Prove that a function $f: A \rightarrow B$ is surjective if $f(f^{-1}(Y)) = Y$ for all $Y \subseteq B$.

Question

My approach:

Suppose $f(f^{-1}(Y)) = Y$ for all $Y \subseteq B$. Now suppose $b \in B$. We want to show that there exists an element $a \in A$ for which $f(a) = b$.

Consider the set $\{b\}$. Since $\{b\} \subseteq B$, it must be that $f(f^{-1}(\{b\})) = \{b\}$.

Now, for the sake of contradiction, suppose that $f^{-1}(\{b\}) = \emptyset$. Then $f(f^{-1}(\{b\})) = \{f(x) : x \in f^{-1}(\{b\})\} = \{f(x) : x \in \emptyset\} = \emptyset$, a contradiction because $f(f^{-1}(\{b\})) = \{b\}$ as stated above. Hence $f^{-1}(\{b\}) \neq \emptyset$.

Then there must exist an element $a \in f^{-1}(\{b\}) = \{x \in A : f(x) \in \{b\}\}$. Therefore $f(a) = b$, so $f$ is surjective.

Is this correct?

Note: This question was already asked in this thread, however, there were only answers given for the direction "$f$ is surjective $\Rightarrow$ $f(f^{-1}(Y)) = Y$" and not the other way around.

Thanks in advance!

I would consider it sufficient to state that $f(f^{-1}({b})) = {b}$. — copper.hat, Sep 01 '20 at 18:22
@TheSilverDoe I meant sufficient for the purposes of proof, of course. — copper.hat, Sep 01 '20 at 18:24
@copper.hat Hmm yeah, but this is not a weaker assumption than $f(f^{-1}(Y))=Y$, so I don't see why it should be prefered... Anyway, both work ;) — TheSilverDoe, Sep 01 '20 at 18:27

score 3 · Accepted Answer · answered Sep 02 '20 at 06:40

Although your proof is completely fine, I'll give you an alternative:

Proof : If $f(f^{-1}(Y)) = Y$ for all $Y \subseteq B$, then $f(f^{-1}(B)) = B$. Now, $$f^{-1}(B) = \{x \in A : f(x) \in B\} = A,$$ which means $f(A) = B$. Therefore, for any $b \in B \ (= f(A))$ there exists $a \in A$ such that $b = f(a)$. $\blacksquare$

Prove that a function $f: A \rightarrow B$ is surjective if $f(f^{-1}(Y)) = Y$ for all $Y \subseteq B$.

1 Answers1