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This category has commutative monoids and their homomorphisms. Pointwise addition makes each hom-set a commutative monoid, and composition is additive in both variables. Finite cartesian products are also coproducts, using the coordinate injections with zero in the other coordinates and summing images. Therefore it is a semi-additive category, but it is not an additive category. The reflexive pair from the submonoid $\{(m,n):m\le n\le 2m\}$ of ${\mathbb{N}}^2$ to $\mathbb{N}$, using the two projections and section $m\mapsto (m,m)$, has an arrow from $1$ to $2$ and none from $2$ to $1$ in its represented graph. Thus a reflexive pair here need not yield a groupoid.

For a reflexive pair $f,g:A\rightrightarrows B$ with common section $r$, form the pushout in a category of $f$ and $g$, with maps $j_1,j_2:B\to P$. The equation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. This common map is a coequalizer of $f,g$: every map equalizing them defines a commuting pair into the pushout and hence factors uniquely through $q$.

For a finite diagram in a category $D:\mathcal{J}\to \mathcal{C}$, set $X={\coprod }_{j}D(j)$ and $Y={\coprod }_{a:i\to j}D(i)$. The two maps $s,t:Y\rightrightarrows X$ on the $a$-summand are the injection of $D(i)$ and the injection of $D(j)$ after $D(a)$. Their coequalizer is the colimit of $D$. The pair $[s,1_X],[t,1_X]:Y⨿X\rightrightarrows X$ is a reflexive pair with the same coequalizer. Thus finite coproduct in a category constructions and coequalizers of reflexive pairs suffice, including the empty coproduct for the empty diagram. Finite products cannot replace coproducts here: the poset with elements $0,a,b,u_0,u_1,\dots$, order $0<a,b<u_{n+1}<u_n$, and incomparable $a,b$ has finite meets and top $u_0$, hence finite categorical products. Every reflexive parallel pair is an equal pair and has its identity as coequalizer, but $a,b$ have no least upper bound and therefore no coproduct.

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.

If $\mathcal{C}$ has finite colimits and a monad $T$ preserves reflexive coequalizers, its Eilenberg-Moore category has finite colimits. The forgetful functor creates reflexive coequalizers preserved by $T$. The coproduct presentation for monad algebras yields binary coproducts, and $F0$ is initial. Any pair $a,b:X\rightrightarrows Y$ can then be replaced by the reflexive pair $[a,1],[b,1]:X+Y\rightrightarrows Y$, with identical coequalizers. Finite coproducts and coequalizers give all finite colimits.

Use the elementary, or weak, topos axioms: finite limits and a power object $PA$ for every object $A$, naturally representing $\mathrm{Sub}(A\times X)$. No arbitrary products or coproducts are assumed. In particular $\Omega =P1$ is the subobject classifier. Inverse image of the membership relation defines $Pf:PB\to PA$ for $f:A\to B$.

Interchanging the factors in a relation gives natural bijectionsThus the contravariant power-object functor $P:{\mathcal{E}}^{\mathrm{op}}\to \mathcal{E}$ has left adjoint $P^{\mathrm{op}}$. Its adjunction unit ${\eta }_A:A\to PPA$ sends a generalized element $a$ to the predicate on $PA$ given by membership of $a$.

This unit is monic. If $x,y:T\to A$ have the same image under ${\eta }_A$, membership of $x$ and $y$ agrees in every $T$-parameterized subobject of $A$. Apply this to the graph of $x$ in $A\times T$: membership of $x$ is true, so membership of $y$ is true, which means $y=x$. This argument uses parameterized relations, not an assumption that global elements detect morphisms.

The functor $P$ reflects isomorphisms. Suppose $Pf$ is invertible. Then $PPf$ is invertible, and naturality $PPf{\eta }_A={\eta }_{B}f$ together with monicity of ${\eta }_A$ shows that $f$ is monic. Regard $f$ as a subobject of $B$. Its inverse image along itself is the whole of $A$, just as for the whole subobject of $B$. Those two subobjects correspond to maps $1\to PB$; since $Pf$ is monic, they coincide. Therefore the subobject $f$ is all of $B$, and $f$ is an isomorphism.

