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An absolute monomorphism is a morphism sent to a monomorphism by every functor out of its category. The absolute monomorphisms are exactly the split monomorphisms.

For a binary product in a category, the diagonal ${\Delta }_A:A\to A\times A$ is the unique map whose two projection composites are $1_A$. It is a split monomorphism, since either projection is a left inverse. The diagonal converts equality of two maps into a lifting problem: $⟨u,v⟩$ factors through ${\Delta }_A$ exactly when $u=v$.

In a complete finitary abelian category, the comparison $j$ with diagonal matrix entries the identity and off-diagonal entries zero is monic. Express its source as the directed colimit of finite biproducts. Their maps into the product have finite-coordinate retractions and are therefore split monomorphisms; the finitary axiom applies to this cocone. If the category is also cofinitary abelian category, duality makes $j$ epic and hence invertible. For countably many copies of any object, the shift-and-fold argument in countable biproducts in an additive category force triviality then forces the identity to be zero, so every object is a zero object.

A coreflexive pair $f,g:B\rightrightarrows A$ has a common retraction $r:A\to B$ with $rf=rg=1_B$. It is the dual of a reflexive pair. Both $f$ and $g$ are split monomorphisms.

Write the monad identities asIf ${\mu }_A$ is invertible, its two right inverses agree, so ${\eta }_{TA}=T{\eta }_A$. Conversely, suppose ${\eta }_T=T\eta$. Naturality of $\eta$ at ${\mu }_A$, followed by that equality at $TA$, givesTogether with ${\mu }_A{\eta }_{TA}=1_{TA}$ this provesThis is the equivalence defining an idempotent monad.

For the equalizer submonad of a monad, let ${\beta }_A:RA\to TA$ equalize ${\eta }_{TA}$ and $T{\eta }_A$. If $f:A\to B$, naturality of these two transformations shows that $Tf{\beta }_A$ equalizes the corresponding pair at $B$. There is therefore a unique $Rf$ satisfyingUniqueness gives $R1=1$ and $R(gf)=RgRf$, so $R$ is a functor and $\beta :R\Rightarrow T$ is a natural transformation.

Take the stipulated factorizations ${\alpha }_A$ and ${\theta }_A$. Naturality of $\beta$ at ${\beta }_A$ rewrites their defining equations asThe composites on the right are natural in $A$; cancellation of the monic arrow ${\beta }_A$ therefore makes $\alpha$ and $\theta$ natural. We now verify every monad law, rather than assuming that restricting the structure automatically preserves them.

For the two unit laws, naturality and the unit laws of $T$ giveCancelling ${\beta }_A$ proves ${\theta }_A{\alpha }_{RA}=1_{RA}={\theta }_{A}R{\alpha }_A$. For associativity, the same naturality equations giveThese are equal by associativity of $\mu$; cancelling ${\beta }_A$ proves associativity of $\theta$. Thus

The remaining claims use two split equalizers associated with a monad. First, ${\eta }_{TA}:TA\to TTA$ is a split equalizer of ${\eta }_{TTA},T{\eta }_{TA}$. Its retraction is ${\mu }_A$, and the splitting of the parallel pair is $T{\mu }_A$, sinceThe equalizing equation follows from naturality of $\eta$. In general, identities $re=1$, $sf=er$, $sg=1$ and $fe=ge$ prove that $e$ is an equalizer: if $fh=gh$, then $h=erh$, with unique factorization $rh$. Hence ${\eta }_{TA}$ and ${\beta }_{TA}$ are two equalizers of the same pair. Their comparison is ${\alpha }_{TA}$, and its inverse is ${\mu }_A{\beta }_{TA}$. In particular

Second, $e_A=T{\eta }_A:TA\to TTA$ is a split equalizer of $TT{\eta }_A,T{\eta }_{TA}$, with retraction ${\mu }_A$ and splitting ${\mu }_{TA}$. Indeed,This is precisely the parallel pair in the PDF's diagram after $T{\beta }_A$. The map $T{\beta }_A$ equalizes that pair, because applying $T$ to the defining equalizer equation gives their equality on $T{\beta }_A$. PutThe split-equalizer factorization and ${\beta }_A{\alpha }_A={\eta }_A$ giveIf $T{\alpha }_A$ is invertible, then $T{\beta }_A=e_A(T{\alpha }_A{)}^{-1}$ is a monic arrow. Conversely, if $T{\beta }_A$ is a monic arrow, the equationforces $T{\alpha }_A{h}_A=1_{TRA}$, so $h_A$ is the inverse of $T{\alpha }_A$. ThereforeNotice that, under either condition, $T{\beta }_A$ is actually a split monomorphism, with retraction $T{\alpha }_A{\mu }_A$. Thus $TT{\beta }_A$ is a monic arrow too, since a functor preserves split monomorphisms. We have not assumed that $T$ preserves arbitrary monomorphisms.

