Thread: 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

I guess you mean = and usually one proves that two sets are equal if each one is contained in the other.
Yes induction is the best way to go about this.
Let
Now you know what you want to prove holds for n =2 and you are assuming by hypothesis that
so
hope this helps

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

for n=2, you proved that for
Let n be such that holds true for .
We want : for
Let
for by hypothesis


for
Now by proof for n = 2 , you know

and we know that for
combining the last two lines you get that for

Thanks so much, i get it now!