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A regular category has finite limits, regular-epimorphism--monomorphism image factorizations, and regular epimorphisms stable under pullback. A relation $A⇝B$ in $\mathbf{Rel}(\mathcal{C})$ is a subobject of $A\times B$; composition forms the pullback over the middle object and then takes its image.

Suppose $R:A⇝B$ has a right adjoint relation $S$, so $1_A\le SR$ and $RS\le 1_B$. In the internal regular logic, the first inequality says that for every $a$ there is a $b$ with $R(a,b)$ and $S(b,a)$. If also $R(a,b^{\prime })$, the second inequality forces $b=b^{\prime }$. Thus $R$ is total and single-valued. Categorically, if $R\hookrightarrow A\times B$ has projections $p:R\to A$ and $q:R\to B$, totality makes $p$ a regular epimorphism and single-valuedness makes it a monomorphism. Hence $p$ is an isomorphism and $R$ is the graph of a morphism as a relation $q{p}^{-1}:A\to B$. Conversely, the graph of any morphism is left adjoint to its converse relation, as the two required inequalities follow directly from equality. This proves the characterization.

Let $L$ be a frame. Composition in the category of matrices valued in a frame isIf $f⊣g$, the diagonal part of $1_A\le gf$ impliesso every row of $f$ joins to $1$. For $b\ne b^{\prime }$, distribute $f(a,b)\wedge f(a,b^{\prime })$ over the displayed join. Every term vanishes by $fg\le 1_B$, first using the factor with column $b$ and then the one with column $b^{\prime }$. HenceConversely, if the rows of $f$ join to $1$ and have pairwise disjoint entries, define $g(b,a)=f(a,b)$. Thenwhile $(fg)(b,b^{\prime })=0$ for $b\ne b^{\prime }$. Thus $1_A\le gf$ and $fg\le 1_B$, proving the stated criterion.

Finally take $L=\Omega (X)$ for a connected topological space $X$. For fixed $a$, the opens $f(a,b)$ are pairwise disjoint and cover $X$. Connectedness forces exactly one of them to be $X$ and all the others to be empty. Hence a left adjoint matrix determines a unique function $\phi :A\to B$ by $f(a,\phi (a))=X$. Conversely every function gives this matrix, and matrix composition agrees with function composition. The left adjoints in $\mathbf{Mat}(\Omega (X))$ therefore form a category isomorphic to $\mathbf{Set}$.

The statement that limits of shape $\mathcal{I}$ commute with colimits of shape $\mathcal{J}$ means that for every $D:\mathcal{I}\times \mathcal{J}\to \mathcal{C}$, the canonical commutation of limits and colimits mapis an isomorphism whenever the iterated limits and colimits exist.

A filtered category is a nonempty category in which every finite diagram has a cocone. Equivalently, any two objects map to a common object and any parallel pair becomes equal after postcomposition. A weakly filtered category requires cocones only for finite connected diagrams; equivalently, each connected component is filtered.

Write a weakly filtered category as the disjoint union ${\coprod }_s{\mathcal{J}}_s$ of its filtered connected components. Its colimit is the coproduct of the filtered colimits over the ${\mathcal{J}}_s$. In sets, a connected finite limit commutes with coproducts: connectedness forces all coordinates of a compatible tuple to lie in the same coproduct summand. By the assumed theorem, the filtered colimit over each ${\mathcal{J}}_s$ commutes with every finite limit. Applying these two facts successively proves that weakly filtered colimits commute with connected finite limits in $\mathbf{Set}$.

The forgetful functor from abelian group to sets creates finite limits and filtered colimits and reflects isomorphisms. The comparison map for a filtered colimit and a finite limit therefore becomes the corresponding isomorphism of sets, so filtered colimits commute with finite limits in $\mathbf{AbGp}$.

The dual claim fails because inverse limit need not preserve epimorphisms. Take the inverse systemswith identity bonding maps on $A_n$, reduction maps on $B_n$, and levelwise epimorphisms $q_n:A_n\to B_n$. Thenand the induced map $\mathbb{Z}\to {\mathbb{Z}}_2$ is not surjective. Since an epimorphism in abelian groups is a finite-colimit cokernel, cofiltered limits do not commute with finite colimits in $\mathbf{AbGp}$.

