# function compositions

• Feb 26th 2008, 10:48 PM
sfitz
function compositions
Here's the problem: Find sets $A, B$ and $C$ and functions $f:A\longrightarrow B, g:B\longrightarrow C$ and $h:B\longrightarrow C$ for which $f\circ h=f\circ g$ but $g\not=h$.
It doesn't even have to be real example functions, just a picture would be good. I just can't think of anything. Thank you!
• Feb 26th 2008, 11:32 PM
Jhevon
Quote:

Originally Posted by sfitz
Here's the problem: Find sets $A, B$ and $C$ and functions $f:A\longrightarrow B, g:B\longrightarrow C$ and $h:B\longrightarrow C$ for which $f\circ h=f\circ g$ but $g\not=h$.
It doesn't even have to be real example functions, just a picture would be good. I just can't think of anything. Thank you!

how about $A = B = C = \mathbb{R}$. with $f(x) = x^2,~g(x) = x, \mbox{ and }h(x) = -x$

can you find another example?
• Feb 26th 2008, 11:50 PM
sfitz
Oh my god, I am so sorry, I typed the problem wrong. I meant $g\circ f$ and $h\circ f$! Can you still help? The part I typed was part a of the problem, and I'm stuck on part b. I'm really sorry.
• Feb 27th 2008, 08:12 AM
Jhevon
Quote:

Originally Posted by sfitz
Oh my god, I am so sorry, I typed the problem wrong. I meant $g\circ f$ and $h\circ f$! Can you still help? The part I typed was part a of the problem, and I'm stuck on part b. I'm really sorry.

how about $g(x) = x,~h(x) = f(x) = 1$

try to find another
• Feb 27th 2008, 10:27 AM
sfitz
ok, I understand, f must not be onto for it to work, so that there are elements in B that are outside the image of f. thank you!
• Feb 27th 2008, 01:38 PM
Jhevon
Quote:

Originally Posted by sfitz
ok, I understand, f must not be onto for it to work, so that there are elements in B that are outside the image of f. thank you!

can you prove that? (Devil)

just kidding... well, you can if you want to
• Feb 27th 2008, 04:08 PM
sfitz
I did prove it! That was the other part of the question, but I couldn't prove it without being able to think of an example
• Feb 27th 2008, 04:23 PM
Jhevon
Quote:

Originally Posted by sfitz
I did prove it! That was the other part of the question, but I couldn't prove it without being able to think of an example

oh, ok. that's fine then