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A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying $\mu T\eta =\mu \eta T=1_T$ and $\mu T\mu =\mu \mu T$. Its Eilenberg-Moore category has objects $(X,a)$ with $a:TX\to X$, $a{\eta }_X=1_X$ and $a{\mu }_X=aTa$. A morphism $f:(X,a)\to (Y,b)$ satisfies $fa=bTf$. An adjunction $F⊣G$, with unit $\eta$ and counit $\epsilon$, induces $T=GF$, $\mu =G\epsilon F$. It is a monadic adjunction if the Eilenberg-Moore comparison functor $K(D)=(GD,G{\epsilon }_D)$ is an equivalence with the Eilenberg-Moore category.

A fork $A\stackrel{p,q}{\rightrightarrows }B\stackrel{r}{\to }C$ is a split coequalizer if $rp=rq$ and there are $s:C\to B$, $t:B\to A$ with $rs=1$, $pt=1$ and $qt=sr$. It is indeed a coequalizer: if $hp=hq$, then $h=hpt=hqt=hsr$, so $h$ factors uniquely through $r$, which is split epic. Every functor preserves these splitting identities, making such coequalizers absolute.

The precise monadicity theorem states: for an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$, the comparison is an equivalence if and only if $G$ reflects isomorphisms and every pair in $\mathcal{D}$ whose $G$-image admits a split coequalizer has a coequalizer which $G$ preserves. Equivalently, the right adjoint creates these split-underlying coequalizers, with creation understood up to the canonical isomorphism of the prescribed underlying coequalizer. Here is a proof.

First consider the forgetful functor from monad algebras. It reflects isomorphisms: the inverse of an invertible algebra morphism also obeys the algebra-morphism identity, by multiplying that identity by the inverses. For an algebra pair with a split underlying coequalizer $r:B\to C$, all powers of $T$ preserve its coequalizer. If $b:TB\to B$ is the algebra structure on $B$, the map $rb$ coequalizes $Tp,Tq$, so there is a unique $c:TC\to C$ withThe unit law follows by composing with the epimorphism $r$: $c{\eta }_{C}r=cTr{\eta }_B=rb{\eta }_B=r$. For associativity, naturality and the algebra law giveCancel the epimorphism $T^{2}r$ to obtain $c{\mu }_C=cTc$. Thus $(C,c)$ is an algebra and $r$ is an algebra morphism. Any algebra morphism coequalizing the pair factors through $r$ on underlying objects, and its factor is an algebra morphism by cancellation of the epimorphism $Tr$. This creates the coequalizer. Transporting these facts through an equivalence proves necessity.

For sufficiency, let $(X,a)$ be a $T$-algebra. In $\mathcal{D}$ consider the free pairIts underlying pair is ${\mu }_X,Ta:T^{2}X\rightrightarrows TX$, with split coequalizer $a:TX\to X$: the splittings are $s={\eta }_X$ and $t={\eta }_{TX}$, using the algebra unit law, the monad unit law and naturality of $\eta$. By hypothesis its coequalizer $q:FX\to D$ exists and is preserved. Identify $GD$ with $X$ by the unique coequalizer isomorphism, so $Gq=a$. Naturality of the counit makes $Kq$ an algebra morphism, hence the comparison structure $c$ on $GD$ satisfies $cTa=a{\mu }_X=aTa$. The map $Ta$ is split epic because $a{\eta }_X=1$, so $c=a$. Every algebra is therefore isomorphic to a comparison algebra.

For an actual object $D$, its counit ${\epsilon }_D:FGD\to D$ is a coequalizer of the analogous free pair. It coequalizes that pair by counit naturality. Compare it with the coequalizer furnished by the hypothesis; applying $G$ gives an isomorphism, since both underlying arrows coequalize the same split pair. Reflection of isomorphisms makes the comparison map an isomorphism. This proves the counit's asserted coequalizer property.

