Characteristics of mappings

Characterize the mappings \(f:A\to B\), for which:

Variant 1
\(\forall M\subseteq A: f^{-1}(f(M))=M\)
Resolution
\(f\) must be injective.
If it is not injective, there exist \(x,x'\in A: f(x)=f(x')\). Then \(f^{-1}(f(\{x\}))\subseteq \{x,x'\} \ne \{x\}\).

In general we have \(\forall M\subseteq A: f^{-1}(f(M))\supseteq M\). If for some \(M\) there exists \(x'\in f^{-1}(f(M))\setminus M\), we can find \(x\in M\) such that \(x'\in f^{-1}(f(x))\) and then \(f(x')=f(x)\), so \(f\) is not injective.
Result
The statement holds when and only when \(f\) is injective.
Variant 2
\(\forall N\subseteq B: f(f^{-1}(N))=N\)
Resolution
\(f\) must be a surjective mapping.
If \(f\) is not surjective, there exists \(y\in B: y\notin f(A)\). Then we have \(f(f^{-1}(\{y\}))=\emptyset\ne \{y\}\).
In general \(\forall N\subseteq B: f(f^{-1}(N))\subseteq N\). If for some \(N\) there exists \(y\in N\setminus f(f^{-1}(N))\), we have \(f^{-1}(y)=\emptyset\), so \(f\) is not surjective.
Result
The statement holds when and only when \(f\) is surjective.