Proving a surjective map iff the map of the inverse image is itself

[lpo]

In the recommended format :)

Homework Statement

First we say that f:S→T is a map. If Y ⊆ T and we define f^-1(Y) to be the largest subset of S which f maps to Y:
f^-1(Y) = {x:x ∈ S and f(x) ∈ Y}

I must prove that f[f^-1(Y)] = Y for every subset Y of T if, and only if, T = f(S).

Homework Equations

f^-1(Y) = {x:x ∈ S and f(x) ∈ Y}

f[f^-1(Y)] = Y for every subset Y of T ⇔ T = f(S)

The Attempt at a Solution

So I understand the idea that f must be a surjective map from S to T because Y can be any subset of T so any inverse image of Y would need to map back to S. I also understand that if the map f of the inverse image of Y equals Y then f must be surjective from S to T. I drew the nice bubbles and arrows to help me out also but am having trouble proving this from a series of logical steps.

(Symbols used: ∀ - for all, ∃ - there is, ∈ - is an element of, ⊆ - is a subset of)

First I attempt to prove from right to left so:
Assume f(S) = T meaning f is surjective
∀ t ∈ T ∃ at least one x ∈ S such that f(x) = t
then by definition x ∈ f^-1(Y)
since the map is surjective then f^-1(Y) = S and f(f^-1(Y)) = f(S) = T = Y
∀ f(f^-1(Y)) = Y

Now I attempt from left to right:
Assume f(f^-1(Y)) = Y
let y ∈ Y so y ∈ f(f^-1(y))
∃ x ∈ f^-1(y) such that f(x) = y ∈ Y ∀ Y ⊆ T
∀ y ∈ Y ∃ at least one x ∈ S such that f(x) = y meaning the map is surjective
f(S) = T

I don't feel like these are fully correct, there feels like there is some idea or connection I am missing. 4. Note

I live about an hour away from school and don't have any study partners to speak of and the professor is very very difficult to understand so any help and guidance would be extremely appreciated. Especially describing your thought process tackling this problem :)

Thanks!

 

[mighty2000]

Thank you for posting your question regarding the proof of f[f-1(Y)] = Y for every subset Y of T if, and only if, T = f(S). I am happy to assist you in understanding this concept.

First, let's clarify the notation used in the question. The symbol "⇔" means "if and only if", which indicates a two-way implication. Also, the symbol "∀" means "for all" and "∃" means "there exists". The symbol "∈" means "is an element of" and "⊆" means "is a subset of".

Now, let's break down the proof step by step. To prove the statement, we need to show that f[f-1(Y)] = Y for every subset Y of T if, and only if, T = f(S).

From right to left:
Assuming T = f(S), we know that f is a surjective map because for every t ∈ T, there exists at least one x ∈ S such that f(x) = t. This means that every element in T has a preimage in S. Now, we can say that for any subset Y ⊆ T, the inverse image of Y, f-1(Y), will be a subset of S. This is because f-1(Y) is defined as the set of elements in S that are mapped to Y by f. Therefore, f[f-1(Y)] will also be a subset of T since it is the image of f-1(Y). However, since T = f(S), we can conclude that f[f-1(Y)] = Y for every subset Y of T.

From left to right:
Assuming f[f-1(Y)] = Y for every subset Y of T, we know that f[f-1(T)] = T. This is because T is a subset of itself and f[f-1(T)] is the image of the inverse image of T, which is T itself. This implies that every element in T has a preimage in S, meaning that f is a surjective map. Therefore, T = f(S).

As you can see, both directions of the proof rely on the fact that f is a surjective map. This is because for every element in T, there must be a preimage in S for the equality to hold. I hope this explanation has helped you understand