Suppose that f is an injection. Show that f-1(f(x)) = x for all x in D

1. Aug 14, 2013

Scienticious

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

Suppose that f is an injection. Show that f-1(f(x)) = x for all x in D(f).

2. Relevant equations

3. The attempt at a solution

Let z be in f-1(f(x)).
Then f(z) is in f(x) by definition of inverse functions.
Since f is injective, z = x for some x in D(f).
Thus z is a subset of x, and therefore
f-1(f(x)) is a subset of x.

Since f(x) = R(f) by definition, we have that f-1(f(x)) = f-1(R(f)).
But the range of a function is equivalent to the domain of its inverse, thus f-1(R(f)) = f-1(D(f-1)).
The range of a function's inverse is the domain of the function.
Thus f-1(D(f-1)) = D(f).
Since x is in D(f), we have that x is in f-1(x(x)).
Therefore x is a subset of f-1(f(x)).

I fairly confident that the logic of my proof works out but there are a few stylistic errors. If you guys could check over this proof I'd appreciate it :3

2. Aug 15, 2013

Staff: Mentor

f-1 should be f-1, otherwise it looks like the difference between f and 1.

How did your course define inverse functions?
f(x) is a single value, not a set.

That does not need injective functions. As z is in the image of f-1, it has to be in D(f), independent of f.
x is not a set, and if you let z be an element of something, it is not a (relevant) set either.

That does not make sense. f(x) is an element of R(f).

I could continue like that... you are confusing sets and elements of sets in such a way that it is hard to understand what you mean.

3. Aug 15, 2013

phoenixthoth

Yes, great question. This will dictate how to proceed as what you are trying to prove may basically be directly derivable from definitions.