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Injective composition: the second function need not be injective. A function is injective (one-to-one) if each possible element of the codomain is mapped to by at most one argument. Equivalently, a function is injective if it maps distinct arguments to distinct images. An injective function is an injection. [1] The formal definition is the ...
In mathematics, an injective function (also known as injection, or one-to-one function [1]) is a function f that maps distinct elements of its domain to distinct elements of its codomain; that is, x 1 ≠ x 2 implies f(x 1) ≠ f(x 2) (equivalently by contraposition, f(x 1) = f(x 2) implies x 1 = x 2).
Inclusion maps are seen in algebraic topology where if is a strong deformation retract of , the inclusion map yields an isomorphism between all homotopy groups (that is, it is a homotopy equivalence). Inclusion maps in geometry come in different kinds: for example embeddings of submanifolds.
In general topology, an embedding is a homeomorphism onto its image. [3] More explicitly, an injective continuous map : between topological spaces and is a topological embedding if yields a homeomorphism between and () (where () carries the subspace topology inherited from ).
A faithful functor need not be injective on objects or morphisms. That is, two objects X and X′ may map to the same object in D (which is why the range of a full and faithful functor is not necessarily isomorphic to C), and two morphisms f : X → Y and f′ : X′ → Y′ (with different domains/codomains) may map to the same morphism in D.
Given a map :, the mapping cylinder is a space , together with a cofibration ~: and a surjective homotopy equivalence (indeed, Y is a deformation retract of ), such that the composition equals f. Thus the space Y gets replaced with a homotopy equivalent space M f {\displaystyle M_{f}} , and the map f with a lifted map f ~ {\displaystyle {\tilde ...
This is not an injective map, as for example every integer is mapped to 0. Nevertheless, it is a monomorphism in this category. This follows from the implication q ∘ h = 0 ⇒ h = 0, which we will now prove. If h : G → Q, where G is some divisible group, and q ∘ h = 0, then h(x) ∈ Z, ∀ x ∈ G. Now fix some x ∈ G.
In the case of maps f : U → C defined on an open subset U of the complex plane C, some authors (e.g., Freitag 2009, Definition IV.4.1) define a conformal map to be an injective map with nonzero derivative i.e., f’(z)≠ 0 for every z in U. According to this definition, a map f : U → C is conformal if and only if f: U → f(U) is ...