Analysis help->Suppose that g1 : A1 → A2 and g2 : A2 → A3. If B ⊂ A3, show that:

[cooljosh2k2]

Analysis help-->Suppose that g1 : A1 → A2 and g2 : A2 → A3. If B ⊂ A3, show that:

Homework Statement

Suppose that g1 : A1 → A2 and g2 : A2 → A3. If B ⊂ A3, show that
a) (g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (B)).

b) Now suppose that n ≥ 2 and gi : Ai → Ai+1 for i = 1, 2, . . . n. If B ⊂ An+1 show that
(gn ◦ gn−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)n (B))))

2. The attempt at a solution

a) first, g^(-1)2(B) = {x$$\in$$A2 : g2(x)$$\in$$B}

--> g^(-1)(g^(-1)2(B)) = g^(-1)1({x $$\in$$ A2 : g2(x) $$\in$$ B})

--> g^(-1)1({x $$\in$$ A2 : g2(x) $$\in$$ B}) = {y $$\in$$ A1 : g1(y) $$\in$$ {x $$\in$$ A2 : g2(x) $$\in$$ B}}

-->g^(-1)1(g^(-1)2(B)) = {y $$\in$$ A1 : (g2 ◦ g1)^(-1)(y) $$\in$$ B}

Which by definition = ((g2 ◦ g1)^(−1)(B)

b) I am not sure how to do part B, can anybody help me. Would i need to use induction? I apologize if this all looks confusing, i tried to make it as clear as possible.

Thanks

 

[mighty2000]

for your help!

Homework Equations

N/A

The Attempt at a Solution

a) Your solution is correct.

b) Yes, you will need to use induction. The key idea is to show that for n=2, the statement is true, and then assume that it holds for n=k and prove that it holds for n=k+1. This will show that the statement holds for all n≥2.

For n=2, we have the statement (g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (B)), which we have already shown to be true in part a.

Now, assume that the statement holds for n=k, i.e. (gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). Now, we want to prove that it holds for n=k+1. Let's call gk+1=g and Ak+1=A. Then we have gk+1 : Ak → Ak+1. Now, we can rewrite the statement as (g ◦ gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). Using the associative property of function composition, we can rewrite this as (gk+1 ◦ gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). Now, using the induction hypothesis, we can rewrite this as (gk+1 ◦ (gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1))^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). But this is just (gk+1 ◦ (gk ◦ gk−1 ◦ . . . ◦