Understanding a projection function from S to B.

Question

Let A and B be sets and $S \subseteq A \times B $ . Let $\pi_{1}$ be the projection function on $S$ to $A$ and $\pi_{2}$ be the projection function on $S$ to $B$. Give an example to show that.

$\pi_{2}$ need not be onto $B.$

Proof:

Let $A = \Bbb R, B = \Bbb R,S = \{(x,y) \in \Bbb R \times \Bbb R ,y=x^2 \}$

Then $\; y \ge 0$.

Would this not be onto because$ y = x,$ and $y=-x$. Then would it not be a proper mapping?

I don't quite understand the question. Can't you just take $S = \varnothing$ or similar? — Caleb Stanford, Jul 17 '16 at 00:22
You already got an answer to your question in the answer below. — Caleb Stanford, Jul 17 '16 at 00:33

score 2 · Accepted Answer · answered Jul 17 '16 at 00:22

2

The projection is always well defined (a proper mapping). It is not clear what "it" is in your last statement, but no, what you are writing has nothing to do with it. $\pi_2$ is not onto because none of the negative values in $B$ are in the image of S.

A much simpler example: take any $A$, a non-empty $B$, and $S$ be the empty subset of $A \times B$. The projection restricted to $S$ has empty image, so it cannot be onto.

answered Jul 17 '16 at 00:22

So to show this symbolically. Let $A \land ; B = \text{nonempty} $ $S = \emptyset$ – Jon Jul 17 '16 at 00:44
Why do you need to show it symbolically? What does that add to anything? Also, $A \ne \emptyset$ is not required. – Jul 17 '16 at 00:46
I am trying to visualize it. – Jon Jul 17 '16 at 00:47

Understanding a projection function from S to B.

1 Answers1