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

[Scienticious]

Homework Statement

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

Homework Equations

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

 

[mfb]

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

How did your course define inverse functions?

Then f(z) is in f(x) by definition of inverse functions.

f(x) is a single value, not a set.

Since f is injective, z = x for some x in D(f).

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.

Thus z is a subset of x

x is not a set, and if you let z be an element of something, it is not a (relevant) set either.

Since f(x) = R(f) by definition

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.

 

[phoenixthoth]

How did your course define inverse functions?

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