We now prove the preservation condition for the crude monadicity theorem. A reflexive pair in ${\mathcal{E}}^{\mathrm{op}}$ is a coreflexive pair $f,g:B\rightrightarrows A$ in $\mathcal{E}$, with $rf=rg=1_B$ for some $r:A\to B$. Let $e:E\hookrightarrow B$ be its equalizer. For any monomorphism $m$, define ${\exists }_m$ on power objects by composing a parameterized subobject with $m$; no general image construction is needed because $m$ is monic. Inverse image followed by this extension obeys $Pm{\exists }_m=1$.

Here both $f,g$ are monic. For a parameterized subobject $S$ of $B$, an element $b$ belongs to $g^{-1}(f(S))$ exactly when $gb=f{b}^{\prime }$ for some $b^{\prime }\in S$. Applying $r$ gives $b=b^{\prime }$, so this condition is precisely $b\in S$ and $fb=gb$. Consequently, as morphisms of power objects,If $h:PB\to Z$ equalizes $Pf,Pg$, these identities giveIt therefore factors uniquely through $Pe$, which is a split epimorphism with section ${\exists }_e$. Since $fe=ge$, $Pe$ also equalizes $Pf,Pg$. Hence $Pe$ is their coequalizer. This proves explicitly that power objects turn coreflexive equalizers into coequalizers.

Finite equalizers in $\mathcal{E}$ give all coequalizers of reflexive pairs in ${\mathcal{E}}^{\mathrm{op}}$, and we have shown that $P$ preserves them and reflects isomorphisms. The crude monadicity theorem now yields

Fix a set $I$ of cardinality $\kappa$. If $\mathcal{E}$ has $I$-indexed products in a category, its Eilenberg-Moore category for the double-power monad has them, since the forgetful functor creates existing limits. The displayed equivalence then gives $I$-products in ${\mathcal{E}}^{\mathrm{op}}$, hence $I$-coproducts in a category in $\mathcal{E}$.

For the converse, suppose those coproducts exist. The diagonal $\Delta :\mathcal{E}\to {\mathcal{E}}^I$ is a logical functor, because all the elementary topos structure in ${\mathcal{E}}^I$ is pointwise, and it has left adjoint ${\coprod }_I$. Apply power-object monadicity to both toposes. Under their monadic identifications, ${\Delta }^{\mathrm{op}}$ is the lift of $\Delta$ to the double-power Eilenberg-Moore categories. The base functor $\Delta$ is a right adjoint to ${\coprod }_I$, so the adjoint lifting theorem for monad algebra functors gives a left adjoint to its algebra lift. Its required reflexive coequalizers exist in ${\mathcal{E}}^{\mathrm{op}}$, since they are opposites of finite equalizers in $\mathcal{E}$. Taking opposites gives a right adjoint to $\Delta$, which is exactly the $I$-product functor. Thus the products and coproducts of fixed cardinality in a topos satisfy

Fix an object $C$. For $x:C\to B$, use $rx:C\to A$ as its identity arrow. For $a:C\to A$, its source is $fa$ and target is $ga$. If $ga=fb$, define the composite of $a$ followed by $b$ byThese expressions use addition and subtraction in the hom-set of the preadditive category. The reflexive pair identities $fr=gr=1_B$ giveso the composite has the required endpoints. Also $f(a^{-1})=ga$ and $g(a^{-1})=fa$.

The unit identities follow directly: $a\circ rfa=rfa+a-rfa=a$ and $rga\circ a=a+rga-rga=a$. If $ga=fb$ and $gb=fc$, either parenthesization of three arrows equals $a+b+c-rga-rgb$, so composition is associative. The inverse identities areTherefore these operations define a groupoid on the specified object and arrow sets. Precomposing by $k:C^{\prime }\to C$ respects each expression, by additivity of composition, so the groupoids are natural in $C$.

