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

## Video transcript

I've got a function f and it's a mapping from the set x to the set y. And let's just say for the sake of argument, let's say that f is invertible. What I want to know is what does this imply about this equation right here. The equation f of x is equal to y. I want to know that for every y that's a member of our co-domains. So for every y, let me write this down, for every y that's a member of my co-domain is there a unique, in all caps UNIQUE, solution x that's a member of our domain. Such that, and I could write such that well, I'll just write it out. I was going to write the mathy way, but I think it's nicer to write in the actual words sometimes. Such that f of x is equal to y. So if we just, let me just draw everything out a little bit. We have our set x right here. This is x. We have our co-domain y here. We know that f, if you take some point here, let's call that a, it's a member of x and you apply the function f to it, it will map you to some element in set y. So that's f of a right there. This is, so far, what this tells us. Now I want to look at this equation here, and I want to know that if I can pick any y in this set any lower-case y in this set y. So let's say I pick something here. Let's say that's b. I want to know is there a unique solution to the equation f of x is equal to b. Is there a unique solution? So one, I guess, you have to think, is there a solution. So is there a solution to saying, look is there some x here that if I apply the transformation f to it that I get there, and I also want to know is it unique. For example, if this is the only one that is unique, but it's not unique if there's some other guy, if there is more than one, solution. If there is some other guy in x that if I apply the transformation I also go to b. This would make it non-unique, not unique. So what I want to concern ourselves with in this video is somehow, is invertibility related to the idea of a unique solution to this for any y in our co-domain. So let's just work through our definitions of invertibility and see if we can get anywhere constructive. So by definition, f is invertible implies that there exists this little backward looking, three-looking thing, this means there exists. I think it's nice to be exposed sometimes to the mathy notations Let me just write that. That means there exists some function, let's call it f inverse, that's a mapping from y to x, such that, and actually the colons are also the shorthand for such that, but I'll write it out. Such that the composition of f inverse with f is equal to the identity on x. So, essentially, it's saying look, if I apply f to something in x then I apply f inverse to that, I'm going to get back to that point which is, essentially, equivalent. Or it isn't just essentially equivalent, it is equivalent to just applying the identity function, So that's i x. So you just get what you put into it. Such that this inverse function, the composition of the inverse with the function, is equal to the identity function. And that the composition of the function with the inverse function is equal to the identity function on y. So if you started y and you apply the inverse, then you apply the function to that, you're going to end up back at y at that same point. And that's equivalent to just applying the identity function. So this is what invertibility tells me, this is how I defined invertibility in the last video. Now we concerned ourselves-- so we are concerned with this equation up here. We're concerned with the equation, I will write it in pink, f of x is equal to y. And what we want to know for any y, or any lower-case cursive y in our big set y, is there a unique x solution to this. So, what we can do is we know that f is invertible. I told you that from the get-go. So given that f is invertible, we know that there is this f inverse function, and I can apply that f inverse function. It's a mapping from y to x. So, I can apply it to any element in y. So for any y, let's say that this is my y right there. So I can apply my f inverse to that y. And I'm going to go over here and, of course, y is equal to f of x. These are the exact same points. So let's apply our f inverse function to this. So if I apply the f inverse function to both sides of the equation, this right here's an element in y, and this is the same element in y. Right? They are the same element. Now, if I apply the mapping, the inverse mapping, to both of that, that's going to take me to some element in x. So let's do that. So, if I take the inverse function on both sides of this equation, where some element over here in y, and I'm taking the inverse function to get to some element in x, what's this going to be equal to? Well, on the right-hand side, we could just write the f inverse of y. That's going to be some element over here. But what does the left-hand side of this equation translate to? The definition of this inverse function is that when you take the composition with f, you're going to end up with the identity function. This is going to be equivalent to-- let me write it this way. This is equal to the composition of f inverse with f of x, which is equivalent to the identity function being applied to x. And then the identity function being applied to x is what? That's just x. This thing right here just reduces to x. This reduces to x. So, we started with the idea that f is invertible. We use the