Inverse image of function

Question

Let $f:A\to B$ be a function such that

$A= \{1,2,3\}, B=\{1,2,3\}$

$f(1)=1 , f(2)=1, f(3)=2$

Then is the inverse image of $f^{-1}(B) = \{1,2,3\}$?

And, $f^{-1}(1) = \{1, 2\}$ but is it okay because it is not a function?

and then the inverse image of the (well-defined)function is always domain of $f$?

I'm confused...

Answer 1

Your first question is no. Since $f$ is not one-to-one, the inverse of $f$ does not exist.

However for your second question is yes.
Since for a function, every object will have an image. So the pre-image of $B$ under function $f$,denoted by $f^{-1}(B)$ is always equal to $A$.