Question

1.            Determine whether the function f from { a, b, c, d } to {a, b, c, d, e} is injective (one-to-one), surjective (onto) and/or bijective (one-to- one correspondence) :

f(a) = a,            f(b) = c,            f(c) = b, f(d) = e

a. Is this function injective?              . surjective?              . bijective?              .

If your answer is no for any of the above, explain:

            b. Is there an inverse for this function?              .

c. Is the composition f ° f always defined?              . Explain:  

2. Determine whether the function from { a, b, c, d } to itself is injective (one-to-one), surjective (onto) and/or bijective (one-to- one correspondence) :

f(a) = a,            f(b) = c, f(c) = b, f(d) = d

a. Is this function injective?              . surjective?              . bijective?              .

If your answer is no for any of the above, explain:

            b. Is there an inverse for this function?              .

c. Is the composition f ° f always defined?              .

3. Determine whether the function from { a, b, c, d } to {a, b, c} is injective (one-to-one), surjective (onto) and/or bijective (one-to- one correspondence) :

f(a) = a,            f(b) = c, f(c) = b, f(d) = a

a. Is this function injective?              . surjective?              . bijective?              .

If your answer is no for any of the above, explain:

b. Is there an inverse for this function?              .

c. Is the composition f ° f always defined?              .

Answer 1

It's an injective function as we can confer from the figure above . It's not surjective because for d in co-domain there is no element in the domain.And for a function to be bijective,it should be injective as well as surjective.

1b.) No,an injective function doesn't have an inverse.If you make it inverse, the current co-domain will be the domain and the current domain will be changed to co-domain. So as there can be no element in the domain of a function without being related to an element of the co-domain, it will not be a function.

1c.)No ,the composition is not always defined because in case f(f(d))=

F(d)=e. (given) .On replacing we get -> f(f(d))=f(e)

And f(e) is not defined.so composition can't be defined.

​​​​​​

It is a bijective function(both injective and surjective).

2b.)Yes ,there is an inverse for a bijective function.

2c.) Yes composition will be defined in this case.

It is a surjective function (onto) as for every element y in co-domain ,there is at least one element in the domain.It is not an injective function because a has more than one element in the domain.

1b.)No it doesn't have a inverse

1c.)Yes composition is defined.