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

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

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))))

My attempt:

a) first, g^(-1)2(B) = {x belongs toA2 : g2(x) belongs to B}

--> g^(-1)(g^(-1)2(B)) = g^(-1)1({x belongs to A2 : g2(x) belongs to B})

--> g^(-1)1({x belongs to A2 : g2(x) belongs to B}) = {y belongs to A1 : g1(y) belongs to {x belongs to A2 : g2(x) belongs to B}}

-->g^(-1)1(g^(-1)2(B)) = {y belongs to A1 : (g2 ◦ g1)^(-1)(y) belongs to B}

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

b) im 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

2. I guess you mean $\ {g_1}^{-1}( {g_2}^{-1} (B))$ = $\{ x \in A_1 : g_1(a) \in {g_2}^{-1} (B) \}$ and usually one proves that two sets are equal if each one is contained in the other.
Let $\ h = f_{n}\circ f_{n-1} .... \circ f_2 \circ f_1$
Now you know what you want to prove holds for n =2 and you are assuming by hypothesis that $\ (f_{n} \circ ... \circ f_1)^{-1} = {f_1}^{-1}(....({f_n}^{-1} (B) )$
so $\forall A \subset A_n ; h^{-1} (A) = {f_1}^{-1}(....({f_n}^{-1} (A) )$
hope this helps

3. Thanks for the reply. Im still not quite getting it. How would you start the induction?

4. for n=2, you proved that $\ (f_{2} \circ f_1)^{-1} (B) = {f_1}^{-1}({f_2}^{-1} (B) )$ for $\ B \subset A_2$
Let n $\ge 2$ be such that $(f_{n} \circ ... \circ f_1)^{-1} = {f_1}^{-1}(....({f_n}^{-1} (B) )$ holds true for $\ B \subset A_n$.
We want : $(f_{n+1} \circ ... \circ f_1)^{-1} = {f_1}^{-1}(....({f_{n+1}}^{-1} (B) )$ for $\ B \subset A_{n+1}$
Let $\ h = {f_n}\circ .... \circ {f_1}$
$\ h^{-1} (A) = {f_1}^{-1}(....({f_n}^{-1} (A) )$ for $\ A \subset A_n$ by hypothesis
$\ f_{n+1} : A_n \rightarrow A_{n+1}$
$\ f_{n+1} \circ ... \circ f_1 (a) = f_{n+1} \circ h (a)$ $\forall a \in A_1$
for $\ B \subset A_{n+1} , (f_{n+1} \circ f_{n} \circ... \circ f_1)^{-1}) (B) = (f_{n+1} \circ h)^{-1} (B)$
Now by proof for n = 2 , you know
$\ (f_{n+1} \circ h)^{-1} (B) = {h}^{-1}({f_{n+1}}^{-1} (B) )$
and we know that $\ h^{-1} (A) = {f_1}^{-1}(....({f_n}^{-1} (A) )$ for $\ A \subset A_n$
combining the last two lines you get that $(f_{n+1} \circ ... \circ f_1)^{-1} (B) = {f_1}^{-1}(....({f_{n+1}}^{-1} (B) )$ for $\ B \subset A_{n+1}$

5. Thanks so much, i get it now!