For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the monad induced by an adjunction is $T=GF$, with unit the adjunction unit and multiplication $\mu =G\epsilon F$. An algebra for a monad is a map $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. The Eilenberg-Moore comparison functor isand sends a morphism $h$ to $Gh$.

Given a $T$-algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is reflexive, with common section $F{\eta }_A$, so it exists by hypothesis. A map $J(A,a)\to D$ is equivalently a map $FA\to D$ equalizing the pair. Under adjunction this is exactly a map $A\to GD$ satisfying the $T$-algebra homomorphism equation. Hencenaturally, and $J⊣K$. Iterating these comparison adjunctions gives the monadic tower; the monadic length is the least number of steps required to reach an equivalence.

For a pair of sets $(X_0,X_1)$, defineA pair of maps $X_0\to A_0$, $X_1\to A_1$ extends uniquely to a commutative square from this inclusion to any injection $A_0\hookrightarrow A_1$, proving that $F$ is left adjoint to the forgetful functor $U:\mathbf{Mono}(\mathbf{Set})\to \mathbf{Set}\times \mathbf{Set}$.

The induced monad sends $(X_0,X_1)$ to $(X_0,X_0\bigsqcup X_1)$. Its algebra unit laws show that an algebra is precisely an arbitrary function $X_0\to X_1$, with no injectivity requirement. Thus its Eilenberg-Moore category is the arrow category $[2,\mathbf{Set}]$, and the comparison functor is the full inclusion of injections into all functions. This inclusion is not an equivalence, so the original adjunction is not monadic. It is reflective: a function $f:A_0\to A_1$ maps to its image inclusion $\mathrm{im}f\hookrightarrow A_1$. The supplied result that reflections are monadic says that the next comparison is an equivalence. Therefore the original adjunction has monadic length $2$.

A categorical limit of $D:\mathcal{J}\to \mathcal{C}$ is a terminal cone: it consists of an object $L$ and compatible maps $L\to D(j)$ through which every other cone factors uniquely. For a finite diagram, take the product $P={\prod }_{j}D(j)$ and the product $Q={\prod }_{u:i\to j}D(j)$. There are two maps $P\rightrightarrows Q$: on the $u:i\to j$ coordinate one uses respectively $D(u)$ after projection to $D(i)$ and direct projection to $D(j)$. Their equalizer is exactly the compatible-cone object. Hence finite products and equalizers construct every finite limit.

In the category of metric spaces and non-expansive maps, give $X\times Y$ the maximum metricThe projections are non-expansive, and a pair of non-expansive maps into $X$ and $Y$ induces a non-expansive map into this product, proving the universal property.

Let two maps $1\rightrightarrows X$ select $a,b\in X$. On the set quotient identifying $a$ and $b$, define the quotient metric by shortest paths that may jump from $a$ to $b$ at zero cost. Explicitly,This is the largest metric making the quotient map non-expansive. A map $X\to Z$ equalizing $a$ and $b$ factors through the set quotient and remains non-expansive by the path formula, so this is the coequalizer. Its underlying set is the set-theoretic coequalizer.

For $X=\{0,1,5,6\}$ with $0\sim 6$, write $\ast =\{0,6\}$. The quotient hasThus in $Q\times Y$, with $Y=\{0,3\}$,After first taking $X\times Y$, the product of the parallel pair identifies $(0,y)$ with $(6,y)$ separately for $y=0,3$. In its quotient metric, every path from $(1,0)$ to $(5,3)$ has length at least $4$, and length $4$ is attained either directly or via one of those identifications. The canonical bijection from this coequalizer to $Q\times Y$ is therefore not an isometry. Hence $-\times Y$ does not preserve this coequalizer. In a Cartesian closed category, $-\times Y$ is a left adjoint and preserves all colimits, so $\mathbf{Met}$ is not cartesian closed.

For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the unit and counit of an adjunction areobtained by transposing identity morphisms. They satisfy the triangular identities ${\epsilon }_{F}F\eta =1_F$ and $G\epsilon {\eta }_G=1_G$.

