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I need help with this proof:

$f: X\rightarrow Y$

$C,D\subseteq Y$

$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$

Thanks.

Martin Sleziak
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mila
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    Presumably you mean $C, D \subseteq Y$. What have you tried so far? – universalset Dec 03 '13 at 14:50
  • Yes, I fixed that. I have no idea where to start. I believe I have to use the definition of preimage. – mila Dec 03 '13 at 15:15
  • Links to several similar question that have already been asked on this site can be found here: http://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages (In this particular case: http://math.stackexchange.com/questions/228711/what-are-the-strategies-i-can-use-to-prove-f-1s-cap-t-f-1s-cap-f ) – Martin Sleziak Dec 03 '13 at 15:36

3 Answers3

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To show set equality,$$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$$ we typically proceed by showing that both the following inclusions hold:

$$f^{-1}(C \cap D) \quad \subseteq \quad f^{-1}(C) \cap f^{-1}(D)\tag{1}$$ $$f^{-1}(C) \cap f^{-1}(D) \quad \subseteq \quad f^{-1}(C \cap D)\tag{2}$$

For each of $(1), (2),$ we can use "element chasing": In general, to show $A \subseteq B$, it suffices to show that $a \in A \implies a \in B$.

For $(2)$, to show $$f^{-1}(C) \cap f^{-1}(D) \subseteq f^{-1}(C \cap D)\tag{2}$$

We start by assuming $x \in f^{-1}(C) \cap f^{-1}(D)$. Then, by the definition of set intersection, we have that $x \in f^{-1}(C)$ AND $x \in f^{-1}(D)$. This means $x \in C$ AND $x \in D$, which means, by definition, that $x \in C\cap D$. Now, $x\in C\cap D$ implies that $x \in f^{-1}(C\cap D).$ Hence, we've shown that $$f^{-1}(C) \cap f^{-1}(D) \subseteq f^{-1}(C \cap D)\tag{2}$$

Use the very same strategy for approaching the first inclusion $(1)$. When you show that inclusion $(1)$ also holds, then you can assert what you set out to prove: $$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$$

amWhy
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To solve these kinds of questions, you want to show that $$f^{-1}(C\cap D) \subseteq f^{-1}(C) \cap f^{-1}(D)$$ and $$f^{-1}(C\cap D) \supseteq f^{-1}(C) \cap f^{-1}(D).$$

I will show $\subseteq$ here; you should try the other direction.

Suppose $x \in f^{-1} (C \cap D)$. This means $f(x) \in (C\cap D)$, which further implies $f(x) \in C$ AND $f(x) \in D$. Thus, $x \in f^{-1}(C)$ AND $x \in f^{-1}(D)$.

angryavian
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1

Just pick an element in $f^{-1}( C \cap D) $ and verify the first inclusion. The same for the second

Maybe this definition will help $ f^{-1}( C) := \{ x \in X \text { such that } f (x) \in C \} $

Just apply it

Riccardo
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