Now prove the specified naturality square is a pullback in a category. Let $x:X\to TRA$ and $y:X\to RTA$ satisfy $T{\beta }_{A}x={\beta }_{TA}y$. Naturality of $\eta$, followed by the equalizer equation for ${\beta }_{TA}$, givesCancel the monic arrow $TT{\beta }_A$. Thus $x$ equalizes ${\eta }_{TRA},T{\eta }_{RA}$ and factors uniquely as $x={\beta }_{RA}z$ with $z:X\to RRA$. Naturality of $\beta$ then givesCancel ${\beta }_{TA}$ to obtain $R{\beta }_{A}z=y$. This is exactly existence and uniqueness in the pullback universal property.

Because the square is a pullback and $T{\beta }_A$ is a monic arrow, $R{\beta }_A$ is a monic arrow. Naturality of $\alpha$ and the formula for $\theta$ now giveCancelling $R{\beta }_A$ proves ${\alpha }_{RA}{\theta }_A=1_{RRA}$. The other composite is already one of the unit laws. Hence ${\theta }_A$ is invertible. If the equivalent monicity conditions hold for every $A$, we conclude

A preadditive category has an abelian-group structure on every hom-set and bilinear composition. One equivalent definition of an additive category is a preadditive category with a zero object and finite products in a category; the following argument proves that these products in a category are automatically coproducts in a category. An abelian category is an additive category with kernels and cokernels in which every monomorphism is a kernel and every epimorphism is a cokernel. Equivalently, for every arrow the canonical map from its coimage to its image is invertible.

Let $P={\prod }_{i=1}^n{A}_i$, with projections $p_i$. Define ${\iota }_i:A_i\to P$ by $p_j{\iota }_i={\delta }_{ji}$, where the diagonal entry is the identity and the others are zero. Since its composites with all projections are identities,Given $f_i:A_i\to B$, the map $f={\sum }_i{f}_i{p}_i$ satisfies $f{\iota }_i=f_i$. If another map has those composites, multiplying the displayed identity by that map forces it to equal $f$. Thus $P$ is also a coproduct in a category. The empty case is the zero object. Reversing arrows proves that a finite coproduct in a category is likewise a product in a category. Hence finite products in a category and coproducts in a category are canonically biproducts in an additive category.

For the category of abelian groups, let $(A_i,D_{ij})$ be a directed diagram, with no injectivity assumption on its transition maps. Its set colimit consists of classes $[i,a]$, withDirectedness proves transitivity of this relation. Define addition by moving representatives to a common upper index:Moving all indices involved to one further upper bound proves independence of representatives and of $k$. Define $-[i,a]=[i,-a]$ and use the common class of the zeros as zero. The abelian-group laws hold at a common index, so descend to the quotient. Every canonical map $A_i\to C$ is a homomorphism, and a compatible cocone induces the homomorphism $[i,a]\mapsto f_i(a)$. This is the unique extension. Moreover the formula for addition is forced by those canonical homomorphisms, proving that the underlying-set functor creates directed colimits. This establishes directed colimits created by the abelian-group forgetful functor explicitly.

If a cocone ${\nu }_i:A_i\to B$ has injective legs, the induced map $C\to B$ is injective. Indeed, equal images of $[i,a]$ and $[j,b]$ give, at a common index $k$,Injectivity of ${\nu }_k$ gives equal representatives at $k$. Since monomorphisms of abelian groups are exactly injective homomorphisms, we have proved that the category of abelian groups is a finitary abelian category in the directed-union sense of the question.

Now let $\mathcal{A}$ be complete, cocomplete and finitary. Index by finite subsets $J\subseteq I$ and put $S_J={⨁}_{j\in J}{A}_j$. For $J\subseteq K$, insert the additional zero coordinates. A compatible cocone on these finite biproducts is exactly a family of maps from every $A_i$, so their directed colimit is $S={\coprod }_{i\in I}{A}_i$.

