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I have to prove the following statement. Let$ f: X\to Y$ be a function and A and B subsets of X. Show:

1) $f(A\cap B) \subset f(A) \cap f(B)$

2) If f is injective, then $f(A\cap B) = f(A) \cap f(B)$

To begin with 1, I started as follows:

y $\in$ f(A$\cap$B) $\iff$ $\exists$x $\in$ A$\cap$B: y = f(x) $\iff$ $\exists$x $\in$ X: x $\in$ A$\cap$B $\land$ y = f(x)

My problem then is, if I write this out like this, I have f(A$\cap$B) = f(A) $\cap$ f(B), which is not correct. Is it maybe possible that my starting point is not good for this statement? Thank you for your help.

OSRCK_
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3 Answers3

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Take any $Y\in f(A\cap B)$, then exist $x\in A\cap B$ such that $f(x)=y$, so $y\in f(A)$ and $y= \in f(B)$ and thus $y\in f(A)\cap f(B)$. Since $y$ was aribtrary we have $$f(A\cap B) \subseteq f(A)\cap f(B)$$

Now if $f$ is injective. Take arbitrary $y\in f(A)\cap f(B)$. Then exist $a\in A$ and $b\in B$ such that $$f(a)=y=f(b)\implies a=b \implies a\in B$$

and thus $y \in f(A\cap B)$. Since $y$ as aribtrary we have $$f(A)\cap f(B) \subseteq f(A\cap B)$$

So if $f$ is injective you have equality.

nonuser
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    I don't understand it. If you take an element y in f(A $\cap$ B) and finally you get that y is also in f(A) $\cap$ f(B) (using double arrow), you can say that f(A $\cap$ B) = f(A) $\cap $f(B), or not? So my issue is, I have to have somewhere an arrow in 1 direction, so I can conclude that f(A$\cap$ B) $\subset$ f(A) $\cap$ f(B). And if f is injective, that arrow would also become a double arrow. Am I wrong somewhere? – OSRCK_ Oct 07 '18 at 10:59
  • Yes, but first is valid always, even if it is not injective. But if it is injective then the other direction also holds , so they are equal. – nonuser Oct 07 '18 at 11:03
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2) $f$ is injective.

Let $y \in f(A) \cap f(B).$

Then $y \in f(A)$ and $y \in f(B)$.

There is an $a \in A$ with $f(a)=y$.

Likewise ;

There is a $b \in B$ with $f(b)=y$.

Since $f$ injective : $a=b$, and

$x:=a=b$ where $x \in A \cap B$, i .e.

$f(x)=y \in f(A \cap B)$.

Peter Szilas
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Let $y\in f(A \cap B)$. Then $y=f(x)$ for some $x\in A\cap B$. Thus $x \in A$ implies $f(x) \in f(A)$ and similarly $x \in B$ implies $f(x) \in f(B)$. Hence, it follows that $y = f(x) \in f(A) \cap f(B)$. This shows $f(A \cap B) \subset f(A) \cap f(B)$.

Further assume that $f$ is injective. Now, let $y \in f(A) \cap f(B)$ and thus, $y \in f(A)$ implies there is an $a\in A$ such that $f(a)=y$ and similarly $y \in f(B)$ implies there is an $b\in B$ such that $f(b)=y$. Since $f$ is injective and $f(a)=f(b)$, it must be that $a=b$. Thus, $a\in A\cap B$ since $a=b$ and $b\in B$. Thus $y= f(a) \in f(A \cap B)$. This shows that $f(A) \cap f( B) \subset f(A \cap B)$ if $f$ is injective. With the first part, you can conclude that $f(A \cap B) = f(A) \cap f(B)$ if $f$ is injective.

ashK
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