نتایج جستجو برای: c closed injective

تعداد نتایج: 1168177  

2017

It is no exaggeration to say that the theory of separably injective spaces is quite different from that of injective spaces. In this chapter we will explain why. Indeed, we will enter now in the main topic of the monograph, namely, separably injective spaces and their “universal” version. After giving the main definitions and taking a look at the first natural examples one encounters, we presen...

Journal: :Transactions of the American Mathematical Society 1968

2017

It is no exaggeration to say that the theory of separably injective spaces is quite different from that of injective spaces. In this chapter we will explain why. Indeed, we will enter now in the main topic of the monograph, namely, separably injective spaces and their “universal” version. After giving the main definitions and taking a look at the first natural examples one encounters, we presen...

1999
MICHAEL FRANK

The local multiplier C*-algebra Mloc(A) of any C*-algebra A can be ∗-isomorphicly embedded into the injective envelope I(A) of A in such a way that the canonical embeddings of A into both these C*-algebras are identified. If A is commutative then Mloc(A) ≡ I(A). The injective envelopes of A and Mloc(A) always coincide, and every higher order local multiplier C*-algebra of A is contained in the ...

2008
JAMES COFFEY

This paper gives a proof that the fundamental group a class of closed orientable 3-manifolds constructed from three injective handlebodies has a solvable word problem. This is done by giving an algorithm, that terminates in bounded time, to decide if a closed curve in the manifold is null-homotopic.

Journal: :Indagationes Mathematicae 1991

2010
BRIAN CRUZ

Proposition 1. M is injective if and only if its singular value decomposition M = UDV H has a V that is square and invertible. In this case, MM is invertible and M = (MHM)−1MH . Proof. Let M be an r × c matrix. Suppose that M is injective, so that rank(M) = c because the kernal is zero. Then D is a c × c matrix and so V H is also c× c. V H must already be injective (lest M not be injective), an...

2005
Marc Culler

Proof. The intersection P of all subfields of F is a field by Exercise 1.4. Consider the ring homomorphism φ : Z → F given by φ(n) = n · 1. Since any subfield contains 1 and is closed under addition, imφ is contained in P . If Char F = p 6= 0 then imφ is isomorphic to Z/pZ = Fp. Since this is a field, we have P = imφ ∼= Fp. If Char F = 0 then φ is injective. Define φ̂ : Q → F by φ̂(m/n) = φ(m)/φ(...

Journal: :bulletin of the iranian mathematical society 2014
leila shahbaz mojgan mahmoudi

‎in this paper some properties of weak regular injectivity for $s$-posets‎, ‎where $s$ is a pomonoid‎, ‎are studied‎. ‎the behaviour‎ ‎of different kinds of weak regular injectivity with products‎, ‎coproducts and direct sums is considered‎. ‎also‎, ‎some‎ ‎characterizations of pomonoids over which all $s$-posets are of‎ ‎some kind of weakly regular injective are obtained‎. ‎further‎, ‎we‎ ‎giv...

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