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# Relating invertibility to being onto and one-to-one

Relating invertibility to being onto (surjective) and one-to-one (injective). Created by Sal Khan.

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• So for a function like f(x)=e^x which is a mapping from ℝ^1 to ℝ^1, Sal's argument would seem to say that e^x is not invertible because the image (range) of e^x is not the entire codomain ℝ^1.
However, we know that e^x does have an inverse, and that inverse is ln(x). Does this go against what Sal is saying?
• You got the co-domain wrong. If the domain of e^x is ℝ, then the codomain is ℝ+ (the positive real numbers).

The domain of ln x is ℝ+ and the codomain is ℝ.
• I still don't understand why f has to be surjective to be invertible.

Assume f is injective.

I understand that if the domain of f^(-1) (let's call it g) is greater than the range of f, then
f(g(y)) = y
where y is a member of g's domain is not always true; so f doesn't satisfy one of the requirements to be fully invertible.

But f being surjective means it's range has to be it's entire codomain. But why does the inverse function g's domain have to be f's codomain? Why can't it just be f's range? If we define g's domain as f's range then codomains and surjectivity (is that a word?) don't even have to enter the picture right?

Does f have to be surjective just bc by definition the domain of f^(-1) has to equal the codomain of f and not necessarily the range?
• Your last statement is correct. In general, if a function has domain A and codomain B, we want its inverse to have domain B and codomain A.

However, if we have a function f(x) with a range R that is not surjective, we can define a new function whose codomain is R, and is otherwise identical to f. These two functions are, in a very strict sense, not the same thing since their codomains are different. But we can treat them as the same for most purposes, since they do the same thing to the entire domain.

The point of this is that the new function is now invertible by this definition.
• I get why a function must be injective to be invertible, as you cannot invert from Y back to X if there are two "paths" you could potentially take. But I don't entirely understand why a function must be subjective to be invertible. It would appear to me that you could invert and revert back and forth between X and Y (using f and f^-1) all day long even if there happen to be elements in Y to which there is no mapping from X. Can somebody explain?
• Another answer Ben is that yes you can have an inverse without f being surjective, however you can only have a left inverse. A left inverse means given two functions f: X->Y and g:Y->X. g is an inverse of f but f is not an inverse of g. So take any element x in X and g(f(x)=x. But when there are elements in Y that are not mapped to from X we need to send them somewhere otherwise g is not a function(definition of a function requires that all elements in the domain map to an element in the range). So we take an element of X and map all extra elements in Y to this element in X. So our function g becomes g(y)=x when f(x)=y, which gives us our left inverse, and for any elements in Y that are not mapped to by y g(y)= our chosen element, say x0. In English g is saying "Whatever f sends us, send it right back, and if we have anything extra send that to f as well". Note how it is not the case that f(g(y))=y since g(y) could send 2 things to one element in f but f cannot send two things back because, again, this would violate the function definition. So there is no right inverse in this case.
• I am confused as to what is the objective of the video "Proof: Invertibility implies a unique solution to f(x)=y" if, as this video clarifies, that unique solution is the very definition of invertibility. Anyone could help?
(1 vote)
• being invertible is basically defined as being onto and one-to-one. theres a difference between this definition and saying that invertibility implies a unique solution to f(x)=y. also notice that being invertible really only applies to transformations in this case.
having a unique solution to f(x)=y would be a consequence of being onto and one-to-one. i didnt look at the video or anything, but im guessing he says this means it's invertible. so unique solution -> invertibility. this does not necessarily mean invertibility -> unique solution. thats what the difference is.
ok i didnt explain that very well, but you should get my idea.
• is is necessary for a function to be bijective so that the inverse exists
(1 vote)
• F(x)= -2x+4
is this one to one function and how
(1 vote)
• There's two ways of looking at whether a function is 1-1.

The easy way is to look at the graph of the function and look for places where multiple different x-values will yield the same y-value. For instance, the function f(x) = x^2 is not one to one, because x = -1 and x = 1 both yield y = 1. If you look at the graph of your function, f(x) = -2x + 4, you'll notice the graph of a function is linear. These functions are one to one by default.

Another way to see if a function is one to one is the evaluate and see if f(m) = f(n) leads to m = n. So, if f(x) = -2x + 4, then f(m) = -2m + 4 and f(n) -2n + 4.

Equating the two functions: -2m + 4 = -2n + 4, if you divide both sides by (-2) and subtract 4 from both sides, this gives you m = n. This implies that the function is indeed one to one.
• Could someone please solve this problem :
Show that the following function is bijective and find the inverse of the function
f:R-{0}->R-{0},defined by f(x)=3/x
• A function is bijective if andi only if it is invertible. Observe that g(x) := 1/x is an involution, i.e. is it's own inverse as g(g(x)) = 1/(1/x) = x. We can write f(x) as two functions - namely one that multiplies by three i.e. h(x):= 3x and g(x) := 1/x. clearly f(x) = h(g(x)) = 3/x. Finally, inverting flips the order of composition (Do you see why?) hence h(g(x))^-1 = g^-1(h^-1(x)) = g(h^-1(x)). The inverse of h(x) is given by x/3, hence the inverse function will simply be g(3x) = 1/3x.

Alternatively, you can just view 3 as multiplication and to undo the multiplication you divide, and the 1/x part is it's own inverse, so the inverse is 1/3x. I leave it as an exercise to you to show that it is indeed the inverse!
(1 vote)
• sir what if x has 2 images in the range like f[x]=x^1/2 then x has two values like for 4 it is -2 and +2
• By definition, a function maps each point in the domain to only one point in the range. √x is not a function if you insist it return the positive and negative square root.
(1 vote)
• So is f: [0, 1] ---> Real numbers ; f(x)=x^2 a injective?
(1 vote)
• Yes, the function is injective. But it would not be injective if we took the domain to include anything in [-1, 0].