Question

Suppose that f is a function from the set A to the set B. Prove that

a) If f is one-to-one, then$S_f$ is a one-to-one function from$P\left(A\right)$ to $P\left(B\right)$.

b) If f is onto function, then$S_f$ is a onto function from$P\left(A\right)$ to $P\left(B\right)$.

c) If f is onto function, then$S^{f^{-1}}$ is a one-to-one function from$P\left(B\right)$ to $P\left(A\right)$.

d) If f is one-to-one, then$S_{f^{-1}}$ is a onto function from $P\left(B\right)$to $P\left(A\right)$.

e) If f is a one-to-one correspondence, then$S_f$ is a one-to-one correspondence from$P\left(A\right)$ to$P\left(B\right)$ and$S_{f^{-1}}$ is a one-to-one correspondence from$P\left(A\right)$ to

Short answer

a)$S_f$ is a one-to-one function.

b)$S_f$ is a onto.

c) ${S_f}^{-1}$is a one-to-one

d) $S_{f^{-1}}$is a onto

e) $S_f$and $S_{f^{-1}}$are one-to-one correspondences.

Step-by-step solution

Step: 1

Suppose$S_f\left(C\right)=S_f\left(D\right)$

This implies $f\left(C\right)=f\left(D\right)$

This gives C=D

Hence$S_f$ is a one-to-one function.

Step: 2

Suppose C is an element that is the union of all elements y which have the image of an element D.

$f\left(C\right)=D$

C is a subset of A because C includes all the elements of A.

So$S_f\left(C\right)=D$

Hence$S_f$ is a onto.

$$\begin{gathered} S_{f^{-1}}\left(C\right)=S_{f^{-1}}\left(D\right) \\ {f}^{-1}\left(C\right)=f^{-1}\left(D\right) \end{gathered}$$

Hence we get

$$\begin{gathered} f^{-1}\left(f\right(E\left)\right)=f^{-1}\left(C\right) \\ {f}^{-1}\left(f\right(E\left)\right)=f^{-1}\left(D\right) \\ {f}^{-1}\left(C\right)=f^{-1}\left(D\right) \\ C=D \end{gathered}$$

$S_{f^{-1}}$is a one-to-one

Step: 3

$$\begin{gathered} X\in P\left(A\right) \\ \Rightarrow f\left(X\right)\subset B \\ \Rightarrow f\left(X\right)\subset P\left(B\right) \\ \mathrm{let}Y=f\left(X\right) \\ {f}^{-1}\left(Y\right)=f^{-1}\left(f\right(X\left)\right) \\ {f}^{-1}\left(Y\right)=X \end{gathered}$$

${S_f}^{-1}$is a onto

By the definition of one-to-one correspondence, we get that both$S_f$ and$S_{f^{-1}}$ are one-to-one correspondences.