Images of intersections and unions

Task number: 2811

For the mapping \(f: A\to B\) and \(M,M'\subseteq A;\ N,N'\subseteq B\) replace \(\Box\) with the appropriate relation \(\subseteq, =\) or \(\supseteq\). If equality does not apply, characterize the mappings for which equality holds.

Variant 1
\(f(M\cup M')\ \Box\ f(M) \cup f(M')\).
Resolution
Evidently \(f(M)\subseteq f(M\cup M')\) and \(f(M') \subseteq f(M\cup M')\), so \(f(M) \cup f(M')\subseteq f(M\cup M')\).

If \(y\in f(M\cup M')\), then there exists \(x\in M\cup M'\) such that \(f(x)=y\). For \(x\in M\) we have \(y\in f(M)\), similarly \(x\in M'\) gives \(y\in f(M')\). Given that at least one of the possibilities \(x\in M\) or \(x\in M'\) must hold, we have \(y \in f(M) \cup f(M')\), or \(f(M\cup M')\subseteq f(M) \cup f(M')\).
Result
\(f(M\cup M') = f(M) \cup f(M')\).
Variant 2
\(f(M\cap M')\ \Box\ f(M) \cap f(M')\).
Resolution
For \(x\in M\cap M'\) we have \(f(x)\in f(M)\) and \(f(x)\in f(M')\), so \(f(M\cap M') \subseteq f(M) \cap f(M')\).

Equivalence does not hold e.g. for \(f(x)=x^2\) and \(M=\{1\}\), \(M=\{-1\}\).

Equivalence holds only for injective mappings.
If \(f\) is injective and \(y\in f(M) \cap f(M')\), there exists exactly one \(x\in A\) such that \(f(x)=y\). Then we must have \(x\in M \land x\in M'\) and so \(y\in f(M\cap M')\).
Conversely, if \(f\) is not injective, we choose distinct \(x,x': f(x)=f(x')\) and we deduce that \(f(\{x\}\cap\{x'\})=f(\emptyset)=\emptyset\ne f(\{x\}) \cap f(\{x'\})\).
Result
\(f(M\cap M') \subseteq f(M) \cap f(M')\).
Equality holds exactly when \(f\) is an injective mapping.
Variant 3
\(f(M\setminus M')\ \Box\ f(M) \setminus f(M')\).
Resolution
For \(y\in f(M)\setminus f(M')\) we have both \(y\in f(M)\) and \(y\notin f(M')\). By the first of these, there exists \(x\in M\) such that \(f(x)=y\), and by the second for all \(x'\in M'\) we have \(f(x')\ne y\). Therefore \(x\in M\setminus M'\) and so \(y\in f(M\setminus M')\).

Equality does not hold for the mapping \(f(x)=x^2\) a \(M=\{1\}\), \(M=\{-1\}\), for example.

Equality holds only for injective mappings.
If equality does not hold, there exists \(y\in f(M\setminus M')\) such that \(y\notin f(M) \setminus f(M')\). Then \(y\) certainly has a preimage in \(M\setminus M'\), but also in \(M'\), and so \(f\) is not injective.
Conversely, if \(f\) is not injective, we may choose distinct \(x,x': f(x)=f(x')\) and now let \(M=\{x\}, M'=\{x'\}\), then \(f(M\setminus M')=f(\{x\})\ne \emptyset=f(M) \setminus f(M')\).
Result
\(f(M\setminus M')\supseteq f(M) \setminus f(M')\).
Equality holds only when \(f\) is injective.
Variant 4
\(f^{-1}(N\cup N')\ \Box\ f^{-1}(N) \cup f^{-1}(N')\).
Resolution
Evidently \(f^{-1}(N)\subseteq f^{-1}(N\cup N')\) and \(f^{-1}(N') \subseteq f(N\cup N')\), so \(f^{-1}(N) \cup f(N')\subseteq f^{-1}(N\cup N')\).

If \(x\in f^{-1}(N\cup N')\), then \(f(x)\in N\cup N'\) and so either \(f(x)\in N\) or \(f(x)\in N'\). In both cases we have \(x \in f^{-1}(N) \cup f^{-1}(N')\), or \(f^{-1}(N\cup N')\subseteq f^{-1}(N) \cup f^{-1}(N')\).
Result
\(f^{-1}(N\cup N')= f^{-1}(N) \cup f^{-1}(N')\).
Variant 5
\(f^{-1}(N\cap N')\ \Box\ f^{-1}(N) \cap f^{-1}(N')\).
Resolution
If \(x\in f^{-1}(N)\) and at the same time \(x\in f^{-1}(N')\), then \(f(x)\in N\cap N'\).
From this we have the inclusion \(f^{-1}(N) \cap f^{-1}(N')\subseteq f^{-1}(N\cap N')\).

If \(x \in f^{-1}(N\cap N')\), we have \(f(x)\in N\) and also \(f(x)\in N'\).
So the reverse holds as well: \(f^{-1}(N\cap N') \subseteq f^{-1}(N) \cap f^{-1}(N')\)
Result
\(f^{-1}(N\cap N')= f^{-1}(N) \cap f^{-1}(N')\).
Variant 6
\(f^{-1}(N\setminus N')\ \Box\ f^{-1}(N) \setminus f^{-1}(N')\).
Resolution
For any \(x\) it is true that \(x\in f^{-1}(N\setminus N')\) exactly when \(f(x)\in N\) and at the same time \(f(x)\notin N'\). This is equivalent to \(x\in f^{-1}(N) \setminus f^{-1}(N')\).
Result
\(f^{-1}(N\setminus N')= f^{-1}(N) \setminus f^{-1}(N')\).