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.

I can spot an analogy, although oversimplistic, of the identity (or 1 in the univariate case) with the wave state of a particle in quantum mechanics.

Both entities have the potential of an infinite variability of instantiations. The identity to an infinite set of f/f-inverse compositions, and the particle to a set of infinite possible trajectories.

Even more, both entities revert to some state under observation. The particle wave collapses to a value; the f-inverse is defined under an assumed f.

big help thx ðŸ™‚

lost myself at the beginning

sex? 20:27

Absolutely pointless