Let $f:KD\to KE$ be an algebra morphism. Its transpose is $h_0={\epsilon }_{E}Ff:FGD\to E$. The two composites of $h_0$ with the free pair have adjunct transposes $G{\epsilon }_{E}Tf$ and $fG{\epsilon }_D$, equal because $f$ is an algebra morphism. Consequently $h_0$ factors uniquely through ${\epsilon }_D$ as $h:D\to E$. Applying $G$ gives $(Gh)G{\epsilon }_D=fG{\epsilon }_D$; the last factor is split epic, so $Gh=f$. Conversely any lift of $f$ must satisfy $h{\epsilon }_D={\epsilon }_{E}Ff$ by naturality and hence is unique. Thus $K$ is full and faithful, and the earlier construction makes it essentially surjective. The comparison is an equivalence, completing the theorem's proof.

Now take $G:\mathbf{CompHaus}\to \mathbf{Set}$. Its left adjoint exists by the permitted topological result; it can be realized as the Stone-Čech compactification of the discrete input set. A bijective continuous map from a compact space to a Hausdorff space is a homeomorphism, so $G$ reflects isomorphisms.

For continuous $f,g:X\rightrightarrows Y$ with a split underlying fork $q,s,t$, let $R$ be the relation of having the same $q$-image. We claimFor the forward inclusion choose $x=t(y)$, $x^{\prime }=t(y^{\prime })$: $ft=1$ and $gt=sq$ supply the equality. For the reverse inclusion use $qf=qg$. The subset $\{(x,x^{\prime }):g(x)=g(x^{\prime })\}\subset X^2$ is closed, because $Y$ is Hausdorff, and compact. Its continuous image $R\subset Y^2$ is compact and hence closed. A closed equivalence relation on a compact Hausdorff space has a compact Hausdorff quotient; its quotient map is a coequalizer here, since it identifies exactly the generated relation. Its underlying set coequalizer is the specified one up to the unique isomorphism. Thus $G$ preserves the needed coequalizers, andThe closedness argument is the closed relation generated by a set-split pair of compact Hausdorff maps lemma; the splitting maps themselves need not be continuous.

Every covariant representable functor preserves categorical limits: arrows from its representing object into a limiting object are exactly compatible cones of arrows into the diagram. Apply the density formula proved in the root solution. For any finite diagram $D:J\to \mathcal{C}$,The middle isomorphism uses exactly the permitted theorem that filtered colimits commute with finite limits in sets; it applies because $E^{\mathrm{op}}$ is filtered. All maps are the natural comparison maps, so $F$ preserves finite limits, including the terminal object. This proves $\boxed{\text{(iii)}\Rightarrow \text{(i)}}$ and completes the equivalence.

Assume $F$ is full and faithful and its image is closed under regular quotients. Its unit is invertible, hence monic. Factor the counit using the hypothesis on $\mathcal{D}$:with $e$ regular epic and $m$ monic. Closure gives $I\cong F{A}^{\prime }$ for some $A^{\prime }$. A right adjoint preserves monomorphisms: postcomposition with $Gm$ on $\mathcal{C}(X,-)$ corresponds under adjunction to postcomposition with $m$ on $\mathcal{D}(FX,-)$, which is injective. Thus $Gm$ is monic.

The triangle identity and invertibility of ${\eta }_{GB}$ give $G{\epsilon }_B=({\eta }_{GB}{)}^{-1}$, so $GmGe$ is invertible. A monomorphism whose composite with another arrow is an isomorphism is itself split epic and hence an isomorphism; therefore both $Gm$ and $Ge$ are isomorphisms. The arrow $e$ is between objects in the essential image of the full and faithful $F$. After choosing $I\cong F{A}^{\prime }$, write it as $Fh$. Naturality of the invertible unit identifies $GFh$ with $h$, so $Ge$ invertible implies $h$ and hence $e$ invertible. Now ${\epsilon }_B=me$ is monic. This proves $\boxed{\text{(ii)}\Rightarrow \text{(i)}}$.

Assume both unit and counit are monic. The unit criterion makes $F$ faithful. The triangle identity is ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Since ${\epsilon }_{FA}$ is monic, this split epimorphism is an isomorphism: cancelling it in ${\epsilon }_{FA}(F{\eta }_A{\epsilon }_{FA})={\epsilon }_{FA}$ gives $F{\eta }_A{\epsilon }_{FA}=1$. Thus $F{\eta }_A$ is invertible. As $\mathcal{C}$ is balanced, the reflection result in the root solution makes ${\eta }_A$ invertible. Hence $F$ is full and faithful.