Every reflexive pair therefore has the required groupoid structure on generalized points. This is the reflexive-pair groupoid formula in a preadditive category. If the composable-arrow pullback in a category $A{\times }_{g,B,f}A$ exists, its projections $p,q$ give the actual internal composition map $p+q-rgp$. The printed generalized-point definition does not assume that this pullback exists, and preadditivity alone would not ensure it.

The crude monadicity theorem in its reflexive-coequalizer form says: if $F⊣G:\mathcal{D}\to \mathcal{C}$, $G$ is a conservative functor, $\mathcal{D}$ has reflexive coequalizers, and $G$ preserves them, then the Eilenberg-Moore comparison functor $K:\mathcal{D}\to {\mathcal{C}}^T$, for $T=GF$, is an equivalence of categories. Requiring all coequalizers to exist and be preserved is a stronger sufficient form.

For an algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is a reflexive pair with common section $F{\eta }_A$: both composites are the identity by the algebra unit law and the triangle identity. An arrow $u:FA\to B$ transposes to $v=Gu{\eta }_A:A\to GB$. The two composites $uFa$ and $u{\epsilon }_{FA}$ transpose to $va$ and $G{\epsilon }_{B}Tv$, respectively. Indeed the latter transpose is $Gu{\mu }_A{\eta }_{TA}=Gu$, while $G{\epsilon }_{B}Tv=Gu{\mu }_{A}T{\eta }_A=Gu$ by counit naturality. Thus $u$ equalizes the pair exactly when $v$ is an monad algebra morphism $(A,a)\to K(B)$. The coequalizer property gives natural bijectionsThese define $L$ on arrows by uniqueness, so $L⊣K$. Its unit has underlying arrow $\beta =Gq{\eta }_A:A\to GL(A,a)$.

By preservation, $Gq$ coequalizes $Ta$ and ${\mu }_A$ in $\mathcal{C}$. The action $a:TA\to A$ is a split coequalizer of that pair: take $s={\eta }_A$ and $t={\eta }_{TA}$, with ${\mu }_{A}t=1_{TA}$ and $Tat={\eta }_{A}a$. Thus there is a unique isomorphism $\alpha :GL(A,a)\to A$ with $\alpha Gq=a$. We have $\alpha \beta =a{\eta }_A=1_A$. AlsoConsequently $\beta \alpha Gq=Gq$, and epimorphic cancellation gives $\beta \alpha =1$. The adjunction unit $\beta$ is an monad algebra morphism; its invertible underlying arrow has an algebra-morphism inverse. Thus the unit of $L⊣K$ is an isomorphism.

For its counit $c_B:LK(B)\to B$, the triangle identity gives $K(c_B){\beta }_{K(B)}=1_{K(B)}$. Hence $K(c_B)$, and therefore $G(c_B)$, is an isomorphism. Since $G$ is a conservative functor, $c_B$ is an isomorphism too. Both unit and counit of $L⊣K$ are invertible, which proves the claimed equivalence. This proof uses only reflexive coequalizers in $\mathcal{D}$ and split coequalizers in $\mathcal{C}$.

For an algebra for a monad $(A,a)$, consider the fork in the Eilenberg-Moore categoryHere ${\mu }_A:F(TA)\to F(A)$ is the counit at $F(A)$, and $Ta=F(a)$. The arrow $a$ is an algebra morphism by $a{\mu }_A=aTa$, which also says that it coequalizes the two arrows.

Let $h:F(A)\to (B,b)$ be an algebra morphism with $h{\mu }_A=hTa$. Define $g=h{\eta }_A:A\to B$. Naturality of $\eta$ at $a$ givesSince $h$ is an algebra morphism,Thus $g:(A,a)\to (B,b)$ is an algebra morphism with $ga=h$. Any other such factorization satisfies $g^{\prime }=g^{\prime }a{\eta }_A=h{\eta }_A=g$. This proves the full coequalizer universal property inside the algebra category.

The pair is moreover a reflexive pair: its common section is $F({\eta }_A)$, with underlying map $T{\eta }_A$, because ${\mu }_{A}T{\eta }_A=1_{TA}$ and $TaT{\eta }_A=T(a{\eta }_A)=1_{TA}$. Hence every monad algebra is a reflexive coequalizer of free algebras. This reflexive free-algebra presentation of a monad algebra needs no general existence theorem for arbitrary algebra-category colimits.

