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Define the free algebra functor byThe monad laws say precisely that ${\mu }_X$ defines an monad algebra structure on $TX$, while naturality of $\mu$ makes $Tf$ an monad algebra morphism. We construct the adjunction explicitly.

For an monad algebra $(A,a)$ and a morphism $f:X\to A$, put $\overline{f}=aTf:TX\to A$. It is an monad algebra morphism becauseConversely, from an monad algebra morphism $h:(TX,{\mu }_X)\to (A,a)$ obtain $h{\eta }_X:X\to A$. These operations are inverse: naturality of $\eta$ and the monad algebra unit law giveand the algebra-morphism equation together with the monad unit law givesThe formulas commute with precomposition in $X$ and composition with monad algebra morphisms in $(A,a)$, so the bijections are natural. Therefore the free-forgetful Eilenberg-Moore adjunction isIts unit is ${\eta }_X$ and its counit at $(A,a)$ is the monad algebra action $a$.

For each algebra for a monad $(A,a)$, consider the specific pair in $\mathcal{D}$Then the exact condition isEach pair is reflexive, with common section $F{\eta }_A$: the two composite identities follow from $a{\eta }_A=1_A$ and ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Thus existence of all reflexive coequalizers is sufficient, but the displayed specified pairs give the necessary-and-sufficient condition.

To prove it, take $h:FA\to d$ and its adjoint transpose $x=Gh{\eta }_A:A\to Gd$. The transpose of $hFa$ is $xa$, by naturality of $\eta$. The transpose of $h{\epsilon }_{FA}$ isMoreover counit naturality and a triangle identity giveBecause transposition is bijective, the coequalizing equation is therefore equivalent toThe right side says exactly that $x:(A,a)\to Kd$ is a monad algebra morphism.

For sufficiency, choose a coequalizer $q_a:FA\to L(A,a)$ for every monad algebra. Its universal property and the equivalence just proved give natural bijectionsTo see explicitly that the chosen objects form a functor, an monad algebra morphism $u:(A,a)\to (B,b)$ makes $q_{b}Fu$ coequalize the pair for $a$: use $ua=bTu$ and naturality of the counit. Hence there is a unique $Lu$ with $Lu{q}_a=q_{b}Fu$. Uniqueness proves its identity and composition laws and the naturality of the displayed bijections. Thus $L⊣K$.

Conversely, suppose $L⊣K$ with unit ${\lambda }_{(A,a)}:(A,a)\to KL(A,a)$. Transpose its underlying arrow $A\to GL(A,a)$ through $F⊣G$ to obtain $q_a:FA\to L(A,a)$. Since the unit is an monad algebra morphism, the equivalence above shows that $q_a$ coequalizes the specified pair. Any coequalizing $h:FA\to d$ transposes to an monad algebra map $(A,a)\to Kd$, which factors uniquely through the unit by $L⊣K$. Transposing back gives a unique factorization of $h$ through $q_a$. Thus $q_a$ is the required coequalizer. This proves both directions of the Left adjoint to the Eilenberg-Moore comparison functor criterion.

Assume the coequalizers from part (b) exist and write them $q_a:FA\to L(A,a)$. The underlying arrow of the comparison-adjunction unit isWe first relate it to the monad algebra action. Naturality of $\eta$, the coequalizer equation and a monad unit law giveIn $\mathcal{C}$, $a:TA\to A$ is itself a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$. Indeed,For an explicit universal-property proof, if $r:TA\to X$ satisfies $rTa=r{\mu }_A$, then $s=r{\eta }_A$ obeysIts uniqueness follows by composing any equation $sa=r$ with ${\eta }_A$.

Now $G{q}_a$ equalizes $Ta$ and ${\mu }_A$. If it is a coequalizer, then $a$ and $G{q}_a$ are coequalizers of the same pair. Their unique comparison map is ${\lambda }_a$, since ${\lambda }_{a}a=G{q}_a$, and is therefore invertible. Conversely, if ${\lambda }_a$ is invertible, $G{q}_a={\lambda }_{a}a$ is an isomorphic copy of the coequalizer $a$, so it is a coequalizer too.

The forgetful functor ${\mathcal{C}}^T\to \mathcal{C}$ reflects isomorphisms: if an monad algebra morphism $u$ has an inverse as an underlying arrow, the equation $ua=bTu$ rearranges to $u^{-1}b=aT(u^{-1})$, making that inverse an monad algebra morphism. Thus underlying invertibility here is exactly invertibility of the unit component. For all monad algebras together, the comparison-adjunction unit criterion isEquivalently, the necessary-and-sufficient conditions for a left adjoint with invertible unit are existence of the coequalizers in part (b) and their preservation by $G$. No preservation of arbitrary coequalizers is required.