Let $q:FA\to B$ be a regular epimorphism. Its adjunct factorization is $q={\epsilon }_{B}Fh$ for $h:A\to GB$. A regular epimorphism is extremal: if a coequalizer $q$ factors through a monomorphism $m$, its other factor also coequalizes the defining pair, descends through $q$, and gives a right inverse for $m$; monicity then makes $m$ invertible. Apply this to the monic ${\epsilon }_B$. More explicitly, if $q$ coequalizes $r,s$, monicity gives $Fhr=Fhs$, so there is $d:B\to FGB$ with $dq=Fh$. Then ${\epsilon }_{B}dq=q$ implies ${\epsilon }_{B}d=1$, and monicity gives $d{\epsilon }_B=1$. Therefore $B\cong FGB$. The image is closed under regular quotients, proving $\boxed{\text{(i)}\Rightarrow \text{(ii)}}$.

A balanced category is one in which every morphism that is both monic and epic is an isomorphism. If a faithful functor sends $f$ to an isomorphism, then $f$ is monic: equality $fg=fh$ becomes $FfFg=FfFh$, cancellation gives $Fg=Fh$, and faithfulness gives $g=h$. The same argument on the other side makes $f$ epic. Balance therefore makes $f$ an isomorphism. Thus a faithful functor out of a balanced category reflects isomorphisms.

Under the adjunction, the transpose of $Ff:FX\to FA$ is ${\eta }_{A}f:X\to GFA$. Hence if $F$ is faithful, ${\eta }_{A}f={\eta }_{A}g$ forces $Ff=Fg$ and then $f=g$, making every ${\eta }_A$ monic. Conversely, if the unit is pointwise monic and $Ff=Fg$, naturality gives ${\eta }_{B}f=GFf{\eta }_A=GFg{\eta }_A={\eta }_{B}g$, whence $f=g$. ThereforeWe also use the standard adjunction fact with its justification: $F$ is full and faithful exactly when $\eta$ is invertible. Indeed the composite hom-set map $\mathcal{C}(X,A)\to \mathcal{D}(FX,FA)\cong \mathcal{C}(X,GFA)$ is postcomposition by ${\eta }_A$. If $F$ is full and faithful these maps are bijective; taking $X=GFA$ produces a right inverse, and taking $X=A$ gives injectivity and therefore a left inverse. Conversely an invertible unit makes these hom-set maps bijective.

For the final example, view the two-element ordered set $0<1$ as a category $\mathcal{C}$, and let $\mathcal{D}$ have one object and only its identity. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to the functor $G$ selecting $1$, since every $\mathcal{C}(x,1)$ is a singleton. Every arrow in an ordered-set category is monic, so the unit arrows $x\to 1$ and the identity counit are monic. But $F$ is not full: $\mathcal{C}(1,0)$ is empty whereas $\mathcal{D}(F1,F0)$ is a singleton. The example does not contradict the upcoming equivalence because $\mathcal{C}$ is not balanced; its arrow $0\to 1$ is both monic and epic but not invertible.

Assume subobjects of projectives are projective. For any object $X$, choose $0\to K\to P\to X\to 0$ with $P$ projective. The kernel in a category $K$ is projective by the assumption. The derived long exact sequence gives a dimension-shifting isomorphismbecause $L_{2}F(P)=L_{1}F(P)=0$. The right side is zero because $K$ is projective. Since $X$ and the right exact functor $F$ were arbitrary, $\boxed{\text{(i)}\Rightarrow \text{(ii)}}$.

An abelian category is an additive category with kernels in a category and cokernels in a category in which the canonical comparison from coimage to image of every morphism is an isomorphism. This is the standard image-coimage form of the definition, equivalently expressed by normality of every monomorphism and epimorphism. Additivity means abelian-group hom-sets with bilinear composition, a zero object and finite biproducts. A sequence is exact at an object when the image of the incoming morphism is the kernel in a category of the outgoing one.

