Proving the Existence of b: B-->A for Tricky Algebra Proof

[Pearce_09]

tricky algebra proof...

hello,

consider the mappings: (R: A-->B)

Suppose T: C-->A and S: C-->A satisfy RT = RS then T = S
prove that there exists a; b: B-->A such that bR = idA (identity of A)

well, I am not sure if I can say that b (inverse) = R
since b maps B to A .. and R maps A to B... and if i can say that...
how do i approach the proof?

I know that RT = RS then T = S.. should i work with this.? using b (inverse)...
or should i try to see if R or b is injective?...

ie. b(inverse)T = b(inverse)S ... ??

 

[matt grime]

The idea of left cancellable (RS=RT => S=T) is exactly the same as R being injective.