- #1
schlynn
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Mod note: Moved from a technical section, so missing the homework template.
Here is what I'm trying to prove.
Let f:A->B. If there are two functions g:B->A and h:B->A such that g(f(a))=a for every a in A and f(h(b))=b for every b in B, then f is bijective and g=h=f^(-1).
I think I have most the proof, I started by showing g(f(a))=a implies g o f(a)=a and that maps A->B->A all exactly once, and a similar argument shows f(h(b))=b implies that the mapping is B->A->B all exactly once again, and this seems like the definition of an inverse, so there for it's bijective? Is that the right way to approach this proof?
Here is what I'm trying to prove.
Let f:A->B. If there are two functions g:B->A and h:B->A such that g(f(a))=a for every a in A and f(h(b))=b for every b in B, then f is bijective and g=h=f^(-1).
I think I have most the proof, I started by showing g(f(a))=a implies g o f(a)=a and that maps A->B->A all exactly once, and a similar argument shows f(h(b))=b implies that the mapping is B->A->B all exactly once again, and this seems like the definition of an inverse, so there for it's bijective? Is that the right way to approach this proof?
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