If $G$ is full and faithful, fullness gives a map $u:B\to FGB$ with $G(u)={\eta }_{GB}$; the second triangular identity and faithfulness show that $u$ is inverse to ${\epsilon }_B$. Conversely, if $\epsilon$ is an isomorphism, then for $h:GB\to G{B}^{\prime }$ the unique arrow whose transpose is $h$ iswhich proves that $G$ is full, and the triangular identities prove faithfulness. Hence (i) and (ii) are equivalent. Condition (ii) immediately gives (iii). Conversely, a natural isomorphism $FG\cong 1_{\mathcal{D}}$ combines with the adjunction bijection to show naturally that $\mathcal{D}(B,B^{\prime })\to \mathcal{C}(GB,G{B}^{\prime })$ is bijective; by naturality and the triangular identities this is the map induced by $G$ up to invertible natural conjugation, so $G$ is full and faithful. Thus all three conditions are equivalent.

Suppose $L⊣F⊣R$. If $L$ is full and faithful, the unit $1\to FL$ is an isomorphism. For $A,B$ in the codomain of $F$, the two adjunctions then give natural bijectionsBy the Yoneda lemma, the counit $FRB\to B$ is an isomorphism, so the fully faithful adjoint criterion makes $R$ full and faithful. The converse is dual.

Now assume $F$ is full and faithful. For $L⊣F$, its counit $ϵ:LF\to 1$ is invertible; for $F⊣R$, its unit $\alpha :1\to RF$ is invertible. DefineApplying $F$, then using naturality and the triangular identities, reduces this composite to the same map asSince $F$ is faithful, the two displayed composites are equal.

For a morphism $u:FB\to A$, let $\hat{u}=Ru{\alpha }_B:B\to RA$ be its transpose under $F⊣R$. The identities just proved giveIf every ${\theta }_A$ is monic and $Lu=Lv$, this equation gives $\hat{u}=\hat{v}$, hence $u=v$; thus $L$ is faithful on arrows whose domains are in the image of $F$. Conversely, if ${\theta }_{A}f={\theta }_{A}g$ for $f,g:B\to RA$, naturality of $ϵ$ and the second formula for $\theta$ giveBoth maps have domain $FB$, so the assumed faithfulness gives ${\beta }_{A}Ff={\beta }_{A}Fg$. These are the transposes of $f$ and $g$, hence $f=g$. Therefore $\theta$ is pointwise monic exactly under the stated faithfulness condition.

The Yoneda lemma states that for $A\in \mathcal{C}$ and $P:\mathcal{C}\to \mathbf{Set}$ there is a natural bijectionIf $P\twoheadrightarrow Q$ is an epimorphism in the functor category, it is pointwise surjective, so $P(A)\to Q(A)$ is surjective. Yoneda identifies this map withThus every representable functor is a projective object in a category.

The colimit form of the Special adjoint functor theorem says that a colimit-preserving functor from a locally small, cocomplete, well-copowered category with a small generating family into a locally small category has a right adjoint functor. For small $\mathcal{C}$, the category $[\mathcal{C},\mathbf{Set}]$ is locally small and has colimits pointwise. Quotients of $P$ are represented by compatible equivalence relations on the sets $P(C)$, so they form a set; hence the category is well-copowered. The set of representables $\{\mathcal{C}(C,-):C\in \mathcal{C}\}$ generates it by the Yoneda lemma. The theorem therefore gives a right adjoint to every small-colimit-preserving functorIn particular, product with a fixed functor $P$ is computed pointwise, and $-\times P(C)$ preserves colimits in the Category of sets. Hence $-\times P$ preserves all small colimits and has a right adjoint $(-{)}^P$. Thus $[\mathcal{C},\mathbf{Set}]$ is a Cartesian closed category.

Now work in $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ and write $yA=\mathcal{C}(-,A)$. If $\mathcal{C}$ has binary products, thenThus exponentiation by $yA$ is precomposition with $-\times A$. Precomposition between functor categories has a right adjoint given by Right Kan extension, so $yA$ is a tiny object.

