An unwanted property of the set $T=\{x,\{x\} \}$ Let $T$ be $T=\{x,\{x\},y \}$ and let $f:A\rightarrow T, \ f(a):=x$, where $A=\{a\}$. Define $B=\{x,y \}$. Now a weird thing happens: We should have that $x \in f(f^{-1}(B))$ by construciton of $f$, but instead we get that  $\{x\} \subseteq f(f^{-1}(B))$, althought $\{x\}$ isn't even in the image of $f$!
The argument goes like this:
$x\in B \Rightarrow f^{-1}(x) \subseteq f^{-1}(B)  $ by definition. Then  $f(f^{-1}(\{x \})) \subseteq f(f^{-1}(B))$. So far, so good. But $$f(f^{-1}(\{x \})) =\{x\} \neq x$$
by definition of the image of a set (the image of a set - even if it contains just one element is also a set - not just the element!), which "proves" the above.
My question are: Hoe to deal with this exotic set ? Should one modifiy the definition of the image to a set with one element to exlude this strange behaviour ? Or is there a better explanation ?
EDIT: I'm terrible sorry for making so many mistakes. I hope I got everything right - but if I didn't, please excuse - it is very tired and I can't focus anymore. I shall correct any mistakes I find tomorrow.

In this problem, there is a problem with the notation we use:
$f^{-1}(\{ x \})$ has two different meanings. The preimage of the element $\{ x \}$ and also the preimage of the set $\{ x \}$. But the two meanings are completely different, the "paradox" comes from the fact that during the proof we use both meanings at different spots...
$f^{-1}(\{x \}) \subseteq f^{-1}(A)$ here $\{x\}$ is a subset of $A$.
$f(f^{-1}(\{x \})) =\{x\} \neq x$ here $x$ is an element of $A$.
