If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Proof: Invertibility implies a unique solution to f(x)=y

Proof: Invertibility implies a unique solution to f(x)=y for all y in co-domain of f. Created by Sal Khan.

## Want to join the conversation?

• The function to square a number cannot be considered to have Invertiblity, if the domain is the set of real numbers, because its inverse, the square root of a number, maps to 2 values, one positive and the other negative. Is it necessary then to restrict the domain to only positive numbers to make it have Invertiblity? • If y=2x+1 is the original function, why is (y-1)/2=x considered the inverse? From where I sit (y-1)/2=x is the same function, just rewritten. Shouldn't the inverse function be y=(x-1)/2? • At , Sal states that Y is the codomain. Shouldn't Y be the range? Couldn't you have a matrix that doesn't span the entire codomain, but it is still invertible, and has unique mappings from domain->range and from range->domain? • For any function f: X -> Y, the set Y is called the co-domain. The subset of elements in Y that are actually associated with an x in X is called the range of f. Since in this video, f is invertible, every element in Y has an associated x, so the range is actually equal to the co-domain. So yes, Y is the co-domain as well as the range of f and you can call it by either name.
I didn't fully understand your second question, but if you're asking that can any element in the co-domain can be left out for an invertible function, then the answer is no by definition.
• Can any one explain F: X → Y is invertible ↔ ∀y∈Y :∃ unique solution to the f(x)=y in a simple layman language...language which everyone can easily relate to?! • At , I think Sal meant to say f(x) = y instead of f(x) = x. Please correct me if I am wrong. • At , why is it that if f(x)=f(a), then x has to be equal to a? If f(x)=abs(x), then both +x and -x could give the same f(x). Did I miss any point here? • Bowen,

The theorem that you quoted in your question only holds for an invertible function, and abs(x) is not invertible. In order to be invertible, every x in the domain must map to a single y in the range, AND every y in the range must be mapped to by only a SINGLE x in the domain. Basically, the function has to be "one-to-one" and "onto." The absolute value function isn't one-to-one, because although every x in the domain maps to only one y in the range, not every y in the range is mapped to by only one x.

But IF the function is invertible, then f(x)=f(a) will imply x=a.

Does that help make it make more sense?
• Just to be clear, does invertibility imply that there is a unique y that satisfies f^-1(y)=x for every x in X as well as vice versa?

Alternatively, would this condition being false imply that f is not a function? • Yes, part of the conditions to be invertible is that the function be ono-to-one, that means that for every element in the domain there is one (and only one) corresponding element in the co-domain, and vice-versa.

But just the failure to fulfil this condition is not enough to disqualify a function. For example, the function `f(x) = x²` is a valid function, but it's not one-to-one, since there are multiple values in the domain that map to the same value in the co-domain, so it's not invertible.
• If f(x)=f(a) =/=> x=a

cos(0) = cos(2pi) ... 0 =/= 2pi

At Sal alludes to this being true   