For $f:X\to Y$, construct $q:X\to \mathrm{coim}f=\mathrm{coker}(\ker f)$ and $j:\mathrm{im}f=\ker (\mathrm{coker}f)\to Y$. The universal properties give $f=j\bar{f}q$: first $f$ kills its kernel in a category and factors through $q$; then the resulting map is killed by $\mathrm{coker}f$ because $q$ is epic, so it factors through $j$. The comparison $\bar{f}$ is invertible by the abelian axiom. Cokernels in a category are epic and kernels in a category monic, directly by uniqueness in their universal properties, so $e=\bar{f}q$ is epic and $j$ monic. Thus $\boxed{f=je\text{ is an epimorphism-monomorphism factorization}}$.

In a short exact sequence $0\to X\stackrel{i}{\to }Y\stackrel{p}{\to }Z\to 0$, we have $i=\ker p$ and $p=\mathrm{coker}i$, up to the canonical isomorphisms. If $r:Y\to X$ retracts $i$, the map $1_Y-ir$ kills $i$ and factors as $sp$ for some $s:Z\to Y$. Composing with $p$ gives $psp=p$, so epicity gives $ps=1_Z$. Conversely if $ps=1_Z$, then $1_Y-sp$ is killed by $p$ and factors as $ir$. Composing with $i$ gives $iri=i$, so monicity gives $ri=1_X$. ConsequentlyIn that case $Y\cong X\oplus Z$.

For the derived-functor requests, use additive right exact functors, as customary in abelian homological algebra. An abelian category with enough projectives permits an augmented projective resolution $\cdots \to P_2\to P_1\to P_0\to X\to 0$, obtained by repeatedly taking an epimorphism from a projective onto the previous kernel in a category. Apply $F$ and defineProjective lifts construct maps of resolutions; the comparison and chain-homotopy results make the construction independent of choices and functorial. For a short exact sequence, the horseshoe construction produces a degreewise split short exact sequence of resolutions, hence of their images under additive $F$. Its homology objects give the long exact sequenceThis supplies the requested sketch; detailed comparison and functoriality proofs are not needed here.

A projective object $X$ has a resolution concentrated in degree zero, so $L_{1}F(X)=0$ for every such $F$. Conversely choose $0\to K\stackrel{i}{\to }P\to X\to 0$ with $P$ projective, and take $F=\mathcal{A}(-,K)$ with target ${\mathbf{Ab}}^{\mathrm{op}}$. Precomposition makes this covariant to the opposite category. The usual left exactness of contravariant Hom, which follows directly from the cokernel in a category universal property, says exactly that this $F$ is right exact in that opposite target. If $L_{1}F(X)=0$, the long exact sequence and $L_{1}F(P)=0$ make $F(K)\to F(P)$ monic in ${\mathbf{Ab}}^{\mathrm{op}}$. Equivalently,is epic, hence surjective, in abelian groups. The identity of $K$ has a preimage, giving a retraction of $i$. The sequence splits, and $X$ is a direct summand of $P$, hence projective: a lifting problem for $X$ extends along the projection $P\to X$, is solved for $P$, and restricts along the summand inclusion. ThereforeThis is projectivity detected by all first left derived functors. The following parts prove the three requested equivalent formulations of a hereditary abelian category with enough projectives.

Use the defining criterion that a flat module $E$ makes $u\otimes 1_E$ injective for every injective module homomorphism $u$. By right exactness of the tensor product, preserving injections is equivalent to preserving every short exact sequence.

First assume $E$ is flat over $A$, and fix a maximal ideal $\mathfrak{m}$. Let $u:U\hookrightarrow V$ be an injection of $A_{\mathfrak{m}}$-modules. Regard it as an injection of $A$-modules. There are natural module isomorphismsFor example, $u_0\otimes (e/s)$ maps to $s^{-1}{u}_0\otimes e$, with inverse $u_0\otimes e\mapsto u_0\otimes (e/1)$. These are well-defined because every $s\notin \mathfrak{m}$ already acts invertibly on $U$. Under these isomorphisms the tensor map is $u{\otimes }_A{1}_E$, which is injective by flatness. Thus $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$.

