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

PhysicsRock
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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.
 
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PhysicsRock said:
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.
 
fresh_42 said:
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 :)
 
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]##
 
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