A rational map $V⇢{\mathbb{P}}^m$ is a morphism on a nonempty open subset of $V$, with two descriptions identified if they agree on a dense open subset. It can be represented there by rational homogeneous coordinate functions not all zero. A point is regular for the map if it extends as a morphism on an open neighbourhood of that point.

For the given projective Cremona transformation, the three products vanish simultaneously exactly at the three coordinate vertices. Away from them they define a regular map. At $(1:0:0)$ consider $[1:t:st]$ as $t\to 0$, with $s\ne 0$. Its image tends to $[0:s:1]$, which depends on $s$. No regular extension could have these different limiting values. Algebraically these curves lie in its domain for $t\ne 0$ and a regular extension must give the same specialization at their common vertex, so the obstruction also works over any infinite characteristic-zero field. The other vertices follow by cyclic symmetry. On the torus $X_0{X}_1{X}_2\ne 0$, applying the map twice givesIt is therefore a birational map and is its own rational inverse.

Use coordinates $[A:B:C]$ on the target. Substitution in the quintic givesThus on the torus its image is the plane quartic $Q=0$. This quartic is nonsingular even over the algebraic closure. Its partial derivatives are $C^3+3A^{2}B$, $A^3+3B^{2}C$ and $3A{C}^2+B^3$. If one coordinate vanishes at a common zero of these partials, all vanish. If none vanishes, multiplying the three equations gives $A^3{B}^3{C}^3=-27A^3{B}^3{C}^3$, impossible in characteristic zero. Hence there is no projective singular point.

A smooth plane curve is geometrically irreducible: if it had two positive-degree components, Bézout theorem over the algebraic closure would give an intersection point, and the product equation has zero gradient there; a repeated factor also makes the gradient vanish along its component. Therefore this quartic is irreducible. Its torus part is dense and is identified by the projective Cremona transformation with the torus part of $V$. No coordinate line is a component of $V$, since restricting $F$ to each coordinate line leaves a nonzero monomial. Consequently every component meets the torus, and its torus part is already irreducible. $V$ is irreducible and birational to the nonsingular quartic $A{C}^3+A^{3}B+B^{3}C=0$.

For a divisor on an algebraic curve $D$, let $L(D)=H^0(V,\mathcal{O}(D))$ and choose a basis of dimension $r+1$. Regarding the basis as sections of the divisor line bundle, evaluation gives ${\varphi }_D:V\to {\mathbb{P}}^r$. If the linear system of divisors has no base points, at least one section is nonzero at each point, and these homogeneous coordinates define a morphism. If the effective divisor has a fixed part $F$, first divide all sections by their common vanishing factor. The moving divisor $E=D-F$ is base-point-free, has the same section space, and gives the uniquely extended morphism on the smooth curve. Thus effectiveness alone does not make the raw evaluations everywhere nonzero, but cancellation supplies the usual map.

A morphism from a projective integral curve is a finite morphism onto its image precisely when it is nonconstant, equivalently $r\ge 1$: a positive-dimensional fiber of a curve would be the whole curve, while a nonconstant proper map with finite fibers is finite. It is an isomorphism onto its image precisely when its moving complete linear system is very ample. In section-space terms, putting $h=h^0(E)$, the criteria arewhere $Q=P$ is allowed. These express absence of base points, separation of distinct points, and separation of tangent directions. In the base-point-free convention $E=D$.

Now suppose the genus is two. The Riemann-Roch theorem gives $\mathrm{deg}K=2$ and $h^0(K)=2$. There cannot be a nonconstant function with only a single simple pole on a positive-genus smooth projective curve: it would give a degree-one map to ${\mathbb{P}}^1$, an isomorphism. Hence $h^0(P)=1$ for every point $P$. Applying Riemann-Roch theorem to $K-P$ gives $h^0(K-P)=h^0(P)=1$, so the canonical linear system has no base point. The resulting map ${\varphi }_K:V\to {\mathbb{P}}^1$ has degree $\mathrm{deg}K=2$.