For the converse, assume every $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$ and take any injection $u:U\hookrightarrow V$ of $A$-modules. By part (c), each $u_{\mathfrak{m}}$ is injective. Since localization commutes with tensor products, the localization of $u{\otimes }_A{1}_E$ identifies withIt is injective by the assumed local flatness. Apply part (c) again, now to $u{\otimes }_A{1}_E$, to conclude that it is injective globally. Since $u$ was arbitrary, $E$ is flat. ThereforeThis proves flatness is local without any finite-generation hypothesis.

Let $S=A\setminus \mathfrak{p}$, let $S$ act on $B$ through $\phi$, and write $\mathfrak{p}B$ for the extension of an ideal generated by $\phi (\mathfrak{p})$. Since $\kappa (\mathfrak{p})=S^{-1}(A/\mathfrak{p})$, the fiber ring has the canonical presentationsThe first isomorphism sends $b\otimes (\bar{a}/s)$ to $\phi (a)b/s$ modulo $\mathfrak{p}B$. Its inverse sends the class of $b/s$ to $b\otimes (\bar{1}/s)$. The tensor balancing relation, reduction modulo $\mathfrak{p}$ and the fraction relations show that these maps are well-defined and inverse. Thus the description is canonical and does not depend on chosen representatives.

Suppose $\mathfrak{q}$ is a prime ideal of $B$ with ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$. Then $\mathfrak{p}B\subseteq \mathfrak{q}$ and $\mathfrak{q}$ is disjoint from $\phi (S)$. Its localization $S^{-1}\mathfrak{q}$ is a proper prime, since the quotient is the localization of the integral domain $B/\mathfrak{q}$. It contains $\mathfrak{p}{S}^{-1}B$, and therefore gives a primeConversely, for a prime $\mathfrak{Q}\subset C$, take its contraction $\mathfrak{q}$ under the ring homomorphism $\psi :B\to C$, $b\mapsto b\otimes 1$. Contraction preserves primality. Every $\phi (a)$ with $a\in \mathfrak{p}$ maps to zero, while every $\phi (s)$ with $s\in S$ maps to a unit and hence cannot belong to a proper prime. Thus ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$.

These constructions are inverse. For a prime disjoint from $\phi (S)$, the fraction $b/1$ belongs to its localized extension exactly when some $s\in S$ has $\phi (s)b\in \mathfrak{q}$; primality then implies $b\in \mathfrak{q}$. In the other direction, every element of $C$ has a representative $b/s$, and membership in $\mathfrak{Q}$ is equivalent to membership of $b/1$ because the denominator is a unit. Hence $\mathfrak{Q}$ is recovered by extending its contraction and quotienting. We have proved the canonical fiber of the map on spectra correspondence:If a denominator maps to zero, the fiber ring is the zero ring and both sets are empty; the same argument includes this case.

The dual space is $V^{\ast }={\mathrm{Hom}}_F(V,F)$, the vector space of all linear functionals with pointwise operations. For finite $d=\dim V$, choose a basis $v_1,\dots ,v_d$ and define its dual basis by ${\epsilon }_j(v_i)={\delta }_{ij}$. Every linear functional is $\ell ={\sum }_j\ell (v_j){\epsilon }_j$, and evaluation on each $v_i$ proves that the ${\epsilon }_j$ are linearly independent. Thus $\dim V^{\ast }=\dim V=d$. This dimension formula is a finite-dimensional assertion.

On the space of polynomials of degree at most $n$, define the evaluation map $E(p)=(p(a_0),\dots ,p(a_n))$. Its kernel is zero: a nonzero degree-at-most-$n$ polynomial cannot have $n+1$ distinct roots. Since domain and codomain both have dimension $n+1$, $E$ is an isomorphism. The evaluations therefore form a basis of the dual space. The linear functional $p\mapsto p^{\prime }(0)$ has a unique expansion in that basis, proving the existence and uniqueness of the requested weights.

Explicitly, use the Lagrange interpolation polynomial basisThe differentiation weights from nodal evaluations are ${\lambda }_j={\ell }_j^{\prime }(0)$. If a node is zero, no singular reciprocal formula is needed: differentiate the finite product directly. For $n=0$ the sole weight is zero.

