Analysis - proof involving 1-1, image and pre image of sets

1. Jan 27, 2008

b0it0i

1. The problem statement, all variables and given/known data

Prove:
If f is injective (1-1), then f^-1 [ f(C) ] = C

2. Relevant equations
f: A -> B
C is a subset of A, and D is a subset of B

note f^-1(D) is the preimage of D in set A
and f(C) is image of C in set B

3. The attempt at a solution

My attempt:

Assume f is injective, WMST f^-1 [ f(C) ] = C

to show set equality, we have to show that the left side is a subset of the right side, and vice versa

I have already shown that C is a subset of f^-1 [ f(C) ]

my problem is showing that f^-1 [ f(C) ] is a subset of C

I assumed x is an element of f^-1 [ f(C) ]

by definition of preimage f^-1(D) = { x element of A | f(x) element of D}
hence, f(x) is an element of f(C)

and by definition of image f(C) = { f(x) | x element of C}

can i conclude that, since f(x) is an element of f(C), hence x is an element of C

therefore the proof is complete

but i believe this is wrong, since i have not used the fact that f is injective (1-1)

can anyone provide any hints?

2. Jan 27, 2008

morphism

This is precisely where you use the fact that f is 1-1!

3. Jan 27, 2008

b0it0i

^^^

just by stating the definition of 1-1

for all x1, for all x2, if x1 does not equal x2, then f(x1) does not equal f(x2)
or the contrapositive form, if f(x1) = f(x2), then x1 = x2

would i need to introduce two points x1 and x2
where x1 is an element of A, and x2 is an element of C?

i'm not really sure how to use the fact f is 1-1

4. Jan 27, 2008

morphism

If f(x) is in f(C), then this means there is an x' in C such that f(x)=f(x').

5. Jan 27, 2008

b0it0i

I see!

that makes sense

i should have looked at the definition of image more closely

thanks a lot