Is the solution to this problem as trivial as I think?

[PhysicsRock]

The problem goes as follows: Let $M, N$ be sets and $f : M \rightarrow N$. Further let $L \subseteq M$ and $P \subseteq N$. Then show that $L \subseteq f^{-1}(f(L))$ and $f^{-1}(f(P)) \subseteq P$.
Obviously, I would simply use the definition of a functions inverse to obtain $f^{-1}(f(L)) = L \subseteq L$ and vice versa for $P$. This seems quite trivial to me though, so am I doing this correctly or is there a mistake in my thoughts?

Thank you everyone and have a great day.

 

[fresh_42]

The problem goes as follows: Let $M, N$ be sets and $f : M \rightarrow N$. Further let $L \subseteq M$ and $P \subseteq N$. Then show that $L \subseteq f^{-1}(f(L))$ and $f^{-1}(f(P)) \subseteq P$.
Obviously, I would simply use the definition of a functions inverse to obtain $f^{-1}(f(L)) = L \subseteq L$ and vice versa for $P$. This seems quite trivial to me though, so am I doing this correctly or }1is there a mistake in my thoughts?

Thank you everyone and have a great day.

You have a typo in the $P$ statement.

$f^{-1}$ is not the inverse function. It is a set, namely $f^{-1}(L)=\{m\in M\,|\,f(m)\in L\}.$

It isn't difficult to prove these statements, but it's not about inverse functions, just preimages.

 

[PhysicsRock]

You have a typo in the $P$ statement.

$f^{-1}$ is not the inverse function. It is a set, namely $f^{-1}(L)=\{m\in M\,|\,f(m)\in L\}.$

It isn't difficult to prove these statements, but it's not about inverse functions, just preimages.

That clears things up, thank you :)

 

[fresh_42]

Here is a list of useful properties:
https://en.wikipedia.org/wiki/List_of_set_identities_and_relations#Functions_and_sets

 

[Office_Shredder]

As an example if you're still confused, $f:\mathbb{,R} \to \mathbb{R}$ defined by $f(x)=x^2$. Try computing $f^{-1}(f(L))$ for $L=[0,1]$