definition of invertibility that there exists this inverse function right there. And then we essentially apply the inverse function to both sides of this equation and say, look you give me any y, any lower-case cursive y in this set y, and I will find you a unique x. This is the only x that satisfies this equation. Remember how do I know it's the only x? Because this is the only possible inverse function. Only one inverse function which is true. I proved that to you in the last video. That if f is invertible, it only has one unique inverse function. We tried before to have maybe two inverse functions, but we saw they have to be the same thing. So since we only have one inverse function and it applies to anything in this big upper-case set y, we know we have a solution. And because it's only one inverse function, and functions only map to one value in this case, then we know this is a unique solution. So let's write this down. So we've established that if f is invertible, I'll do this in orange, then the equation f of x is equal to y for all. That little v that it looks like it's filled up with something, for all y, the member of our set y, has a unique solution. And that unique solution, if you really care about it, is going to be the inverse function applied to y. It might seem like a bit of a no brainer, but you can see you have to be a little bit precise about it in order to get to the point you want. Let's see if the opposite is true. Let's see if we assume-- let's see if we start from the assumption, that for all y that is a member of our set Y, that the solution, that the equation f of x, is equal to y has a unique solution. Let's assume this and see if it can get us the other way. If given this, we can prove invertibility. So let's think about the first way. So we're saying that for any y-- let me draw my sets again. So this is my set X and this is my set Y right there. Now we're working for the assumption that you can pick any element in Y right here, and then the equation right here has a unique solution. Let's call that unique solution. Well, we could call it whatever. But a unique solution x. So you can pick any point here, and I've given you, we're assuming now, that, look, you pick a point in Y, I can find you some point in X such that f of x is equal to y. And not only can I find that for you, that is a unique solution. So given that, let me define a new function. Let me define the function s. The function s is a mapping from y to x. It's a mapping from y to x and s of, let's say s of y, where, of course, y is a member of our set capital-Y. s of y is equal to the unique solution in x to f of x is equal to y. Now, you're saying, hey, Sal, it looks a little convoluted. But think about it. This is a completely valid function definition. Right? We're starting with the idea that you give me any y here. You give me any member of this set, and I can always find you a unique solution to this equation. Well OK, so that means that any guy here can be associated with a unique solution in the set X, where the unique solution is the unique solution to this equation here. So, why don't I just define a function that says, look I'm going to associate every member y with its unique solution to f of x is equal to y. That's how I'm defining this function right here. And, of course, this is a completely valid mapping from y to x. And we know that this only has one legitimate value because this, any value y, any lower-case value y, in this set has a unique solution to f of x is equal to y. So this can only equal one value. So it's well defined. So let's apply, let's take some element here, let me do a good color, let's say this is b. And b is a member of y. So let's find, so let's just map it using our new function right here. So let's take it, and map it, and this is s of b right here. s of b which is a member of x Now, we know that s of b is a unique solution by definition. I know it seems a little circular, but it's not. We know that s of b is a solution. So we know that s of b is the unique solution to f of x is equal to b. Well, if this is the case, if this is good, we just got this because this is what this function does. It maps every y to the unique solution to this equation. Because we said that every y has a unique solution. So, if this is the case, then what happens if I take f of s of b. Well, I just said this is the unique solution to this. So if I put this guy in here, what am I going to get? I'm going to get b. Or other way of saying this is that the composition of f with s applied to b is equal to b. Or another way to say it, is that we take the composition of f with s, this is the same thing, because if I apply s to b, and then I apply f back to that, that's the composition, I just get back to b. That's what's happening here. So this is the same thing as the identity function on y being applied to b. So it's equal to b. So we can say that the composition-- we can say that there exists, and we know that this function exists, or that we can always construct this. So we already know that this exists. This existed by me constructing it, but I've, hopefully, shown you that this is well defined. That from our assumption that this always has a unique solution in x for any y here, I can define this in a fairly reasonable way. So it definitely exists. And not only does it exist, but we know that the composition of f with this function that I just constructed here is equal to the identity