# Onto and one-to-one composite functions

1. Feb 7, 2004

### Claire84

I was wondering if someone here could help me with onto and one-to-one composite functions. I get the meanings of one-to-one and onto, but I'm just finding it hard applying them to composite functions. For instance if A,B and C are sets and f:A-B and g:B-C then if f isn't onto then is gof onto or not? Also, how can you prove this if it is true? I would have said it wouldn't be onto but I'm not sure how to prove this.

Also, if you use the same sets and g is not one-to-one, then is gof one-to-one or not? I would have said that is wouldn't be one-to-one, but I'm doubting that's correct and I'm unsure of how to prove it. Any help would be much appreciated. Thanks.

2. Feb 7, 2004

### Hurkyl

Staff Emeritus
What I do in such a situation is I try generating examples. Lots of them.

Use small sets of varying sizes for A, B, and C. (maybe, 1 to 3 elements each). Try lots of possibilities for the functions f and g. You should be able to get the whole idea after a few minutes.

3. Feb 7, 2004

### HallsofIvy

"For instance if A,B and C are sets and f:A-B and g:B-C then if f isn't onto then is gof onto or not?"

gof "Onto" ,means that, for every element, y, of C, there exist some x in A so that gof(x)= g(f(x))= y. Of course, for that to be true there would have to be some z= f(x) in B so that g(z)= y. If g is NOT "onto" then that is not true so: if g is not onto C, then gof cannot be. By the way, if g IS onto C but f is NOT onto B, gof may not be onto.

"if g is not one-to-one, then is gof one-to-one or not?"
"One to one" means: if gof(x1)= gof(x2) then x1= x2. If g is not one-to-one, there exist z1 and z2, not the same, so that g(z1)= g(z2). But now we run into a problem. If f is not ONTO, there may be no x in A so that f(x)= z1. There may only be one x that gives f(x)= z2 so that g(z1) doesn't matter.

Example: f:{a, b, c}-> {m,n,p,q} defined by f(a)= m, f(b)= n, f(c)= p. f is one-to-one but not onto.
g:{m,n,p,q}-> {x,y,z} defined by g(m)= x, g(n)= y, g(p)= z, g(q)= z. g is onto but not one-to-one.

gof:{a,b,c}->{x,y,z) is both one to one and onto:

gof(a)= g(m)= x, gof(b)= g(n)= y, gof(c)= g(p)= z.

4. Feb 7, 2004

### matt grime

Just from knowing what ONE of f and g is either a surjection or injection does not tell you anything about its composite - just let f say but the identity map, doesn't say what g is does it?

What is true, and left as an easy exercise, is that:

if f and g are BOTH injections, so is their composite.

Similarly :

if f and g are BOTH surjective so is the composite.

other things:

if g is not injective then fg is not injective.

if f is not surjective the fg is not surjective.

5. Feb 7, 2004

### phoenixthoth

this is really useful stuff when you want to start counting infinities so keep at it.

you can do things like prove N is the smallest infinity...

6. Feb 10, 2004

### Claire84

Thanks very much for all your help- it has been much-appreciated. The ideas are now a lot clearer in my head and now I'm not too sure why I was co confused in the first place. So thanks again for helping a very confused girl! :)

7. Feb 10, 2004

### matt grime

That's pretty much maths all over - it seems hard, you figure it out and you can't see how you ever thought it difficult. Just rest assured that all us other mortals feel the same thing. Don't let it put you off doing more maths.

8. Feb 10, 2004

### phoenixthoth

remember your own advice when you hit a wall yourself, as will i try to remember it. it is good advice and the way it works.

9. Feb 10, 2004

### Claire84

I'll remember that. Sometimes you just end up feeling so stupid, but I suppose that's to be expected when you're only learning something. I'm doing a Maths and Physics degree and Pure Maths is definitely the most challenging because it's a completely different way of thinking, but I'm coming round to it more and more.

One more question about inverse functions. I know about the exponential function being the inverse of the logarithmic function, but the question posed on our homework is slightly different. We have the bijection f: (o, +infinity)-(0,+infinity) defined by the expression log (1+ sqrtx). We've to find the inverse of this. So I've said we'd have exp(1+sqrtx) but for this to be bijective would the range of the inverse not have to be changed from (0, +infinity) to (e, +infinity) to make it bijective? I've tried drawing graphs of what it should be like but we haven't covered anything like it in the lectures. So far all we've been told is to find the inverser eflect the grpah in the line y=x. However, since we're not working with a function such as sinx here, do we reflect the graph in the line y=sqrtx? Btw, sqrt just means square root- I've a habit of confusing people!

10. Feb 10, 2004

### phoenixthoth

set y=log(1+srtx).
switch x and y:
x=log(1+sqrty)
solve for y.
y=the inverse.

the domain of the inverse is the range of the function and the range of the inverse is the domain of the function. this is precisely because of your switching the x and the y.

11. Feb 10, 2004

### Claire84

Oh I see where you're coming from now! Sorry to sound so dozey but we'd never had an eqt where we'd anything else than x, like we have 1+sqrtx here. That really does explain a lot!