nLab inverse

Redirected from "inverse morphism".
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Definition

Definition

(inverse morphisms)
An inverse of a morphism f:XYf \colon X \to Y in a category or unital magmoid (or an element of a monoid or unital magma) is another morphism f 1:YXf^{-1} \colon Y \to X which is both a left-inverse (a retraction) as well as a right-inverse (a section) of ff, in that

ff 1=id Y:YY f \circ f^{-1} \,=\, id_Y \colon Y \to Y

equals the identity morphism on YY

and

f 1f=id X:XX f^{-1} \circ f \,=\, id_X \colon X \to X

equals the identity morphism on XX.

Definition

A morphism that has an inverse morphism (Def. ) is called an isomorphism.

Remark

A (small) category in which all morphisms have inverses is called a groupoid.

Properties

Proposition

(inverse morphisms are unique)
If ff is an isomorphism (Def. ) with inverse morphism f 1f^{-1} (Def. ), then any left inverse as well as any right inverse is already an actual inverse morphism and in fact is equal to f 1f^{-1}.

In particular, inverse morphisms are unique when they exist.
Proof

Let gg be a left inverse, hence such that gf=idg \circ f \,=\, id. Write f 1f^{-1} for the actual inverse, hence such that f 1f=idf^{-1} \circ f \,=\, id and ff 1=idf \circ f^{-1} \,=\, id.
Then the following sequence of equalities implies that g=f 1g = f^{-1}:

g =gid =g(ff 1) =(gf)f 1 =idf 1 =f 1. \begin{aligned} g & \;=\; g \circ id \\ & \;=\; g \circ ( f \circ f^{-1} ) \\ & \;=\; (g \circ f) \circ f^{-1} \\ & \;=\; id \circ f^{-1} \\ & \;=\; f^{-1} \,. \end{aligned}

Here all steps use just the definitions of the various morphisms, except the third step, which uses associativity of composition in any category.

An analogous argument applies to right inverses; and either argument applies to actual inverses.

Proposition

Let ff be an isomorphism (Def. ). Then the inverse morphism (f 1) 1\left(f^{-1}\right)^{-1} of an inverse morphism f 1f^{-1} (Def. ) exists and is equal to the original morphism:

(f 1) 1=f. \left(f^{-1}\right)^{-1} \;=\; f \,.

Proof

By the uniqueness of inverses (Prop. ) for f 1f^{-1}.

Examples

Example

(identity morphisms are their own inverse morphisms)
Any identity morphism is its own inverse morphism (Def. ):

id 1=id. id^{-1} \,=\, id \,.

Example

In a balanced category, such as in a topos (in particular in Sets) every morphism that is both a monomorphism and well as an epimorphism is actually an isomorphism and thus has an inverse morphism.

To see that this is not generally the case, notice that any partial order is an (necessarily unbalanced) category where every morphism is both a monomorphism as well as an epimorphism, but only its identity morphisms have inverse morphisms (as they must, by Exp. ).

In non-unital contexts

In a magmoid or semicategory (or an element of a semigroup or magma), a morphism f:abf:a \to b has a unique retraction f 1:baf^{-1}:b \to a if

  • for every morphism g:bcg:b \to c, g(f 1f)=gg \circ (f^{-1} \circ f) = g,

  • for every morphism g:cag:c \to a, (f 1f)g=g(f^{-1} \circ f) \circ g = g,

and a morphism f:abf:a \to b has a unique section f 1:baf^{-1}:b \to a if

  • for every morphism g:acg:a \to c, g(ff 1)=gg \circ (f \circ f^{-1}) = g,

  • for every morphism g:acg:a \to c, (ff 1)g=g(f \circ f^{-1})\circ g = g,

A morphism f:abf:a \to b has a unique inverse if it has a retraction that is also a section.

Last revised on September 16, 2023 at 10:32:10. See the history of this page for a list of all contributions to it.