# Image and Inverse Images for Projection Functions

• Apr 20th 2010, 01:05 PM
onemore
Image and Inverse Images for Projection Functions
Let A and B be nonempty sets and $\displaystyle \pi_1 : A \times B \rightarrow A$ be the projection function $\displaystyle \pi_1 (x,y) = x$ for $\displaystyle (x,y) \epsilon A \times B$.

(a) For $\displaystyle C \subset A$, determine the inverse image set $\displaystyle \pi^{-1} (C)$.

(b) For $\displaystyle C \subset A, D \subset B$ and $\displaystyle D \neq \emptyset$, determine the image $\displaystyle \pi_1 (C \times D)$.
• Apr 20th 2010, 09:56 PM
kompik
Quote:

Originally Posted by onemore
Let A and B be nonempty sets and $\displaystyle \pi_1 : A \times B \rightarrow A$ be the projection function $\displaystyle \pi_1 (x,y) = x$ for $\displaystyle (x,y) \epsilon A \times B$.

(a) For $\displaystyle C \subset A$, determine the inverse image set $\displaystyle \pi^{-1} (C)$.

(b) For $\displaystyle C \subset A, D \subset B$ and $\displaystyle D \neq \emptyset$, determine the image $\displaystyle \pi_1 (C \times D)$.

You probably mean $\displaystyle \pi_1^{-1} (C)$, not $\displaystyle \pi^{-1} (C)$. Have you at least tried to solve this problem?

I thinkt the if you understand the definitions and if you draw a picture or think for a while, you see immediately that:
$\displaystyle \pi_1^{-1} (C) = C\times B$and
$\displaystyle \pi_1 (C\times D) = C$.

One question, to see whether you understand the problem - can you explaine why $\displaystyle D \ne \emptyset$ is needed in (b)?