2020
DOI: 10.1007/s11117-020-00780-8
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Tensor product of f-rings

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“…It is straighforward to prove that πB× πA con πC is a Boolean space. 3 Summarizing we have that the space X = πB× πA con πC is a Boolean space. Then q ∈ X is an element of the form q = (q 1 , q 2 ) ∈ πB ×πC such that f (q 1 ) = g(q 2 ); where f : πB → πA con and g : πC → πA con are continuous and surjective functions.…”
Section: Now Let's See Thatmentioning
confidence: 95%
“…It is straighforward to prove that πB× πA con πC is a Boolean space. 3 Summarizing we have that the space X = πB× πA con πC is a Boolean space. Then q ∈ X is an element of the form q = (q 1 , q 2 ) ∈ πB ×πC such that f (q 1 ) = g(q 2 ); where f : πB → πA con and g : πC → πA con are continuous and surjective functions.…”
Section: Now Let's See Thatmentioning
confidence: 95%