Each finite biproduct has a map $k_J:S_J\to P={\prod }_{i\in I}{A}_i$ using its coordinates in $J$ and zeros outside $J$. Projection to the finite set of coordinates in $J$ is a retraction, so $k_J$ is a split monomorphism. These maps form a cocone of monomorphisms. The finitary property makes its induced map a monic arrow, and that induced map is exactly the canonical infinite identity matrix:This proves the coproduct-to-product comparison in a finitary abelian category. The PDF's use of finite products in a category in the hint is valid because these are the same finite biproducts.

If $\mathcal{A}$ is also a cofinitary abelian category, apply this argument in its opposite category. The same canonical comparison $j$ is then an epic arrow in $\mathcal{A}$. In an abelian category an arrow that is both a monomorphism and an epimorphism is invertible, so $j$ is an isomorphism.

For countably many copies of $A$, put $S={\coprod }_{n\ge 0}A$ and let $q=j^{-1}\Delta :A\to S$, where $\Delta$ is the product in a category diagonal. Let ${\iota }_n$ denote the coproduct in a category injections, let $s:S\to S$ shift them by $s{\iota }_n={\iota }_{n+1}$, and let $\nabla :S\to A$ be the codiagonal. Comparing every product in a category coordinate givesSince $j$ is a monic arrow, $q={\iota }_0+sq$. Also $\nabla {\iota }_0=1_A$ and $\nabla s=\nabla$. Therefore, for $z=\nabla q$,Hom-sets are abelian groups, so cancellation gives $1_A=0$. Any arrow to or from $A$ is consequently zero by composition with its identity. Thus every object is both initial and terminal, andThis is the countable shift-and-fold argument behind countable biproducts in an additive category force triviality.

A monomorphism is a morphism $m:X\to Y$ such that $mu=mv$ implies $u=v$ for every pair $u,v:Z\to X$. Dually, an epimorphism $e:X\to Y$ satisfies $ue=ve\Rightarrow u=v$. A regular epimorphism is a coequalizer of some parallel pair. An isomorphism has a two-sided inverse.

For a product in a category, write ${\Delta }_A:A\to A\times A$ for its categorical diagonal. Since ${\pi }_1{\Delta }_A=1_A$, equality ${\Delta }_{A}u={\Delta }_{A}v$ gives $u={\pi }_1{\Delta }_{A}u={\pi }_1{\Delta }_{A}v=v$. Thus the diagonal is a monomorphism, indeed a split monomorphism.

Assume $F$ is a full and faithful functor and its essential image is closed under strong quotients. First its unit is invertible. To prove this directly, fullness gives $k:GFA\to A$ with $Fk={\epsilon }_{FA}$. The triangle identity and faithfulness give $k{\eta }_A=1_A$. Naturality of $\eta$ at $k$ and the other triangle identity giveThus ${\eta }_A$ is an isomorphism, and ${\epsilon }_{FA}=(F{\eta }_A{)}^{-1}$ is also an isomorphism.

Factor ${\epsilon }_B$ as a strong epimorphism followed by a monomorphism. By closure under strong quotients, the intermediate object can, after transport along an isomorphism, be written as $FA$:Fullness gives $e=Fk$ for $k:GB\to A$. Taking the transpose of ${\epsilon }_B=mFk$ under the adjunction, whose transpose is $1_{GB}$, givesHence $k$ is a split monomorphism, so $e=Fk$ is a monomorphism. A strong epimorphism which is monic is invertible: apply its lifting property to the square with that same map on both sides and identity top and bottom edges. Thus $e$ is invertible and ${\epsilon }_B=me$ is monic. We have proved the pointwise-monic unit-and-counit criterion:For a counterexample without the balanced hypothesis, use the pointwise-monic adjunction over a non-balanced poset. Let $\mathcal{C}=\{0<1\}$ and let $\mathcal{D}$ have one object and only its identity. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to $G$ selecting $1$, since both $\mathcal{D}(Fc,\ast )$ and $\mathcal{C}(c,1)$ are singletons. All morphisms in a poset viewed as a category are monomorphisms, so the unit and counit are monic. However $F$ is not full: the identity $F1\to F0$ has no preimage $1\to 0$. The arrow $0\to 1$ is both monic and epic but not invertible, so $\mathcal{C}$ is not a balanced category. The balanced hypothesis cannot be dropped.

