Welcome to a lesson
on determining whether a function is only injective,
only surjective, bijective
or neither injective nor surjective.
For a quick review, a function is injective or one-to-one
provided every element of the codomain
is the image of at most one element from the domain,
which means injective functions do not have
repeated elements from the codomain in the range,
but may have missing elements.
A function is surjective or onto,
provided every element of the codomain
is the image of at least one element from the domain,
which means surjective functions do not miss elements
from the codomain in the range,
but may have repeated elements.
Looking at the first example,
the domain is a set with elements
one, two, three, four, five,
and the codomain is a set with elements
one, two, and three.
We are given the function in two-line notation,
where in the first row of the matrix we have
the elements from the domain or the inputs.
And the second row we have the range or outputs.
Remember, the range is a set of actual outputs
and the codomain is a set of possible outputs.
So first notice all the elements in the codomain
do appear in the range.
And therefore the function is surjective,
since surjective functions do not miss elements
from the codomain in the range,
but may have repeated elements.
Notice in this case the elements of one and two
from the codomain are repeated,
which is allowed for a surjective function.
However, the function is not injective
because injective functions do not have repeated elements
from the codomain in the range.
So again, the function is not injective
because the elements of one and two
from the codomain are repeated,
or more formally, the function is not injective
because every element of the codomain
is not an image of that most one element from the domain,
and that's because of the elements of one and two,
and the function is surjective
because every element of the codomain is an image
of at least one element from the domain.
The elements of one and two
are images of two elements from the domain,
and the element of three from the codomain
is an image of one element from the domain.
The function is only surjective.
The second function, the domain is a set with elements
one, two, three and four, and the codomain
is a set with elements, one, two, three, four and five.
Because there are more elements in the codomain
than there are in the domain,
we are going to have missing elements
from the codomain in the range.
Notice the range is missing the element of three
from the codomain, which means the function
is not surjective, again because surjective functions
do not miss elements from the codomain in the range,
or more formally, the function is not surjective
because every element of the codomain
is not the image of at least one element from the domain.
Again, because the element of three from the codomain
is missing from the range.
However, the function is injective
because none of the elements from the codomain
in the range are repeated.
And again, injective functions do not have repeated elements
from the codomain in the range,
but may have missing elements.
In our case, the range is missing the element of three,
but it does still satisfy the requirement
for an injective function.
The given function is only injective.
Again, every element in the codomain is the image
of at most one element from the domain.
The elements of one, two, five and four from the codomain
are images of exactly one element from the domain,
and the element of three from the codomain
is an image of zero elements from the domain,
or is not an image at all.
Next, we have a function where the domain is a set
with elements one, two, three and four.
And the codomain is a set with elements
one, two, three, four and five.
Again, because there are more elements in the codomain
than there are in the domain,
there's no way the function can be surjective
because for a surjective function,
every element of the codomain is the image
of at least one element from the domain.
Notice the elements of four and five from the codomain
are missing from the range,
and therefore the elements of four and five in the codomain
are not an image of at least one element from the domain.
Or again, it's not surjective
because subjective functions do not miss elements
from the codomain in the range.
And again, we do have the missing elements
of four and five from the codomain in the range.
And this function is also not injective
because notice how the element of two from the codomain
is repeated in the range.
And injective function do not have repeated elements
from the codomain in the range, or more formally,
it's not injective because every element
of the codomain is not an image
of at most one element from the domain,
again because of the element of two from the codomain.
So this function is neither injective nor subjective.
And for the last example, the domain is a set with elements
one, two, three and four, and the codomain is the same.
Notice all the elements in the codomain do appear
in the range, and therefore the function is surjective.
Again, surjective functions do not miss elements
from the codomain in the range, or more formally,
every element of the codomain is an image
of at least one element from the domain.
In this case, all the elements from the codomain
are images of exactly one element from the domain.
And because none of the elements in the codomain
are repeated in the range, the function is also injective.
Again, injective functions do not have repeated elements
from the codomain in the range, or more formally,
the function is injective because every element
of the codomain is an image
of at most one element from the domain.
So because this function is injective and surjective,
we say the function is bijective.
A function is bijective when every element of the codomain
is an image of exactly one element from the domain.
I hope you found this helpful.