A preadditive category has an abelian group structure on every hom-set, with composition additive in each variable. Neither a zero object nor biproducts are part of this definition.

Fix $C$ and use the reflexive pair $f,g:A\rightrightarrows B$, $r:B\to A$, with $fr=gr=1_B$. Take objects $x:C\to B$ and arrows $a:C\to A$, with source $fa$, target $ga$, and identity at $x$ equal to $rx$. For composable $a,b$, so $ga=fb$, defineThe preadditive category axioms giveThus the formula has the required endpoints. The identities satisfy $(rga)\circ a=a$ and $a\circ (rfa)=a$, using $gr=fr=1_B$.

For $ga=fb$ and $gb=fc$, both ways of composing three arrows equalIndeed $g(b\circ a)=gb$, while expanding $c\circ b$ and then composing with $a$ gives the same expression. Hence composition is associative.

Every arrow has inverseIts source is $ga$ and its target is $fa$. Substituting in the composition formula gives $a^{-1}\circ a=rfa$ and $a\circ a^{-1}=rga$. Therefore this is a groupoid.

For $u:C^{\prime }\to C$, precomposition by $u$ preserves sources, targets, identities, composition and inverses by bilinearity. Thus the construction is natural in $C$, giving the requested internal groupoid structure in its hom-set formulation. If the composable-arrow pullback in a category exists, the same formula defines its internal composition morphism. The reflexive-pair groupoid formula in a preadditive category requires no extra additive-category hypotheses.

For an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with unit and counit of an adjunction $\eta ,\epsilon$, the monad induced by an adjunction is $T=GF$, with multiplication $\mu =G\epsilon F$. Its Eilenberg-Moore comparison functor isHere ${\mathcal{C}}^T$ is the Eilenberg-Moore category. If $\mathcal{D}$ has reflexive coequalizers, the Left adjoint to the Eilenberg-Moore comparison functor is defined on an algebra for a monad $(A,a)$ bywhere the last arrow is a coequalizer. The parallel pair has common section $F{\eta }_A$, so it is a reflexive pair. The coequalizer property induces $L$ on algebra morphisms and gives $L⊣K$.

Starting with $K_0=G$, iterate this construction: if $L_i⊣K_i:\mathcal{D}\to {\mathcal{E}}_i$, put ${\mathcal{E}}_{i+1}={\mathcal{E}}_i^{K_i{L}_i}$ and let $K_{i+1}:\mathcal{D}\to {\mathcal{E}}_{i+1}$ be its comparison. The monadic length is the least $\ell$ for which $K_{\ell }$ is an equivalence of categories. Thus an equivalence has length zero, and an already monadic adjunction whose right adjoint is not an equivalence has length one. The assumed reflexive coequalizers ensure the needed left adjoints; in the example we identify them explicitly.

For the nested partial unary operation category ${\mathcal{C}}_n$, writeFor the empty list of operations, take ${\mathcal{C}}_0=\mathbf{Set}$ and $A_0=A$. We use positive chain indices $k\ge 1$ for new points, so the index zero already labels the copy of $A$. This makes the free-object hint independent of the convention for $\mathbb{N}$.

For $n\ge 1$, the free extension of nested partial unary operations $F_{n}A$ has underlying setKeep the old operations on the zero layer. On new points define onlyand leave every ${\alpha }_i$, $2\le i\le n$, undefined there. Define the new top operation byand leave it undefined elsewhere. These are precisely the permitted domains: the first unary operation on every new point moves it, so none of the higher operations is defined there. When $n=0$, take $F_{0}A=A\times {\mathbb{N}}_{\ge 0}$ with ${\alpha }_1(a,k)=(a,k+1)$. In either case the unit is $a\mapsto (a,0)$.

