# Thread: topology - the lift of a function f

1. ## topology - the lift of a function f

Let X,Y be subsets of the universal set S. Suppose f:X to Y is a function. Define the lift of f to 2^X , F:2^X to 2^Y by F(A)=f(A), A in 2^X. Show the following:

a) F is one-to-one if and only if f is one-to-one.
b)F is onto if and only if f is onto.

I know that 2^X is the power set of X, meaning that 2^X is the set of all subsets of X. Similarly, 2^Y is the set of all subsets of X. But I have no experience with lifts at all. To prove a), I will first assume that F is one-to-one. Then for a1,a2 in 2^X F(a1)=F(a2) implies a1=a2. But I don't understand how the lift is relating to f, so I cannot complete the proof. Thank you for your help.

2. Originally Posted by poXGxxi
Let X,Y be subsets of the universal set S. Suppose f:X to Y is a function. Define the lift of f to 2^X , F:2^X to 2^Y by F(A)=f(A), A in 2^X. Show the following:

a) F is one-to-one if and only if f is one-to-one.
Let $A\subseteq X$ define $f[A]=\{ y\in Y | y=f(x) \text{ for some }x\in A \}$. I do not like to use $2^X$ because it does not usually mean the power set, instead I shall use $\mathcal{P}(X)$ for the power set. Now define $F: \mathcal{P}(X) \to \mathcal{P}(Y)$ by $F(A) = f[A]$. Assume that $f$ is one-to-one and $F(A) = F(B)$ then $f[A] = f[ B ]$. To show $F$ is one-to-one we need to show $A=B$. Thus, let $a\in A$ then $f(a)\in f[A]$. Since $f[A] = f[ B ]$ it means $f(a) \in f[ B ]$ and so by definition $f(a) = f( b )$ for some $b\in B$. However, $f$ is one-to-one so $a=b\in B$. Thus, $A\subseteq B$, repeating the mirror argument shows $B\subseteq A$, which means $A=B$. Let us summarize what we did: we showed that if $f$ is one-to-one then $F$ is one-to-one.

Conversely, say $F$ is one-to-one, i.e. if $F(A)=F(B)$ then $A=B$. We want to show $f$ is one-to-one. Say that $f(a) = f(b)$. Now form the sets $\{ a \}$ and $\{ b \}$ which are subsets of $A$. Note $F( \{a\}) = \{ f(a)\}$ and $F(\{b\}) = \{ f(b) \}$. Therefore $F( \{a \}) = F(\{ b\})$ but by above assumptions it means $\{ a \} = \{ b \}$ and so $a=b$. Let us summarize what we did: we showed that if $F$ is one-to-one then $f$ is one-to-one.

Putting these together it means $f$ is one-to-one if and only if $F$ is one-to-one.

You try the second part.