Set $D=K+P_1+P_2$, of degree four. The Riemann-Roch theorem gives $h^0(D)=3$ and $h^0(D-P)=2$ for every $P$, since the residual divisors have negative degree. Thus ${\varphi }_D$ is a base-point-free map to ${\mathbb{P}}^2$ with nondegenerate image $C$. Degree of the pullback of a general line givesThe image cannot be a line because the three sections are independent. The only alternatives are a birational quartic or a double cover of a conic. In the latter case identify the smooth conic with ${\mathbb{P}}^1$. Its hyperplane bundle has degree two, so $D\sim 2A$, where $A$ is a degree-two divisor with $h^0(A)\ge 2$. Again Riemann-Roch theorem implies $h^0(K-A)=h^0(A)-1\ge 1$. Since $K-A$ has degree zero, it must be linearly equivalent to zero; hence $A\sim K$. This would force $P_1+P_2\sim K$, contrary to the hypothesis. Therefore the image is a plane quartic and the map is birational. A smooth plane quartic has genus $(4-1)(4-2)/2=3$, while its normalization here has genus two. Consequently $\boxed{{\varphi }_D\text{ is birational onto a singular plane quartic}}$.

The general adjoint functor theorem says that, for a functor $U:\mathcal{A}\to \mathcal{B}$ between locally small categories, with $\mathcal{A}$ a complete category, $U$ has a left adjoint if and only if it preserves small limits and satisfies the solution-set condition. The solution-set condition means that for each $B\in \mathcal{B}$ there is a set of arrows $b_i:B\to U{A}_i$ such that every $b:B\to UA$ factors as $U(a)b_i$ for some $i$ and some $a:A_i\to A$.

Here is the smallness argument behind the theorem. A complete category that is locally small and has a weakly initial set $(W_i)$ has an initial object. Form the product in a category $W={\prod }_i{W}_i$. It is weakly initial, since a projection $W\to W_i$ followed by a chosen arrow $W_i\to X$ gives an arrow $W\to X$. Take the simultaneous equalizer $e:E\to W$ of all endomorphisms of $W$ with $1_W$; the family is a set by local smallness. The object $E$ is also weakly initial. For parallel arrows $a,b:E\rightrightarrows X$, take their equalizer $j:Y\to E$ and a map $t:W\to Y$. Since $ejt$ is an endomorphism of $W$, the defining property of $e$ gives $ejte=e$. Cancellation of the monomorphism $e$ gives $jte=1_E$. Thus the monomorphism $j$ has a right inverse and is an isomorphism, so $a=b$. Existence of arrows out of $E$ was already established, proving the initial-object lemma for complete categories with a weakly initial set.

If $U$ preserves small categorical limits, the comma category $(B\downarrow U)$ is complete: take the limiting object in $\mathcal{A}$, and use preservation to assemble the arrows from $B$. It is locally small, and the solution-set condition provides a weakly initial set. Its initial object is an arrow ${\eta }_B:B\to ULB$ with a natural bijectionFor a morphism $v:B\to B^{\prime }$, define $Lv$ by $U(Lv){\eta }_B={\eta }_{B^{\prime }}v$. Uniqueness proves functoriality and naturality, giving a left adjoint $L⊣U$. Conversely, a right adjoint preserves categorical limits, because applying the adjunction converts a limiting cone over a diagram into the corresponding limiting cone of hom-sets. Its adjunction unit at $B$ is a singleton solution set. This proves both directions of the general adjoint functor theorem.

Apply it to the inclusion $J:\mathbf{KHaus}\hookrightarrow \mathbf{Top}$, where $\mathbf{KHaus}$ is the full subcategory of compact Hausdorff spaces. A small product in a category of compact Hausdorff spaces is again compact Hausdorff by the Tychonoff theorem. An equalizer of two continuous functions into a Hausdorff space is closed, hence compact Hausdorff. Products and equalizers construct all small categorical limits, and the inclusion preserves these constructions. Both categories are locally small.

Fix a topological space $X$ and a continuous function $f:X\to K$ into a compact Hausdorff space. The closure $K_f=\overline{f(X)}$ is compact Hausdorff and has a dense subset of cardinality at most $|X|$. The permitted cardinal estimate givesFor $X=\varnothing$, use $K_f=\varnothing$. Choose a set of representatives of all compact Hausdorff topologies on underlying sets of cardinality at most this bound, and all continuous functions from $X$ to those representatives. Every $f$ factors through one of them, using the inclusion $K_f\hookrightarrow K$. This is the solution-set condition. Consequently the inclusion of compact Hausdorff spaces has a left adjoint, the compact Hausdorff reflection:The universal arrow $X\to RX$ has dense image: factoring it through its closed image and applying uniqueness supplies an inverse to that image inclusion. It need not be an embedding when $X$ is not sufficiently separated; the assertion is a compact Hausdorff reflection, rather than a claim that every topological space embeds in a compact Hausdorff space.