function on y. Now let's do another little experiment. Let's take a particular-- let me just draw sets again. This is our set X. Let me take some member of set X, call it a. Let me take my set Y right there. And so we can apply the function to a and we'll get a member of set Y. Let's call that right there. Let's call that f of a right there. Now, if I apply my magic function here that always, I can give you any member of set Y and I'll give you the unique solution in X to this equation. So, let me apply that to this. Let me apply s to this. So, if I apply s to this it'll give me the unique solution. So, let me write this down. So if I apply s to this, I'm going to apply s to this-- and maybe I shouldn't point it back at that, I don't want to imply that it necessarily points back at that. So, let me apply s that, s to this. So what is this going to point to? What is that point going to be right there? So that's going to be s of this point, which is f of a, which we know is the unique solution. So, this is equal to the unique solution to the equation f of x is equal to this y right here. Or this y right here is just called f of a. Right? Remember, the mapping s just mapped you from any member of a to the unique solution to the equation f of x is equal to that. So this is the mapping from f of a to the unique, so this s of f of a is going to be a mapping to the-- or this right here, is going to be the unique solution to the equation f of x is equal to this member of y. And what's the remember y I called? It's called f of a. Well, you could go say this in a very convoluted way, but if I were to just, before you learned any linear algebra, if I said look if I have the equation f of x is equal to f of a. What is the unique solution to this equation? What does x equal? x would have to be equal to a. So the unique solution to the equation f of x is equal to f of a is equal to a. And we know that there's only one solution to that because that was one of our starting assumptions. So this thing is equal to a. Or we could write s of f of a is equal to a. Or that the composition of s with f is equal or applied to a is equal to a. Or that the composition of s with f is just the identity function on the set x. Right? This is a mapping right here from x to x. So we could write that the composition of s with f is the identity on x. So, what have we done so far? We started with the idea that you pick any y in our set capital-Y here and we're going to have a unique solution x such that this is true. Such that f of x is equal to y. That's what the assumption we started off with. We constructed this function s that immediately maps any member here with its unique solution to this equation. Fair enough. Now from that, we said this definitely exists. Not only does it exist, but we figured out that the composition of f with our constructed function is equal to the identity on the set y. And then we also learned that s-- the composition of s with f is the identity function on x. Let me write this. So we learned this, and we also learned that the composition f with s is equal to the identity on y. And s clearly exists because I constructed it, and we know it's well defined because every y-- for every y here there is a solution to this. So given that I was able to find for my function f, I was able to find a function that these two things are true. This is by definition. What it means to be invertible. Remember, so this means that f is invertible. Remember f being invertible, in order for f to be invertible, that means there must exist some function from, so if f is a mapping from to x to y, invertibility means that there must be some function f inverse that is a mapping from y to x such that, so I can write there exists a function, the inverse function composed with our function should be equal to the identity on x. And the inverse and the function in the composition of the function, with the inverse function, should be the identity on y. Well, we just found a function. It exists, and that function is s. Where both of these things are true. We can say that s is equal to f inverse. So f is definitely invertible. So, hopefully, you found this satisfying. This proof is very subtle and very nuanced because we keep bouncing between our sets X and Y. But what we've shown is that if f, in the beginning part of this video, we show that if f is invertible then there is for any y a unique solution to the equation f of x equals y. And the second part of the video, we showed that the other way is true, That if-- let me put it this way, that if for all Y, a member of capital Y, there is a unique solution to f of x is equal to x, then f is invertible. So the fact that both of these assumptions imply each other, we can write our final conclusion of the video. That f being invertible, if f, which is a mapping from x to y is invertible, this is true if and only if, and we could write that either as a two-way arrow. Or we could write if for if and only if so both of these statements imply each other. If and only if for all y, for every y that is a member of our set y there exists a unique-- I can actually write that like that means there's exists a unique x for the-- or let me write this way there exists a unique solution to the equation f of x is equal to y. So that was our big takeaway in this video. That invertibility of a function implies there's a unique solution to this equation for any y that's in the co-domain of our function.