prove or disprove
If $f : X → Y$ is a injective function and $f(X) = Y$, then $f^{−1}(f(\{u\})) = \{u\}\quad\forall u∈ X$.
After I worked in this statement I find it is true statement since $f(x)$ is one to one but I am not sure since I find that is difficult to prove it .
For all $x\in X$ we have $$x\in f^{-1}(f(\{u\}))\iff f(x)\in f(\{u\})\iff$$ $$\iff f(x)\in \{f(u)\}\iff f(x)=f(u)\iff $$ $$\iff x=u.$$ The last equivalence holds because $f$ is injective.
You have stated that f is injective and $f(X)=Y $ hence it is bijective. Hence, $\exists f^{-1}: Y \to X$ Satisfying, $f^{-1} \circ f= f \circ f^{-1} = Id$. Thus, you can conclude $\forall x \in X$, $f^{-1} \circ f(x) =x$ and $\forall y \in Y$, $f \circ f^{-1}(y) =y$.
This is true even without assuming $f$ is onto!
$f : X \rightarrow Y$ is injective if and only if $f : X \rightarrow f(X)$ is invertible .Therfore for all $x \in X$ we have $\{y\}=f(\{x\}) \subseteq f(X)$ equivalently $f^{-1}(\{y\}) = \{x\}$ which means $f^{-1}(f(\{x\}))=\{x\}$
