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

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The discussion focuses on proving the existence of a mapping b: B-->A such that bR equals the identity mapping idA, given the condition that RT = RS implies T = S for mappings T and S from C to A. Participants express uncertainty about whether b can be considered an inverse of R, as they map in opposite directions. The proof approach is debated, with suggestions to explore the implications of injectivity for R or b. The concept of left cancellability is highlighted as being equivalent to R being injective. The conversation emphasizes the need for a clear strategy to tackle the proof effectively.
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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 ... ??
 
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The idea of left cancellable (RS=RT => S=T) is exactly the same as R being injective.
 

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