Help with Algebra Question: Proving T = T* with Surjectivity

[Pearce_09]

Hello,

Im having a bit of trouble with this one question, a little help would be great.

this is a mapping question involving surjectivity.

S: A --> B T: B --> C

T*: B --> C satisfies TS = T*S
And S is surjective

Show that T = T*

Now i know the defenition of surjective. I am just having trouble showing what i have to show

let x be in A and y be in B
There exists an x in A, for every y in B such that S(x) = y
--thats surjective--
but where do i go now??

 

[HallsofIvy]

let x be in A and y be in B
There exists an x in A, for every y in B such that S(x) = y
--thats surjective--

If would phrase if differently: "For every y in B there exist x in A such that S(x)= y."
Do you see the difference? Your phrasing implies you can just pick any x and y you like in advance. You can pick y as you like, but not x!

Try proof by contradiction:
If T and T* are not equal, then there exist some y in B such that $Ty<br /> \ne T*y$. Since S is surjective, there exist x in A such that Sx= y. Then what is TSx?
What is T*Sx