Direct Image of Intersection of Sets under an Injective Function Proof

[@TheMathSorcerer]

[@krishnasingh662]

thanks man... I watch all preimage and direct image videos .... helped a lot... thanks once again

[@TheMathSorcerer]

Np glad it helped 😄

[@rikyikhwan9739]

Thanks man, super clear, highly appreciated. One question, what if the f is not injective?

[@owenallen2215]

You're doing Gods work.

[@TheMathSorcerer]

Hehe thanks

[@reinhardwilmer]

Wait so does this work for every function or only injective functions?

[@TheMathSorcerer]

equality works for injective only:)

[@wontpower]

If f were not injective, then you can only show that f(A∩B) ⊆ f(A) ∩ f(B).
For instance, consider the function R -> R f(x) = x^2, which is not injective, (i.e. you can find two different values of x that map to the same number, for example, -1 and 1). Also consider the sets A = { -1 } and B = { 1 }.
The intersection of A and B is empty, meaning f(A∩B) is also empty.
However, f(A) = { 1 } and f(B) = { 1 }. So, the intersection of f(A) and f(B) is { 1 }.
So in this case, you have a strict subset. f(A∩B) = φ ⊂ { 1 } = f(A) ∩ f(B)

[@mattetor6726]

Thank you!

[@Anteater23]

What about the converse to this?

[@tumelotlhaodi8577]

bro you just help me with part a of my problem which says, Let f : A→B be a mapping. a) Prove that if f is injective then f(C∩D)=f(C)∩f(D) for all the subsets C,D of A. b) Prove that f is injective if and only if for each subset C of A f(C)∩f(A-C)=∅. can you help me how i can approach b)

[@冈本近平]

thanks！

[@fvs3189]

Thank u