Image of Pre-Image is Subset of Set

For \(f : X \to Y\) and \(B \subseteq Y\)

\[ f(f^{-1}(B)) \subseteq B\]

and in particular

\[ f(f^{-1}(B)) = B \quad \forall B \subseteq Y \iff f \ \text{is surjective}\]

Intuitively, if we take the set of things that map into \(B\) (\(f(f^{-1}(B))\)), and apply the map to them, those things will all be in \(B\). However if \(B\) contains elements which have no corresponding input under \(f\), then those elements will not be in \(f(f^{-1}(B))\).

Proof

For the first part, we have that

\[\begin{align*} x \in f(f^{-1}(B)) &\implies \exists y \in f^{-1}(B), f(y) = x \\ &\implies \exists y, f(y) \in B, f(y) = x \\ &\implies x \in B \\ \end{align*}\]

We will now prove the second result, applied to a function \(f : X \to Y\) where \(B \subseteq Y\). First, consider, by unravelling definitions that because

\[ f^{-1}(B) = \{x \in X : f(x) \in B\}\]

and

\[ f(f^{-1}(B)) = \{f(y) : y \in f^{-1}(B)\}\]

we have by substitution of the first into the second and some simplification that

\[ f(f^{-1}(B)) = \{f(y) : y \in \{x \in X : f(x) \in B\}\} = \{f(y) : y \in X \quad \text{and} \quad f(y) \in B\} = \{f(y) \in B : y \in X \}.\]

Now, if \(f\) is surjective then clearly this is equal to \(B\), as for every element in \(B\) there is a \(y\) such that \(f(y) = B\).

On the flipside, if the above is equal to \(B\) for any \(B\), then for \(B = Y\) we have that

\[ Y = \{f(y) \in Y : y \in X \} = \{f(y) : y \in X\} = f(X)\]

which is the definition of surjectivity.