Image of Pre-Image is Subset of Set

Image of Pre-Image is Subset of Set

For f:X→Y and B⊆Y

f(f−1(B))⊆B

and in particular

f(f−1(B))=B∀B⊆Y⟺f 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

x∈f(f−1(B))⟹∃y∈f−1(B),f(y)=x⟹∃y,f(y)∈B,f(y)=x⟹x∈B

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

f−1(B)={x∈X:f(x)∈B}

and

f(f−1(B))={f(y):y∈f−1(B)}

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

f(f−1(B))={f(y):y∈{x∈X:f(x)∈B}}={f(y):y∈Xandf(y)∈B}={f(y)∈B:y∈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)∈Y:y∈X}={f(y):y∈X}=f(X)

which is the definition of surjectivity.