# Thread: total, injective, surjective, and bijective functions

1. ## total, injective, surjective, and bijective functions

hello all!

i have a question here..its an exercise question from the usingz book. the question is:

We may categorise functions of
{0; 1} -> {0; 1} according to whether they are injective, surjective both. State functions which are:

(a) total functions

(b) injective but not surjective

(c) surjective but not injective

(d) bijective

for this question, how do i answer it? i don't understand wht the question is asking for. thanks in advance!!

2. Originally Posted by amaryllis
hello all!

i have a question here..its an exercise question from the usingz book. the question is:

We may categorise functions of
{0; 1} -> {0; 1} according to whether they are injective, surjective both. State functions which are:

(a) total functions

(b) injective but not surjective

(c) surjective but not injective

(d) bijective

for this question, how do i answer it? i don't understand wht the question is asking for. thanks in advance!!
Injective is also called one-to-one

A function f is said to be one-to-one, or injective, iff f(a) = f(b) implies that a=b for all a and b in the domain of f.

A function f from A to B in called onto, or surjective, iff for every element b $\epsilon$ B there is an element a $\epsilon$ A with f(a)=b.

Bijective means it's both injective and surjective.

With the iff you have to be able to prove it both ways.

I'm a little confused with the total function but I think it means that every element in the domain is defined on the function.

Hope this helps

3. ohk..so far i can understand.. so my next question is..wht is the meaning of this?
{0; 1} -> {0; 1}

4. Originally Posted by amaryllis
ohk..so far i can understand.. so my next question is..wht is the meaning of this?
{0; 1} -> {0; 1}

I'm a little confused by the semicolons, I usually have just seen commas but you should have the set {0,1} mapped to the set {0,1}.

An injection would be f(0)-->0 and f(1)-->1 but this would also be onto so this would be your bijection the way I understand this.

I had a lousy instructor and a lousy book last fall so I'm not an expert.

An example of an injection that is not onto would be the following:

{0,1}-->{0,1,2} with f(0)-->0 and f(1)-->1 so that 2 in the range does not have an element of the domain mapped to it. I'm not sure how to do this with your original problem.

5. so in the case of my example..assuming that my question is {0, 1} -> {0, 1} (comma),

my elements are only 0 and 1? and i have to answer them using these elements and define how surjective or bijective looks like?

6. Originally Posted by amaryllis
so in the case of my example..assuming that my question is {0, 1} -> {0, 1} (comma),

my elements are only 0 and 1? and i have to answer them using these elements and define how surjective or bijective looks like?

Correct, I gave you the bijective but that was the easiest one. Remember your original problem said injective and not surjective; I don't know how to do that one. Also you need surjective and not injective so what maps the first set to the second set but is not one-to-one, and every element of the range has something mapped to it?

7. Originally Posted by amaryllis
hello all!

i have a question here..its an exercise question from the usingz book. the question is:

We may categorise functions of
{0; 1} -> {0; 1} according to whether they are injective, surjective both. State functions which are:

(a) total functions

(b) injective but not surjective

(c) surjective but not injective

(d) bijective

for this question, how do i answer it? i don't understand wht the question is asking for. thanks in advance!!
I think it's pretty clear that you mean {0,1}. Right?

Going with that fix and the fact that (a) asks about total functions, most likely the question is in the context of partial functions
(which will include total functions as a subset).
Additionally, if it's about total functions exclusively, the categories (b) and (c) would be laughable.

So in this context, you should be able to show that there are 9 partial functions. And of those, 4 of them are total.

Of those 4, 2 are bijective and the other 2 are neither injective nor surjective.

Now what about the remaining 5? Clearly, none of them can be surjective. But at least four of them are injective.
What about the remaining map? Remember that, typically, the key phrase in the definition of partial function is: "at most one".
So now we have to account for a function that isn't willing to entertain sending 0 to a member of {0,1} and similarly for 1.
Where would you place this function?

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