We wrote these two statements in class:
$A \subseteq f^{-1} \circ f(A)$
$ f \circ f^{-1}(B) \subseteq B$
where $A$ and $B$ are sets and $f(A)= \lbrace f(x):x \in A \rbrace $ and $f^{-1}(B)= \lbrace x:f(x) \in B \rbrace $
My question is shouldn't we have equality for both statements? And why? I drew to gain intuition but it didn't work
Any help would really be appreciated.
Thank you!
Both of these inclusions are correct, but neither can be replaced by an equality in the general case. For a counterexample, consider $f: \mathbb{N} \to \mathbb{N}$ defined as $f(n) = 1$ for every $n \in \mathbb{N}$, $A=\{1\}$ and $B=\mathbb{N}$. Both inclusions in this example are proper.
UPDATE: drawing can actually help you in a situation like that. But it also helps to keep in mind that counterexamples are more likely to be found when $f$ is not injective and/or surjective. The first inclusion would in fact be an equality if $f$ were injective, but when injectivity fails, many elements outside of $A$ can map into $f(A)$ under $f$.
In a similar fashion, if $f$ were surjective, then the second inclusion would be an equality. But if $f$ isn't surjective then it's possible to have a situation when some element in $B$ doesn't have a preimage, and then $f(f^{-1}(B))$ just cannot cover all of $B$.
UPDATE 2: one more thing that can help gain some intuition. It may be helpful to formulate what sets $f^{-1}(f(A))$ and $f(f^{-1}(B))$ are in "human" terms. Namely, $f^{-1}(f(A))$ is actually "all the elements in $A$ and also all those who are glued with them under $f$". The second set $f(f^{-1}(B))$ can be described as "all those elements in $B$ that have a preimage". If you convince yourself that these human-readable definitions are equivalent to the formal ones, then it will also become clear why exactly the inclusions in the original question are only one way and not the other.