# Prove that $(\pi_1)^{−1}(A) = A × Y$ and $(\pi_2)^{−1}(B) = X × B.$

Let $$X$$ and $$Y$$ be sets, let $$A \subset X$$ and $$B \subset Y$$ be subsets and let $$\pi_1: X\times Y \to X$$ and $$\pi_2: X\times Y \to Y$$ be projection maps. Prove that $$(\pi_1)^{−1}(A) = A \times Y$$ and $$(\pi_2)^{−1}(B) = X \times B.$$

So I know to solve something is equal to something else, I need to show that something is in both of these things. I'm just not sure how to do this. Could someone also explain what a projection map is? Thanks!

• Instead of "be projection maps" your teacher should have written "be the corresponding projection maps". You have $\pi_1(x,y) = x$ and $\pi_2(x,y) = y$. It's as simple as that. Now, you should be able to prove the claims. – amsmath Mar 24 at 2:59

The projection maps are defined as $$\pi_1: X \times Y \to X, (x,y) \mapsto x$$ and $$\pi_2: X \times Y \to Y, (x,y) \mapsto y$$
The preimage of $$\pi_1(A)$$ is $$\pi_1^{-1}(A) := \{(x,y) \in X\times Y \mid \pi_1(x,y)\in A\}$$
which is equivalent to $$A\times Y$$ given the definition of $$\pi_1$$.
Proving $$\pi_2^{-1}(B) = X \times B$$ works equally.