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Definition

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

equals the identity morphism on $Y$ and

equals the identity morphism on $X$. \end{definition}

\begin{definition}\label{Isomorphism} A morphism that has an inverse morphism (Def. \ref{InverseMorphism}) is called an isomorphism. \end{definition}

\begin{remark} A (small) category in which all morphisms have inverses is called a groupoid. \end{remark}

Properties

\begin{prop}\label{InversesAreUnique} (inverse morphisms are unique) \linebreak If $f$ is an isomorphism (Def. \ref{Isomorphism}) with inverse morphism $f^{-1}$ (Def. \ref{InverseMorphism}), then any left inverse as well as any right inverse is already an actual inverse morphism and in fact is equal to $f^{-1}$. \end{prop} In particular, inverse morphisms are unique when they exist. \begin{proof} Let $g$ be a left inverse, hence such that $g \circ f \,=\, id$. Write $f^{-1}$ for the actual inverse, hence such that $f^{-1} \circ f \,=\, id$ and $f \circ f^{-1} \,=\, id$.
Then the following sequence of equalities implies that $g = f^{-1}$:

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. \end{proof}

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

\end{prop} \begin{proof} By the uniqueness of inverses (Prop. \ref{InversesAreUnique}) for $f^{-1}$. \end{proof}

Examples

\begin{example}\label{IdentityMorphismsAreTheirOwnInverses} (identity morphisms are their own inverse morphisms) \linebreak Any identity morphism is its own inverse morphism (Def. \ref{InverseMorphism}):

\end{example}

\begin{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. \ref{IdentityMorphismsAreTheirOwnInverses}). \end{example}

In non-unital contexts

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

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

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

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

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

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

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

Related concepts

• inverse morphism, inverse function

• weak inverse

• bi-invertible morphism

• inversion involution

• reversible computation