A composition series is a subnormal chain whose nontrivial successive quotients are simple. The Jordan–Hölder theorem asserts that any two composition series of a finite group have the same length and the same multiset of simple factors, up to isomorphism. Existence follows by successively choosing maximal proper normal subgroups until the identity is reached.

For uniqueness, induct on the group order. Let $A,B$ be the penultimate terms of two series. If $A=B$, apply induction to that subgroup and add the common top factor. If $A\ne B$, their normality and maximality give $AB=G$. Put $C=A\cap B$. The isomorphism theorems giveBoth are simple. Take any composition series of $C$. Appending $A$ to it gives a series for $A$, so induction identifies the factors of the original series inside $A$ with the factors of $C$ together with $G/B$. Similarly the factors inside $B$ are those of $C$ together with $G/A$. Adding the respective top factors produces the same multiset in both cases. This proves uniqueness, including multiplicities and length.

A normal subgroup $K$ of order two is central because its automorphism group is trivial. By Jordan–Hölder, a composition series through $K$ leaves the simple quotient $G/K\cong A_5$. The image of $Z(G)$ in this quotient is central and hence trivial, since $A_5$ is nonabelian simple. ThusThe symmetric group $S_5$ has trivial group center: conjugation of each transposition by a central permutation must preserve its two-element support, forcing the permutation to fix every point. Since its Jordan–Hölder factors are $A_5$ and $C_2$, a normal subgroup of order two would equal that trivial group center, a contradiction.

Identify $C_2=\{1,-1\}$ and put $D=1\times A_5$ in $C_2\times S_5$. Its three maximal normal subgroups areThey are the three kernels of nonzero maps from the abelianization $C_2^2$ to $C_2$. There is no quotient $A_5$: a central $C_2$ must map trivially, while $S_5$ has no quotient $A_5$. The latter follows from its normal subgroups $1,A_5,S_5$, which can themselves be obtained by intersecting with the simple subgroup $A_5$ and using its trivial centralizer. The complete list of composition series isWithin $M_1$ the two direct factors are the only maximal normal subgroups; within each of $M_2,M_3\cong S_5$ the only such subgroup is its copy of $A_5$. Hence there are exactly four series.

For $S_5\times S_5$, put $D=A_5\times A_5$, $X=A_5\times 1$, $Y=1\times A_5$ andAgain these are exactly the maximal normal subgroups: they are the three cyclic quotient kernels, and there is no simple quotient $A_5$ of the whole product. All series aretogether withTo verify exhaustiveness, $M_L$ has maximal normal subgroups $D$ and $1\times S_5$, and $M_R$ has $D$ and $S_5\times 1$. A quotient $A_5$ of a direct product can receive a nontrivial image from only one simple factor, since the images commute and $A_5$ has trivial group center. The remaining maximal-normal possibility is $M_D$. Its derived subgroup is $D$, because both alternating factors are perfect. A hypothetical quotient $A_5$ would project one alternating factor isomorphically and kill the other, but conjugation by an odd diagonal element induces an outer automorphism on that surviving factor, impossible inside $A_5$. Odd conjugation is outer because otherwise an odd permutation times an even one would centralize $A_5$; a permutation commuting with every three-cycle fixes every three-element support and is the identity. Thus $D$ is the only maximal normal subgroup of $M_D$. Finally the only maximal normal subgroups of $D$ are $X,Y$: normal subgroups of a product of two centerless simple groups are products of the factors, as follows by commutating with each factor. This proves exactly eight series and the classification of composition chains in products with S5.

For $S{L}_2(5)$, commuting with both elementary upper and lower transvections forces a central matrix to be scalar. A scalar $\lambda I$ has determinant ${\lambda }^2$, soIf $g^2=I$, characteristic five makes $g$ diagonalizable with eigenvalues in $\{1,-1\}$. Determinant one rules out one of each, so the unique nonidentity involution is $-I$. An index-two subgroup would have order $60$ and, by the Cauchy theorem for groups, contain that involution and hence the group center. Its image in $PS{L}_2(5)$ would have index two. The simplicity and identification $PS{L}_2(5)\cong A_5$ are established by the independent matrix argument in Solution 4; a simple group has no proper index-two subgroup. Therefore $S{L}_2(5)$ has no subgroup of index two.