# function compositions

• Feb 26th 2008, 10:48 PM
sfitz
function compositions
Here's the problem: Find sets \$\displaystyle A, B\$ and \$\displaystyle C\$ and functions \$\displaystyle f:A\longrightarrow B, g:B\longrightarrow C\$ and \$\displaystyle h:B\longrightarrow C\$ for which \$\displaystyle f\circ h=f\circ g\$ but \$\displaystyle 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 \$\displaystyle A, B\$ and \$\displaystyle C\$ and functions \$\displaystyle f:A\longrightarrow B, g:B\longrightarrow C\$ and \$\displaystyle h:B\longrightarrow C\$ for which \$\displaystyle f\circ h=f\circ g\$ but \$\displaystyle 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 \$\displaystyle A = B = C = \mathbb{R}\$. with \$\displaystyle 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 \$\displaystyle g\circ f\$ and \$\displaystyle 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 \$\displaystyle g\circ f\$ and \$\displaystyle 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 \$\displaystyle 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