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

## Video transcript

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.