nLab
endomorphism

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Definition

An endomorphism of an object xx in a category CC is a morphism f:xxf : x \to x.

An endomorphism that is also an isomorphism is called an automorphism.

Properties

Given an object xx, the endomorphisms of xx form a monoid under composition, the endomorphism monoid of xx:

End C(x)=Hom C(x,x), End_C(x) = Hom_C(x,x) ,

which may be written End(x)End(x) if the category CC is understood. Up to equivalence, every monoid is an endomorphism monoid; see delooping.

An endomorphism monoid is a special case of a monoid structure on an end construction. Let d:DCd:D\to C be a diagram in CC, where CC is a monoidal category (in the case above the monoidal structure is the cartesian product and dd is a constant diagram from the initial category). One defines End(d)End(d) as an object in CC, equipped with a natural transformation a:End(d)dda: End(d) \otimes d \to d which is universal in the sense that for all objects ZCZ \in C, and any natural transformation f:Zddf: Z \otimes d \to d there exists a unique morphism g:ZEnd(d)g: Z \to End(d) such a(gd)=f:Zdda \circ (g \otimes d) = f: Z \otimes d \to d.

Proposition

If the universal object (End(d),a)(End(d),a) exists then there is a unique structure of an internal monoid μ:End(d)End(d)End(d)\mu: End(d) \otimes End(d) \to End(d), such that the map a:End(d)dda: End(d) \otimes d \to d is an action.

Proposition

In a cartesian monoidal category CC, if an endomorphism monoid End(c)End(c) for an object c:1Cc: 1 \to C exists and is commutative, then cc is a subterminal object.

Proof

Let k:cEnd(c)k: c \to End(c) correspond to first projection π 1:c×cc\pi_1: c \times c \to c. Then the composition

c×ck×kEnd(c)×End(c)compEnd(c)c \times c \stackrel{k \times k}{\to} End(c) \times End(c) \stackrel{comp}{\to} End(c)

(where compcomp denotes internal composition) may be computed to be kπ 1k \pi_1, corresponding to first projection π 1:c×c×cc\pi_1: c \times c \times c \to c. Thus, assuming commutativity of End(c)End(c) and letting σ\sigma generally denote a symmetry map, consideration of the diagram

c×c k×k End(c)×End(c) comp End(c) σ σ id c×c k×k End(c)×End(c) comp End(c)\array{ c \times c & \stackrel{k \times k}{\to} & End(c) \times End(c) & \stackrel{comp}{\to} & End(c) \\ \mathllap{\sigma} \downarrow & & \mathllap{\sigma} \downarrow & & \downarrow \mathrlap{id} \\ c \times c & \stackrel{k \times k}{\to} & End(c) \times End(c) & \stackrel{comp}{\to} & End(c) }

leads to the conclusion that kπ 1=kπ 1σk \pi_1 = k \pi_1 \sigma, or π 1=π 2:c×c×cc\pi_1 = \pi_2: c \times c \times c \to c. We easily conclude π 1=π 2:c×cc\pi_1 = \pi_2: c \times c \to c, which forces equality f=gf = g for any two maps f,g:dcf, g: d \to c, so that the unique map !:c1!: c \to 1 is a monomorphism.

Last revised on July 4, 2018 at 06:51:45. See the history of this page for a list of all contributions to it.