Assume (c), and take an epimorphism $p:M\twoheadrightarrow h_A$ with $M$ a monofunctor. Since $h_A$ is a projective object in a category, there is a natural transformation $s:h_A\to M$ with $ps=1_{h_A}$. A retract in a category of a monofunctor is a monofunctor: if $h_A(f)(u)=h_A(f)(v)$, then naturality of $s$ gives $M(f)s(u)=M(f)s(v)$, injectivity of $M(f)$ gives $s(u)=s(v)$, and applying $p$ gives $u=v$. Thus (c) implies (b), completing

The stronger assertion that every set-valued functor is a monofunctor holds exactly when $\mathcal{C}$ is a groupoid. Sufficiency follows because a functor preserves inverses, so sends every arrow to a bijection.

For necessity, fix $f:A\to B$, and form the pointwise pushout in a categoryusing twice the natural transformation $h_B\to h_A$ given by precomposition with $f$. At an object $X$, $P(X)$ consists of two copies of $\mathcal{C}(A,X)$, with the two copies of $kf$ identified for every $k:B\to X$. No different underlying elements are identified: each generating relation simply joins the two copies of one element. In $P(B)$, the two copies of $f$ coincide, since $f=1_{B}f$. Hence $P(f)$ sends the two copies of $1_A$ in $P(A)$ to the same element. If $P$ is a monofunctor, those copies of $1_A$ coincide. The description of the pushout implies $rf=1_A$ for some $r:B\to A$. Thus every morphism is a split monomorphism.

Apply the same conclusion to $r$: there is $t:A\to B$ with $tr=1_B$. Then $t=trf=f$, and so $fr=1_B$ as well. Every arrow is invertible, proving

For a locally small category $\mathcal{C}$, an object $A$, and a functor $F:\mathcal{C}\to \mathbf{Set}$, the covariant Yoneda lemma is the natural bijectionIts inverse sends $x\in F(A)$ to the natural transformation whose component at $B$ maps $f:A\to B$ to $F(f)(x)$.

Suppose now that $\mathcal{C}$ is a small category. For $F:\mathcal{C}\to \mathbf{Set}$, form the coproduct in a categoryThe Yoneda lemma associates to every summand the natural transformation determined by $x$, and these transformations combine to a map $p:P\to F$. At an object $B$, the element $y\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,y)$, so $p$ is a pointwise epimorphism in a functor category. Each representable functor $\mathcal{C}(A,-)$ is a projective object in a category, sinceand evaluation preserves pointwise epimorphisms. A coproduct of projectives is projective, so $P$ is the required projective object. This is the projective cover of a set-valued functor by representables.

We next prove the three equivalent conditions. If every morphism of $\mathcal{C}$ is a monomorphism, then for $f:B\to C$ and every $A$, postcompositionis injective. Thus every covariant representable functor is a monofunctor. Conversely, taking $A=B$ shows that injectivity for every representable implies that $fg=fh$ forces $g=h$, so every $f$ is monic.

If all representables are monofunctors, the object $P$ above is a monofunctor because a coproduct of injective functions is injective. Hence every $F$ is an epimorphic image of a monofunctor. Conversely, suppose every functor is an epimorphic image of a monofunctor and apply this to a representable $R$. Choose an epimorphism $q:M\twoheadrightarrow R$ with $M$ a monofunctor. Since $R$ is projective, $1_R$ lifts to $s:R\to M$ with $qs=1_R$. Thus $R$ is a retract in a category of $M$. Every retract of a monofunctor is a monofunctor: if $R(f)x=R(f)y$, then injectivity of $M(f)$ applied to $s(x)$ and $s(y)$ gives $s(x)=s(y)$, and applying $q$ gives $x=y$. This completes the equivalence.

Finally, every functor $\mathcal{C}\to \mathbf{Set}$ is a monofunctor exactly when every morphism of $\mathcal{C}$ is a split monomorphism. The forward implication is immediate because every functor preserves a left inverse. For the converse, fix $f:A\to B$ and form a quotient of $\mathcal{C}(A,-)\bigsqcup 1$ by identifying the distinguished point with every arrow of the form $kf:A\to X$, where $k:B\to X$. In the resulting functor $Q$, the two elements $[1_A]$ and $\ast$ of $Q(A)$ have equal images under $Q(f)$ because $[f]=\ast$. If every functor is a monofunctor, $Q(f)$ is injective, so $[1_A]=\ast$. By construction this means $1_A=rf$ for some $r:B\to A$. Thus $f$ is split monic. Equivalently, every morphism of $\mathcal{C}$ must be an absolute monomorphism.

An object $A$ is a retract of $B$ when there are morphisms $s:A\to B$ and $r:B\to A$ with $rs=1_A$. The map $s$ is a split monomorphism and $r$ is a split epimorphism.