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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!

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  • $\begingroup$ 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. $\endgroup$ – amsmath Mar 24 at 2:59
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As amsmath mentioned, it basically follows from the definition of the corresponding projection maps.

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.

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