $f^{-1} (A \cap B) =f^{-1} (A) \cap f^{-1} (B)$

Is that correct?

Thanks again for your help ...

Peter

fresh_42
Mentor
Is that correct?
Yep, correct. It does not generally work for $f$, where we have $f(A\cap B) \subseteq f(A) \cap f(B)$ but not the other direction.