Inverse Image(Preimage) of Intersection of Sets Proof

Inverse Image(Preimage) of Intersection of Sets Proof. Given a function from X to Y and subsets A and B of Y, in this video we prove that f^(-1)(A n B) = f^(-1)(A) n f^(-1)(B). For a subset A of Y, inverse image or preimage of A under f is the set of all elements of X which get mapped to A; i .e., f^(-1)(A) = {x in X | f(x) in A}. I hope this video helps.

[oingomoingo]

very good video
there are very few ppl who explain this with a easy speech pls make more math videos !!

[noisjang]

You've made a beautiful job... Your country must be proud of you!

[alexanderzieschang]

Why do I have to write it as x∈f(A∩B) first, to show the equality only to change it back again. Couldn't I do the same thing with f^(-1)(A∩B)?

[dariustan]

Just curious, what's stopping me from substituting your f^-1 with f and f wth f^-1 such that it proves the image of Intersection of sets is the intersection of images of sets?

[laykefindley]

What lets you just replace intersect with and? Is their a definition that is unmentioned?

[shivaprasadbs]

in this converse not true but here you show equal how is this explain sir please