surjective mapping

surjective mapping

[sər′jek·tiv ′map·iŋ]
(mathematics)
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(2) Suppose, [phi] is surjective mapping. Here, [[([phi], [psi])([M.sub.1], E)].sup.c] = [[([phi]Mi)).sup.c], [psi](E)] and ([phi], [psi])[([M.sub.1],E).sup.c] = [[phi]([M.sup.c.sub.1]), [psi](E)].
We recall that a surjective mapping f: [C.sub.1] [right arrow] [C.sub.2] is a dual isomorphism if for all [X.sub.1], [X.sub.2] [member of] [C.sub.1] it is true that
If f : (X, [tau]) [right arrow] (Y, [sigma]) is an [[alpha].sub.([gamma],[gamma]')]-continuous surjective mapping and E is an [[alpha].sub.[gamma]]D-set in Y, then the inverse image of E is an [[alpha].sub.[gamma]]D-set in X.
A topological space (X, [tau]) is [[alpha].sub.[gamma]][D.sub.1] if for each pair of distinct points x, y [member of] X, there exists an [[alpha].sub.([gamma],[gamma]')]-continuous surjective mapping f : (X, [tau]) [right arrow] (Y, [sigma]), where Y is an [[alpha].sub.[gamma]],D1space such that f(x) and f(y) are distinct.
This shows that (xi)[theta] = (a, i), and thereby [theta] is a surjective mapping.
Assume also that [V.sub.0] is a surjective mapping such that for any x, y [member of] [S.sub.1](E) and r > 0, we have
All of above works only considered the surjective mappings between the unit spheres of two normed spaces of the same type.
for every M [subset or equal to] [A.sub.j] By the construction, A[right arrow][A.sub.j], a [??] j(a) is a surjective mapping that preserves joins.
Let [psi] : B(H) [right arrow] B(H) be a continuous linear surjective mapping in the weak operator topology.
It is the aim of us to prove that every weak continuous linear surjective mapping is an automorphism if and only if it is multiplicative at I.
We give an example (inspired by the proof of [18, Theorem 9]) of a unitary Morita context with surjective mappings where the semirings are non-isomorphic.