The image of the preimage of a set is contained in the set

up:: Image of Function

For a given set $B\subseteq Y$, not all of its points are reached by $f:X\to Y$. Thus, premapping from $B$ then mapping back via $f$ may yield a smaller set than $B$.

Let $f:X\to Y$ be a function, and $B\subseteq Y$ a subset.

Let $y\in f(f^{-1}(B))$. Then, by the definition of Image of Function, we have that

$$\exists x\in f^{-1}(B)\mid f(x)=y$$

By the definition of Preimage of Function, we have that this implies that $f(x)=y\in B$.

Counterexample of equality

Note that $y\in B$ need not imply that $f^{-1}(y)\subseteq X$: for all points $c\in Y\setminus f(X)$ we have $f^{-1}(c)=\varnothing$. That is why The surjective image of the preimage of a set is equal to the set.

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References

• Notas para um Curso de Física-Matemática, João Carlos Alves Barata. Cap. 01>)