For completeness, the alternative Special adjoint functor theorem can be proved from the same initial-object argument. Its limit form assumes that $\mathcal{A}$ is complete, locally small and well-powered, with a small cogenerating family $(Q_i{)}_{i\in I}$; then $U:\mathcal{A}\to \mathcal{B}$, with $\mathcal{B}$ locally small, has a left adjoint precisely when it preserves small limits. To prove sufficiency, start with $b:B\to UA$ and intersect all subobjects $M\hookrightarrow A$ through which $b$ factors after applying $U$. Well-poweredness makes this a small intersection of subobjects, and preservation of limits gives a smallest supporting object $(M,b_M)$. For maps $s,t:M\rightrightarrows Q_i$, equality $U(s)b_M=U(t)b_M$ makes their equalizer another supporting subobject. Minimality forces that equalizer to be invertible, so $s=t$. Thereforeis an injective function. The evaluation embedding into cogenerator products embeds $M$ in ${\prod }_i{Q}_i^{S_i}$, where $S_i$ is the realized subset of the fixed set $\mathcal{B}(B,U{Q}_i)$. There are only a set of choices of these subsets, only a set of subobjects of each resulting product, and only a set of arrows from $B$ to each $U$-image. They give a solution set, so the already proved general adjoint functor theorem applies. Necessity is again preservation of limits by a right adjoint. For $\mathbf{KHaus}$, monomorphisms are embeddings onto closed subspaces, and $[0,1]$ is a coseparator because continuous functions to it separate points. Thus the Special adjoint functor theorem gives the same reflection directly.

A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A strong monomorphism $m:A\to B$ is a monomorphism right orthogonal to all epimorphisms: whenever $e:X\to Y$ is epic and $a:X\to A$, $b:Y\to B$ satisfy $ma=be$, there is a unique $d:Y\to A$ with $de=a$ and $md=b$. A regular monomorphism is an equalizer of some parallel pair of morphisms. Its equalizer property supplies the indicated lift after epic cancellation, so every regular monomorphism is strong.

If every monomorphism is strong and $m:A\to B$ is also epic, apply strongness to the square whose left and right sides are both $m$, with horizontal arrows $1_A$ and $1_B$. The lift $d:B\to A$ satisfies $dm=1_A$ and $md=1_B$. Thus $m$ is invertible and the category is balanced.

Conversely, suppose the category is balanced and has pullbacks in a category. Given the lifting square above, form $P=A{\times }_{B}Y$ with projections $p:P\to Y$, $q:P\to A$. The pair $(a,e)$ gives $h:X\to P$ with $ph=e$, $qh=a$. The pullback of the monomorphism $m$ is monic, so $p$ is monic. It is also epic: $rp=sp$ implies $re=se$, hence $r=s$ because $e$ is epic. Balancedness makes $p$ invertible. Now $d=q{p}^{-1}$ satisfies both required equations, and monicity of $m$ proves uniqueness. Hence in a balanced category with pullbacks, every monomorphism is strong.

For the finite example, choose a strong monomorphism $m:A\to B$ that is not regular. It is not an isomorphism, since an isomorphism is the equalizer of an identical pair. It is not epic either, since an epic strong monomorphism is invertible by the first argument. Consequently there are distinct arrows $r,s:B\rightrightarrows D$ with $rm=sm$. Since $m$ is not the equalizer of this pair, there is a map $h:X\to B$ with $rh=sh$ that does not factor through $m$. Monicity rules out failure of uniqueness, so failure of existence is the only possible failure of its equalizer property.

Define ${\mathcal{C}}_0$ to have four formally distinct objects $a,b,x,d$ and exactly the following six nonidentity arrows:Their only nontrivial composites areThere are no longer composable strings of nonidentity arrows, so these rules define an associative category. Sending $a,b,x,d$ to $A,B,X,D$, respectively, and the six arrows to $m,h,r,s,rm,rh$, defines a functor. It is a faithful functor: every hom-set has at most one arrow except ${\mathcal{C}}_0(b,d)=\{r_0,s_0\}$, and their images are distinct. Faithfulness does not require the four image objects to be distinct.

The map $m_0$ is monic, by direct inspection of arrows into $a$. Neither $m_0$ nor $h_0$ is epic, since the distinct pair $r_0,s_0$ agrees after either. Every other nonidentity arrow has target $d$ and is epic, because there is only one arrow out of $d$. A lifting square against $m_0$ whose left side is one of these epimorphisms would require an arrow $d\to b$, which does not exist. Squares whose left side is an identity have the evident unique lift. Thus $m_0$ is a strong monomorphism.

