Relating invertibility to being onto and one-to-one

Linear algebra

Inverse functions and transformations

Introduction to the inverse of a function
Proof: Invertibility implies a unique solution to f(x)=y
Surjective (onto) and Injective (one-to-one) functions
Relating invertibility to being onto and one-to-one
Determining whether a transformation is onto
Exploring the solution set of Ax=b
Matrix condition for one-to-one trans
Simplifying conditions for invertibility
Showing that Inverses are Linear

Relating invertibility to being onto and one-to-one Relating invertibility to being onto (surjective) and one-to-one (injective)

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Relating invertibility to being onto and one-to-one

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A couple of videos ago, we learned that a function that
is a mapping from the set x to the set y is invertible if and
only if-- and I'll write that as a if with two f's --if and
only if for every y-- so let me write this down.
I'll do this in yellow.
For every y that is a member of our co-domain, there exists
a unique-- and I'll make that a little bit bold --a unique x
that is a member of our domain, such that f of this x
is equal to this y.
So that's just saying that if I take my domain right here,
that's x, and then I take a co-domain here, that is y, we
say that the function f is invertible.
And we know what invertibility means.
It means that there's this other function called the
inverse that can essentially, if you take that in
composition with f, it's like taking the identity on x.
Or if you take f in composition with it, it's like
taking the identity on y.
And we've done that multiple times so I
won't repeat that there.
We know what invertibility means, but we learned that
it's invertible, if and only if, for every y here, so you
take any y here, any y that's a member of your co-domain,
there exists a unique x, such that f of x is
equal to that y.
Let me write it this way.
If this is an x, let's say that's an x not, f of x not
would be equal to y.
So this y would be equal to f of x not.
You apply the function here.
It's going to map it to this point here.
It wouldn't be invertible if you had this.
If you had two members of x mapping here.
That would break invertibility if you had this situation,
because then you wouldn't have the unique condition.
You have to have a unique x that maps to this thing.
And what I just drew here, this other pink mapping, we
don't have a unique x that maps to y, we have two x's
that map to y.
Now, based on what I just told you on that last video, what
does this mean?
If we have a unique x that maps to each y?
That means that we have to have a one-to-one mapping.
That f has to be one-to-one.
Let me write that.
Another way of saying this, is that f is
one-to-one, or injective.
If we have two guys mapping to the same y, that would break
down this condition.
We wouldn't be one-to-one and we couldn't say that there
exists a unique x solution to this equation right here.
Now, the other part of this is that for every y -- you could
pick any y here and there exists a unique x
that maps to that.
There cannot be some y here.
Let's say that there's some y here, and no one maps to that.
If that's the case, then we don't have our conditions for
invertibility.
So that would be not invertible.
Everything in y, every element of y, has to be mapped to.
All of these guys have to be mapped to.
And they can only be mapped to by one of the elements of x.
Everything here has to be mapped to by a unique guy.
Now, in the last video, what did we call it went a function
maps to every element of your co-domain?
So this every here, what is another way of saying that?
That a function maps to every element of your co-domain?
On the last video I explained that notion is called a
surjective or onto function.
So the whole reason why I'm doing this video is because I
really just want to restate this condition for
invertibility using the vocabulary that I introduced
to you in the last video.
So given that the statement for every y that's a member of
our co-domain, there exists a x that maps to it.
We could just say that f is surjective.
If we just say that f is surjective, that means that
everything here is mapped to.
But it could be mapped to, maybe this person right here,
could be mapped to by more than one.
Surjective doesn't, by itself, make the condition that
there's a unique mapping from a member of x to
that element of y.
So in order to get that, in order to satisfy the unique
condition of this condition for invertibility, we have to
say that f is also injective.
And obviously, maybe the less formal terms for either of
these, you call this onto, and you could call this
one-to-one.
So using the terminology that we learned in the last video,
we can restate this condition for invertibility.
We can say that a function that is a mapping from the
domain x to the co-domain y is invertible, if and only if --
I'll write it out -- f is both surjective and injective.
Or we could have said, that f is invertible, if and only if,
f is onto and one-to-one.
And these are really just fancy ways of saying for every
y in our co-domain, there's a unique x that f maps to it.
There isn't more than one and every y does get mapped to.

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