Maps in Discrete Structures?

Question

Let S,T, and U be sets. Let e be a mpa from S to T, let f be a map from T to U.

Show that, if e and f are injective, then so is ef.
Show that, if e and f are surjective, then so is ef.
Show that, if e and f are bijective, then so is ef.

I have no clue where to start. I do know that if e and f are injective, then so is ef because its a rule, but I'm not sure how to prove that.

thank you for that encouraging comment.

Yes, I do know what injective, surjective and bijective means. I was asking for a little help. If you can't help out, skip over my questions.

If it's so trival, please enlighten me

I am not a math major. Im a CIS major and are required to take this class. We do not have a professor to show up the problems or to ask questions to.

I know that ef is injective, surjective and bijective because e and f are injective, surjective and bijective. I need to write a proof about it.

Andy, I praise you for your help.

**I need to explain why ef is injective, surjective and bijective. I know it is as per definition. but why

Answer 1

*sigh* The poster above me is an idiot. You need to make use
of the definitions...

Volter the point of asking a question is to learn. I'm here to help him learn, what exactly is it you are here for? Whatever it is you're here for, you're missing the point and that's cuz you're an idiot as I stated previously. At the very least I'm a well-intentioned idiot.

What does it mean for e to be injective?
e : S→T a,bεS and e(a),e(b)εT
e(a)=e(b) implies a=b

What does it mean for e to surjective?
e : S→T aεS and bεT
For every bεT there exists an aεS such that e(a)=b

What does it mean for e to bijective?
It means it is both injective and surjective (above)

ef is invalid e S→T and f T→U you need a chain S→T→U
you must mean fe.

Suppose e and f are injective...

e(a)=e(b) implies a=b
f(c)=f(d) implies c=d
So if g,hεS
f(e(g))=f(e(h))
Implies e(g)=e(h) since f is injective
Implies g=h since e is injective
Thus fe is shown to be injective

Suppose e and f are surjective...

For every bεT there exists an aεS such that e(a)=b
For every dεU there exists an cεT such that f(c)=d

Now suppose gεU there exists an hεT such that f(h)=g since f is surjective
There also exists an iεS such that e(i)=h since e is surjective
So there exists an iεS such that f(h)=f(e(i))=g
Thus fe is shown to be surjective

The third statement is trivial (note the proper use of the word trivial here Volter) from the first two statements.

Answer 2

Do you know what injective surjective and bijective mean?

Your clearly in a math class... but i wonder why since you have no friggin clue if you cant answer a trivial question like this.

EDIT

Andy your the idiot. If he cant do these questions hes out of his place. This is a TRIVIAL question asking one to use the definition

EDIT

The intention is to learn. This isnt an elementary school question. Based on the question he is at the point where he should be doing this himself. I think hes getting you to do his homework.

I do help people who are obviously getting homework help... but why is he in this class at all when he clearly has no clue? Perhaps my tone was too strong. But the point stands... at the university level (1st or 2nd year i presume depending on the nature of the course) he should be doing this himself.

IT IS SO FRIGGIN OBVIOUS THAT TO CHECK IT HE SIMPLY HAS TO CHECK WHETHER THE COMPOSITION IS IN/SUR/BI -JECTIVE AND USE THE PROPERTY THAT THE ORIGINALS ARE.

Again i ask why is he in this course?!?