For $B\in {\mathcal{C}}_{n+1}$, let its operations be ${\beta }_i$. Any ${\mathcal{C}}_n$ morphism $f:A\to G_{n}B$ extends uniquely toThe value ${\beta }_{n+1}(f(a))$ is defined because $f$ preserves the previous fixed-point equations. The extension respects the top operation at old fixed points and respects the first operation along each new chain; there are no other defined operations on those chains to check. Uniqueness follows from the same forced formulas. For $n=0$ the formula is $\bar{f}(a,k)={\beta }_1^k(f(a))$. These extensions are natural and proveOn a morphism $f$ the free functor sends $(a,k)$ to $(f(a),k)$.

Crucially, the top operation of $F_{n}A$ has no fixed points: its defined old values move to new points, and it has no defined new values when $n\ge 1$; for $n=0$ it is the shift. The next free extension therefore adds no points and equips the object with an everywhere-undefined next operation. The same remains true at every subsequent free extension. Hence, for $m>n$, the composite left adjoint $F_{m-1}\cdots F_n$ has the same first $n+1$ operations and underlying set as $F_n$.

This agreement includes the monad structure, not just its underlying endofunctor. The units are the same zero-layer inclusions. The composite counit on an object of ${\mathcal{C}}_m$ is the same forced extension formula, depending only on its first $n+1$ operations. Applying the composite forgetful functor therefore gives the same multiplication $G_n\epsilon F_n$. Thus

Next prove creation of the required split coequalizers. Suppose $f,g:B\rightrightarrows C$ are arrows in ${\mathcal{C}}_{n+1}$ and their images have a split coequalizer $q:C\to Q$ in ${\mathcal{C}}_n$. Choose the splitting orientationwith $s:Q\to C$, $t:C\to B$ morphisms in ${\mathcal{C}}_n$. Denote the top operations on $B,C$ by $b,c$. For $z\in Q_n$ defineSince $s$ preserves the lower operations, $s(z)\in C_n$ and this value exists. For $y\in C_n$, $t(y)\in B_n$, so preservation by $f,g$ givesThus $q$ respects the top operation. The splitting $s$ and $qs=1$ also show that this is the unique operation on its prescribed domain $Q_n$ for which $q$ is a morphism. For $n=0$ all these domains mean the whole underlying set.

If $u:C\to D$ is a ${\mathcal{C}}_{n+1}$ morphism equalizing $f,g$, its unique factor $v:Q\to G_{n}D$ in ${\mathcal{C}}_n$ satisfies, on $Q_n$,where $d$ is the top operation of $D$. Hence $v$ is a ${\mathcal{C}}_{n+1}$ morphism. This proves $G_n$ creates coequalizers of $G_n$-split pairs.

Moreover, $G_n$ reflects isomorphisms: an isomorphism of the first $n$ operations reflects their fixed-point conditions, hence the domain of the next operation; the inverse of a bijective top-operation-preserving map then preserves that operation too. The Beck monadicity theorem now givesUnder this equivalence, the first Eilenberg-Moore comparison functor for the long composite is exactly forgetting all operations after the $(n+1)$st: its algebra structure is the top-operation extension formula already computed. Repeating the argument identifies its monadic tower with the successive categoriesAt each stage the corresponding comparison has the remaining composite of the already constructed free extensions as a left adjoint.

No earlier stage can be an equivalence of categories. For $j<m$, take the two-point set with its first $j$ operations all identities. One extension to ${\mathcal{C}}_m$ has every later operation the identity; another has its $(j+1)$st operation interchange the two points, and all subsequent operations undefined. Both have the same image in ${\mathcal{C}}_j$, but the identity function between those images is not a morphism between the extensions. Thus forgetting from ${\mathcal{C}}_m$ to ${\mathcal{C}}_j$ is not full. The final comparison at $j=m$ is an equivalence. This proves the monadic tower for nested partial unary operations has

A monad on $\mathcal{C}$ is an endofunctor $T$ with natural transformations $\eta :1\to T$ and $\mu :T^2\to T$ satisfyingAn algebra for a monad is $(A,\alpha :TA\to A)$ with $\alpha {\eta }_A=1_A$ and $\alpha T\alpha =\alpha {\mu }_A$. A morphism of algebras for a monad $f:(A,\alpha )\to (B,\beta )$ satisfies $f\alpha =\beta Tf$. These form the Eilenberg-Moore category ${\mathcal{C}}^T$. Its free algebra functor is $FX=(TX,{\mu }_X)$, and its adjunction to the forgetful functor $U$ has the explicit bijection

