Question

f: A→B, g :B→A

Is g ° f defined? If so, what is its domain and range?

Any suggestions?

Answer 1

Since
f: A->B, the domain of f is A, its range is B.
g: B->A, the domain of g is B, its range is A.

g°f = g(f) : A -> A

To illustrate, draw the arrow diagrams of f and g, with the range of f and domain of g coinciding a B.

Alternatively, see the arrow diagram of the following link, where X=A, Y=B, and Z is again A.
http://en.wikipedia.org/wiki/Function_composition

Answer 2

Can you give an example using numbers?

Answer 3

f: A{2,4,6} -> B(21,42,52)
(2,21),(4,42),(6,52)

g: B(21,42,52) -> B(2,4,6)
(21,4),(42,2),(52,6)

(g°f)(2)=g(f(2))=g(21)=4
(g°f)(4)=g(f(4))=g(42)=2
(g°f)(6)=g(f(6))=g(52)=6

So both f and g are one-to-one and onto (i.e. injective and surjective), therefore bijective, so g°f exists.

However, you will notice that g does not have to be the inverse of f for g°f to exist. In the example given, g is <i>not</i> the inverse of f.

Answer 4

Sorry, typos in the definitions of f and g, although still understandable, should read:
f: A{2,4,6} -> B{21,42,52}
mapping: (2,21),(4,42),(6,52)

g: B{21,42,52} -> A{2,4,6}
mapping: (21,4),(42,2),(52,6)

Answer 5

So, if A = (2,4,6) and B = (21,42,52)

g : B -> A, wouldn't g = {(21,2), (42,4),(52,6)}? Is g ° f defined because it is one-to-one?

Answer 6

There is no limit on what g can be. It could be
g = {(21,4),(42,2),(52,6)}, or
g = {(21,2), (42,4),(52,6)}
In the latter case, it "happens" to be the inverse of f, but it does not have to be.

However, if g is to be the inverse of f, then f <i>must</b> be bijective.</i>

Answer 7

So, if A = (1, 2, 3) and B = (4, 5, 6)

f: A -> B => {(1, 4), (2, 5), (3, 6)}
g: B ->A => {(4,1), (5,2), (3,6)}

g ° f = {(4,4), (5,5), (6,6), so g ° f = B right? And the domain and range are equal. But I'm still sure how it is defined?

When finding the composite of a function you are suppose to match the range to the domain. If you go to photobucket and type in flutegirl516 in the search bar there is a picture that explains it better. It's the photo with the colorful lines. . .

Answer 8

I believe we have to set something straight: namely the definition of g°f, which is the main cause of confusion.

At the end of this post, there are two references which define the composition operator:
(g°f)(x)=g(f(x)), or equivalently
(f°g)(x)=f(g(x))

They amount to the same thing: evaluate the second function first, feed the result to the first and evaluate.

In your example:
f: A -> B => {(1, 4), (2, 5), (3, 6)}
g: B ->A => {(4,1), (5,2), <i>(6,3)</i>}
(note correction for g.)
where A={1,2,3}, and B={4,5,6}.
so f:A->B, and g: B->A
therefore
(g°f)(x)=g(f(x)) => g°f: A->A
as in:
(g°f)(1)=(g(f(1))=g(4)=1
(g°f)(2)=(g(f(2))=g(5)=2
(g°f)(3)=(g(f(3))=g(6)=3
(and not B->B)

I do not know the source of the aforementioned picture (with coloured lines) in which the bottom line is incorrect: it should read f°g. This is probably why you get confused.

For the given functions,
f: {1,2,3,4}->{1,2,4}={(1,2),(2,1),(3,1),(4,4)}
g: {1,2,3,4}->{1,2,3,4}={(1,2),(2,4),(3,1),(4,3)}

Note that
dom f={1,2,3,4} range f={1,2,4}
dom g={1,2,3,4} range f={1,2,3,4}
so it is not exactly comparable to the example of
g: A->B, and f:B->A.

Let's evaluate the function g°f:
(g°f)(1)=g(f(1))=g(2)=4
(g°f)(2)=g(f(2))=g(1)=2
(g°f)(3)=g(f(3))=g(1)=2
(g°f)(4)=g(f(4))=g(4)=3

and f°g:
(f°g)(1)=f(g(1))=g(2)=1
(f°g)(2)=f(g(2))=g(4)=4
(f°g)(3)=f(g(3))=g(1)=2
(f°g)(4)=f(g(4))=g(3)=1

which clearly shows how the last line of the picture should read.


http://en.wikipedia.org/wiki/Function_composition
http://mathworld.wolfram.com/Composition.html

Answer 9

I received that picture from my teacher. . .that's what she says is correct. That's why I am so confused. . .I didn't just make that up. . .So I really don't know what to follow, but I guess if I want to get a good grade I better just go along with the teacher. The way you explain it makes sense though!

Answer 10

Look up your teacher's notes to see if she defines the composition of functions the same way as the two references I gave you. If she defines otherwise, you will have no choice but to follow her definition.

If she defines the operator as in the two references, you have the option of politely asking her to explain the picture to you. If it was a typo, which can happen because it's just the inversion of two letters, she'll be glad to offer a correction.

Good luck!

Answer 11

Okay thank you