Proving One-to-One Function f: X -> Y

[mathboy]

[SOLVED] One-to-one function

Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?

 

[NateTG]

Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?

You need to clean up your notation. What are $X$ and $Y$? Sets?

 

[mathboy]

X is the domain of f and Y is the range of f, so they are sets. A is any subset of X. So the question is asking to prove that if f(complement of A)= complement of f(A), then f is one-to-one.

 

[morphism]

Let a be in X and set A=X-{a}. What's f(X-A) now?

 

[HallsofIvy]

You might want to try proving the contrapositive: if f is not one to one, then there exist a set A such that f(complement of A) is not complement of f(A).

If f is not one to one, then there exist x₁ and x₂ such that f(x₁)= f(x₂). What if A= {x₁}?

 

[NateTG]

Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?

In situations like this, you can always try assuming that $a \neq b$ by contradiction. The proof might not require it, but you can sometimes find direction that way.