Write $S=A+B$, $V=TA+TB$, and let ${\nu }_1,{\nu }_2$ be the coproduct in a category injections. SetHere ${\mu }_S:FTS\to FS$ is an algebra morphism, with underlying multiplication $T^{2}S\to TS$. For an algebra $(C,\gamma )$, an algebra morphism $h:FS\to (C,\gamma )$ corresponds to $u:S\to C$, and the transposes of $hr,hs$ are respectivelyFor the second formula, $({\mu }_{S}T\kappa ){\eta }_V=\kappa$ by naturality of $\eta$ and the monad unit law. Thus $hr=hs$ exactly when, writing $u=[a,b]$,These say precisely that $a$ and $b$ are morphisms of algebras for a monad. Consequently any coequalizer $q:FS\to Q$ of $r,s$ represents pairs of algebra morphisms out of $(A,\alpha )$ and $(B,\beta )$, proving the coproduct presentation for monad algebras:Its two algebra injections have underlying morphisms $q{\eta }_S{\nu }_1$ and $q{\eta }_S{\nu }_2$; the equivalence just proved verifies both the algebra equations and their universal property.

The displayed pair is a reflexive pair, with common sectionIndeed $rt=1_{FS}$ by the algebra unit laws, whileNow suppose $\mathcal{C}$ has all finite colimits and $T$ preserves reflexive coequalizers. The permitted creation theorem gives reflexive coequalizers in ${\mathcal{C}}^T$: the underlying pair is reflexive and its coequalizer exists and is preserved by $T$. The preceding construction therefore gives binary algebra coproducts in a category. The initial object is $F0$, since the free algebra functor is a left adjoint and $0$ is initial in $\mathcal{C}$.

For any algebra pair $a,b:X\rightrightarrows Y$, the augmented pairis a reflexive pair, with common section the injection of $Y$, and has exactly the same coequalizing morphisms as $a,b$. Hence all coequalizers exist in ${\mathcal{C}}^T$. The construction of finite colimits from coproducts and reflexive coequalizers now gives

Use the category of commutative monoids, which is a semi-additive category: hom-sets have pointwise addition, composition is additive in each variable, and finite cartesian products are also coproducts. Explicitly, maps $M_i\to N$ extend from the coordinate injections by $(m_i)\mapsto {\sum }_i{f}_i(m_i)$.

Take the submonoidwith coordinatewise addition, and the two projections $f(m,n)=m$, $g(m,n)=n$. The homomorphism $r(m)=(m,m)$ satisfies $fr=gr=1_B$, so this is a reflexive pair. For $C=\mathbb{N}$, evaluation at $1$ identifies $\mathcal{C}(C,B)$ with $B$ and $\mathcal{C}(C,A)$ with $A$.

The element $(1,2)$ gives an arrow from $1$ to $2$ in the prescribed graph. An inverse would require an element $(2,1)$ of $A$, which does not exist. Therefore no groupoid structure with these source and target maps is possible, regardless of the proposed composition rule:

Let $r:B\to A$ split the reflexive pair $f,g$. Fix $C$, and take objects $u:C\to B$ and arrows $a:C\to A$ with source $fa$ and target $ga$. Since $\mathcal{C}$ is an additive category, its hom-sets are abelian groups, and composition is additive.

The identity at $u$ is $ru$. For $a:u\to v$ and $b:v\to w$, defineIts source is $fa-v+fb=u-v+v=u$ and its target is $ga-v+gb=v-v+w=w$. The two identity laws follow from $a-rv+rv=a$ and $ru-ru+a=a$. For $c:w\to x$, both parenthesizations of the triple composite are $a-rv+b-rw+c$, so composition is associative.