Conversely, suppose $\mathcal{C}$ has a terminal object $1$ and $F$ is tiny. The representable $y1$ is the terminal presheaf, and the exponential adjunction plus Yoneda givesSince $F$ is tiny, $(-{)}^F$ is a left adjoint and preserves all colimits; evaluation at $1$ also preserves pointwise colimits. Therefore the hom functor $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}](F,-)$ preserves coproducts and epimorphisms. Preservation of epimorphisms makes $F$ projective, while preservation of coproducts makes it indecomposable.

A Riemannian metric $g$ on a complex manifold is a Kähler metric when its complex-linear extension is a Hermitian form on each tangent space and its fundamental two-formis closed. Equivalently, $\omega$ is a positive real closed $(1,1)$-form, making $X$ a Kähler manifold.

The Lefschetz operator of a Kähler manifold and its adjoint areBecause $d\omega =0$ and $\omega$ has type $(1,1)$, both $\partial \omega$ and $\bar{\partial }\omega$ vanish. The graded Leibniz rule therefore gives $[L,\partial ]=[L,\bar{\partial }]=0$.

Writing formal adjoints with stars, define the three Laplacians byThe supplied Kähler identities identity $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$ gives, by complex conjugation and taking adjoints,Expanding these commutators and using $\partial \bar{\partial }+\bar{\partial }\partial =0$ shows that the mixed terms in $\Delta$ vanish and that ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Since $d=\partial +\bar{\partial }$, it follows that

The adjoint of $[L,\partial ]=0$ gives $[\Lambda ,{\partial }^{\ast }]=0$. Hencewhere the last equality is the adjoint of $\partial \bar{\partial }+\bar{\partial }\partial =0$. Thus $\Lambda$, and therefore every power ${\Lambda }^k$, commutes with ${\Delta }_{\partial }$. It follows that ${\Lambda }^k$ sends every $\partial$-harmonic $(p,q)$-form to a $\partial$-harmonic $(p-k,q-k)$-form whenever $0\le k\le \min (p,q)$.

Choose holomorphic coordinates centered at $p$. Locally, the blowup of a complex manifold at a point iswith $\sigma (z,[\ell ])=z$; away from $p$ this is an isomorphism, so it glues to $X\setminus \{p\}$. The exceptional divisor is $E={\sigma }^{-1}(p)\cong {\mathbb{CP}}^{n-1}$.

The proper transform is the closure of ${\sigma }^{-1}(Y\setminus \{p\})$. If $p\notin Y$, it is isomorphic to $Y$. If $p\in Y$, the holomorphic implicit function theorem supplies coordinates in which $Y=\{z_1=0\}$. In the blowup chart with $z_i=t$ and $z_j=t{u}_j$ for $j\ne i$, every chart with $i\ne 1$ describes the proper transform by $u_1=0$, while the chart $i=1$ does not meet it. These are smooth coordinate hypersurfaces, so $\tilde{Y}$ is smooth.

For a divisor $D$, the line bundle associated to a divisor $\mathcal{O}(D)$ consists locally of meromorphic functions $f$ such that $(f)+D\ge 0$. Pulling back a local defining function for $Y$ shows that its divisor iswhere $m$ is the order of vanishing at $p$ of a local defining function for $Y$. ThereforeBecause $Y$ is smooth, $m=1$ when $p\in Y$ and $m=0$ when $p\notin Y$.

Applying the definition with $D=E$ gives directlyThe map sends a section to its local meromorphic coefficient relative to the canonical meromorphic section of $\mathcal{O}(E)$; the divisor inequality is exactly the condition that these coefficients define a holomorphic section, and the inverse construction is local multiplication by that canonical section.

The restriction of $J$ to the $J$-invariant bundle $TY$ is an almost complex structure. Its Nijenhuis tensor is the restriction of the ambient Nijenhuis tensor because vector fields tangent to an embedded submanifold have tangent Lie bracket. The ambient tensor vanishes since $X$ is a complex manifold, so the Newlander-Nirenberg theorem makes the induced structure on $Y$ integrable. The inclusion has complex-linear differential and is therefore holomorphic; hence $Y$ is a complex submanifold.

