Are these statements correct? $A \subseteq f^{-1} \circ f(A)$ and $ f \circ f^{-1}(B) \subseteq B$ 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!
 A: 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. 