Define the inverse byIts source is $u+v-u=v$ and its target is $u+v-v=u$. Moreover $a^{-1}\star a=ru$ and $a\star a^{-1}=rv$. Thus these data form a groupoid, proving the represented form of reflexive pair in an additive category is an internal groupoid:

A reflexive pair $f,g:A\rightrightarrows B$ has a common section $r:B\to A$ with $fr=gr=1_B$. Form the pushout in a category of $f$ and $g$, and write its two maps from $B$ as $j_1,j_2$. Its relation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. Thus $qf=qg$. Any $h:B\to C$ with $hf=hg$ gives the compatible pair $(h,h)$ for this pushout and therefore factors uniquely through $q$. This proves the coequalizer of a reflexive pair from a pushout construction.

There is a genuine transcription difference: the original PDF says finite products in the finite-colimit assertion, while the TeX says finite coproducts. The PDF assertion is false with products. Even retaining a pushout-existence assumption, the full subcategory of the Category of sets on nonempty sets has finite products, pushouts, and all coequalizers, but no initial object, hence no empty colimit. Regard the poset with elements $0,a,b,u_0,u_1,\dots$, orderand incomparable $a,b$ as a category. It has finite meets and top $u_0$, hence finite products in a category. Every reflexive pair in a poset is an equal pair and has its identity as a coequalizer. But $a,b$ have no least upper bound, so they have no coproduct in a category.

For the corrected construction using finite coproducts, let $D:\mathcal{J}\to \mathcal{C}$ be a finite diagram in a category. PutDefine $s,t:Y\rightrightarrows X$ on the summand indexed by $a:i\to j$ as ${\nu }_i$ and ${\nu }_{j}D(a)$. Maps $h:X\to Z$ with $hs=ht$ are exactly cocone under a diagram data on $D$. Hence their universal coequalizer is its colimit. Replace $s,t$ by the reflexive pairwhose common section is the second injection and whose coequalizer is unchanged. This proves construction of finite colimits from coproducts and reflexive coequalizers, including the empty diagram via the empty coproduct:

A semi-additive category is a category whose hom-sets are commutative monoids and whose composition is additive in each variable, with finite products and coproducts.

Suppose first that $P=A\times B$ is a binary product in a category, with projections $p_1,p_2$. The zero morphisms and the product property define mapsbyThe two projections of $i_1{p}_1+i_2{p}_2$ equal those of $1_P$, soFor $f:A\to C$ and $g:B\to C$, the mapsatisfies $h{i}_1=f$ and $h{i}_2=g$. If $k$ has the same restrictions, thenThus $(P,i_1,i_2)$ is also the binary coproduct. The dual argument starts from a coproduct and makes it a product. Hence binary products and coproducts coincide canonically as biproducts.

Let $f,g:A\rightrightarrows B$ be a reflexive pair in an additive category, with $fr=gr=1_B$. For every object $C$, regard $x:C\to A$ as an arrow from $fx$ to $gx$ between objects of $\mathcal{C}(C,B)$. The identity at $b:C\to B$ is $rb$.

If $gx=fy$, define the composite byIts source and target areThe identities follow fromand associativity follows immediately by expanding both iterated composites and using the matching equations. The inverse of $x$ iswhose source is $gx$, whose target is $fx$, and whose two composites with $x$ are the appropriate identity arrows. These formulas are natural in $C$, so the Yoneda lemma identifies them with structure morphisms in $\mathcal{C}$. The pair is therefore an internal groupoid, proving that every reflexive pair in an additive category is an internal groupoid.

This fails for semi-additive categories. In the category of commutative monoids, letunder coordinatewise addition. The two projections $f,g:R\rightrightarrows \mathbb{N}$ have the common splitting $r(n)=(n,n)$, so they form a reflexive pair. Its underlying reflexive graph is the usual order category on $\mathbb{N}$: there is an arrow $m\to n$ exactly when $m\le n$. If it were an internal groupoid, the arrow $0\to 1$ would have an inverse $1\to 0$, but $(1,0)\notin R$. Therefore this reflexive pair is not an internal groupoid, and “additive” cannot be weakened to “semi-additive.”