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  1. 66.1.3

66.1.3 #

Exercise 66.1.3 (On requiring surjectivity of 𝑝). Let 𝑝:𝐸→𝐵 be satisfying this definition, exceptthat 𝑝 need not be surjective. If 𝐵 is connected and 𝐸 is nonempty, then 𝑝 is surjective.Proof. Let 𝑝:𝐸→𝐵 be a continuous map such that every point 𝑏∈𝐵 has an open neighborhood 𝑈 of 𝑏 which isevenly covered by 𝑝. We claim that 𝑝(𝐸) and 𝐵∖𝑝(𝐸) are both open.•Let 𝑥∈𝑝(𝐸). Then there is an open neighborhood 𝑈 of 𝑥 which is evenly covered by 𝑝. That is,𝑝−1(𝑈)=⨄𝛼∈𝐼𝑉𝛼for some(𝑉𝛼)𝛼∈𝐼𝑝|𝑉𝛼:𝑉𝛼→𝑈is a homeomorphism for all𝛼∈𝐼Since 𝑥∈𝑝(𝐸), there is 𝑦∈𝑝−1(𝑥)⊆𝑝−1(𝑈), which implies that there is 𝛼0∈𝐼 such that 𝑦∈𝑉𝛼0. Then 𝑝|𝑉𝛼0:𝑉𝛼0→𝑈 is a homeomorphism. Then 𝑈=𝑝(𝑉𝛼0)⊆𝑝(𝐸). Thus 𝑝(𝐸) is open.•Let 𝑦∈𝐵∖𝑝(𝐸). Then there is an open neighborhood 𝑈 of 𝑦 which is evenly covered by 𝑝. Suppose 𝑥∈𝑝(𝐸) forsome 𝑥∈𝑈. Then 𝑈⊆𝑝(𝐸) by the same argument of the above. But then 𝑦∈𝑈⊆𝑝(𝐸), hence contradiction. Thus𝑥∉𝑝(𝐸) for all 𝑥∈𝑈. That is, 𝑈⊆𝐵∖𝑝(𝐸). Thus 𝐵∖𝑝(𝐸) is open.Now suppose that 𝐵 is connected and 𝐸 is nonempty. Then 𝑝(𝐸) is nonempty. If 𝐵∖𝑝(𝐸) is also nonempty, 𝑝(𝐸) and𝐵∖𝑝(𝐸) separates 𝐵 into two disjoint nonempty open sets, which is contradiction. Hence 𝐵∖𝑝(𝐸) must be empty.i.e. 𝑝 is surjective.