Suppose $f: B->C$ is a homomorphism between Boolean Algebras such that $ker(f)=0$. Is $f$ necessarily a monomorphism? Thank You.
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If $f(a)=f(b)$, with $a\neq b$, then $a\nleq b$ or $b\nleq a$.
Suppose, without loss of generality, that $a \nleq b$.
Thus $a \wedge b' \neq 0$.
Now, taking $c = a \wedge b'$, we have
$$f(c) = f(a) \wedge f(b') = f(a) \wedge (f(b))' = f(a) \wedge (f(a))'=0.$$
amrsa
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Thanks a lot but I don't see immediately why $a^b'=0$ forces $a<=b$ – Paul Epstein Jun 19 '22 at 12:28
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Ok, I think I can fill in the gaps. Thanks a lot! a <= 1 -> a <= (b v b') -> ab + ab' = a which means that ab=a if a^b' = 0 contrary to a not <= b. – Paul Epstein Jun 19 '22 at 13:11
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@PaulEpstein That's correct. And you see that this answers positively the question you posted, right? – amrsa Jun 19 '22 at 13:15
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Yes, definitely. Thanks. The text I'm reading talks about a - b which doesn't really make sense. – Paul Epstein Jun 19 '22 at 13:22
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$a-b$ might be a notation for $a \wedge b'$, somehow... – amrsa Jun 19 '22 at 13:23
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in sets, $A\setminus B = A \cap B'$, and that could explain the notation – amrsa Jun 19 '22 at 13:24