For a complex submanifold, the holomorphic normal bundle isIf $Y$ is a smooth hypersurface, taking top exterior powers in the holomorphic conormal sequencegivesThe normal bundle of a hypersurface is $N_{Y/X}\cong {\mathcal{O}}_X(Y){|}_Y$, so the adjunction formula is

On $X={\mathbb{CP}}^n\times {\mathbb{CP}}^m$, a bihomogeneous polynomial of bidegree $(d_1,d_2)$ is a section of the holomorphic line bundle $\mathcal{O}(d_1,d_2)$. Its zero locus is smooth precisely when the section is transverse to the zero section, equivalently when $F$ and all of its homogeneous first partial derivatives have no common projective zero. Sincethe adjunction formula yieldsThus a smooth $F$ with $d_1=n+1$ and $d_2=m+1$ defines a complex submanifold with trivial canonical bundle.

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.

For a nonnegative measurable $f$, defineEach $f_n$ takes only finitely many values, its level sets are measurable, and $f_n(x)\to f(x)$ pointwise. Thus measurable nonnegative functions are pointwise limits of simple functions.

If $f$ is integrable, the same functions satisfy $0\le f_n\le f$ and converge pointwise. The dominated convergence theorem givesThey are integrable because they are bounded above by $f$.

If $0\le f\le M$, instead setThis has finitely many values andeverywhere, so $g_n\to f$ uniformly.

The map $\varphi$ is an immersion when $d\varphi$ has rank two, equivalently ${\varphi }_u$ and ${\varphi }_v$ are linearly independent. It is isothermal when its first fundamental form hasIn general the mean curvature isThe surface is minimal when $H=0$. In isothermal coordinates the Laplace-Beltrami operator identity isThus $H=0$ exactly when each coordinate function $x,y,z$ is harmonic.

Forthe isothermal conditions areand harmonicity givesConsequentlyand the isothermal equation reduces toThese are all the solutions for which $f$ never vanishes. When $AB>0$ they parameterize a catenoid, up to translations, reflection and scaling. The limiting cases $AB=0$, $c=0$, with exactly one of $A,B$ nonzero, parameterize a punctured plane.

Inside ${\mathbb{A}}^{n^2}$,a nonempty Zariski-open set. Affine space is smooth and irreducible; smoothness passes to open subsets, and every nonempty open subset of an irreducible space is irreducible.

The mapidentifies $G{L}_n$ with the closed hypersurfaceIn these coordinates multiplication is $(A,s)(B,t)=(AB,st)$, whose entries are polynomial functions, so it is a morphism.

A morphism ${\mathbb{A}}^n\to {\mathbb{A}}^1\setminus \{0\}$ is a polynomial with no zero. Over an algebraically closed field every nonconstant polynomial in $n$ variables has a zero: specialize all but one variable so that a nonconstant coefficient remains, then use algebraic closure. The morphism is therefore constant.

A morphism ${\mathbb{P}}^n\to {\mathbb{P}}^1$ is represented by homogeneous polynomials $F,G$ of the same degree with no common projective zero. If the degree is positive and $n\ge 2$, the Projective dimension theorem forces the two hypersurfaces $V(F)$ and $V(G)$ to meet. Thus the degree is zero and the morphism is constant.

An analytic continuation along $\alpha$ is a chain of function elements $(f_t,U_t)$ covering the path, beginning with the germ of $(f,U)$ at $\alpha (0)$, such that neighboring elements agree on the component of overlap met by the path. The terminal element is $(g,V)$.

The monodromy theorem says that if continuation is possible along every path in $D$, then continuations along endpoint-fixed homotopic paths have the same terminal germ. It follows by subdividing a homotopy square into small rectangles on which the identity theorem identifies neighboring germs. Hence

The requested final example with homotopic paths and different terminal values cannot exist under these hypotheses; it contradicts the theorem just proved. With “non-homotopic” in place of “homotopic”, take the germ of $\log z$ near $1$ in ${\mathbb{C}}^{\ast }$. Continuation along the constant path gives $0$, whereas continuation once counterclockwise around zero gives $2\pi i$.