(inverse morphisms)
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 $X$.
A morphism that has an inverse morphism (Def. ) is called an isomorphism.
(inverse morphisms are unique)
If $f$ is an isomorphism (Def. ) with inverse morphism $f^{-1}$ (Def. ), then any left inverse as well as any right inverse is already an actual inverse morphism and in fact is equal to $f^{-1}$.
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.
Let $f$ be an isomorphism (Def. ). Then the inverse morphism $\left(f^{-1}\right)^{-1}$ of an inverse morphism $f^{-1}$ (Def. ) exists and is equal to the original morphism:
(identity morphisms are their own inverse morphisms)
Any identity morphism is its own inverse morphism (Def. ):
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 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.
Last revised on July 17, 2021 at 08:08:21. See the history of this page for a list of all contributions to it.