It is not a regular monomorphism. The only distinct parallel pair out of $b$ is $r_0,s_0$, and $h_0$ equalizes it but has no factorization through $m_0$, since there is no arrow $x\to a$. An identical pair has $1_b$ as its equalizer and cannot have the noninvertible $m_0$ as its equalizer. Therefore

Let $U_n:{\mathcal{C}}_{n+1}\to {\mathcal{C}}_n$ forget the last partial unary operation. For an object $A$ of the nested partial unary operation category ${\mathcal{C}}_n$, setThe nesting rule implies that every earlier operation is defined and fixes such an $x$. To construct a free extension of nested partial unary operations, adjoin a disjoint infinite chain for each $x\in S_n(A)$:Keep the old operations on $A$. On each new chain put ${\alpha }_1(x,k)=(x,k+1)$; every old operation ${\alpha }_i$ with $2\le i\le n$ is undefined there, because the first operation has no fixed points there. Define the new operation only on $S_n(A)$, by ${\alpha }_{n+1}(x)=(x,0)$. These are exactly the fixed points of the old top operation, so the required domain rule is satisfied. The inclusion $A\hookrightarrow U_n{L}_{n}A$ preserves all old defined operations.

For a morphism $f:A\to U_{n}B$, preservation of the old operations makes $f(x)\in S_n(B)$ whenever $x\in S_n(A)$. Any extension to a morphism $\hat{f}:L_{n}A\to B$ must haveThe operation ${\beta }_{n+1}$ is defined at $f(x)$, and ${\beta }_1$ is total, so this formula exists. It preserves the new operation and the chain operation; there are no other defined operations on new points to check. It therefore gives the unique extension. Naturality and functoriality follow from uniqueness. ThusNotice the useful extra fact: the new top operation has no fixed points, since every value $(x,0)$ is new and every input $x$ is old. Hence freely adding any further operations adds no more points; those operations have empty domains.

Write $\eta :1_{\mathcal{C}}\to GF$ and $\epsilon :FG\to 1_{\mathcal{D}}$ for the unit and counit of an adjunction. The induced monad hasThe Eilenberg-Moore comparison functor sends $D\in \mathcal{D}$ to the algebra for a monadand sends $v:D\to D^{\prime }$ to $Gv$. The triangle identities for an adjunction give the algebra unit law, and naturality of the counit gives the algebra multiplication law and the algebra-morphism equation for $Gv$.

If $\mathcal{D}$ has coequalizers, define the Left adjoint to the Eilenberg-Moore comparison functor at an algebra $(A,a:TA\to A)$ byThe parallel pair is reflexive through $F{\eta }_A$. A morphism $L(A,a)\to D$ is equivalently an arrow $b:FA\to D$ with $bFa=b{\epsilon }_{FA}$. Under $F⊣G$, its transpose is $\bar{b}=Gb{\eta }_A:A\to GD$. The transpose of $bFa$ is $\bar{b}a$, while the transpose of $b{\epsilon }_{FA}$ is $G{\epsilon }_{D}T\bar{b}$. Hence the coequalizer condition is exactlythe equation for a morphism of algebras for a monad $(A,a)\to K(D)$. This supplies the natural bijection proving $L⊣K$, and also defines $L$ on algebra morphisms by the coequalizer universal property.

Let $F:\mathcal{C}\to \mathcal{D}$ be a final functor, and let ${\lambda }_c:GFc\to V$ be a cocone under a diagram $GF$. For $d\in \mathcal{D}$, choose an object $(c,a:d\to Fc)$ of the nonempty comma category $(d\downarrow F)$ and defineIf $t:(c,a)\to (c^{\prime },a^{\prime })$ is a morphism in that comma category, then $Fta=a^{\prime }$, so the cocone equation gives ${\lambda }_{c}G(a)={\lambda }_{c^{\prime }}G(Ft)G(a)={\lambda }_{c^{\prime }}G(a^{\prime })$. Equality also holds along a reversed arrow, and hence along any zigzag. Connectedness makes the definition independent of the chosen object. For $v:d\to d^{\prime }$, use $(c,a:d^{\prime }\to Fc)$ and $(c,av:d\to Fc)$ to obtain ${\bar{\lambda }}_d={\bar{\lambda }}_{d^{\prime }}Gv$. Thus $\bar{\lambda }$ is a cocone under $G$. Taking $(c,1_{Fc})$ shows that it extends $\lambda$.

Every extension must satisfy ${\bar{\lambda }}_d={\lambda }_{c}G(a)$ by its cocone equation, proving uniqueness. The construction commutes with postcomposition by a morphism $V\to V^{\prime }$. Hence restriction gives a natural bijection between the sets of cocones of $G$ and $GF$ with any fixed vertex. A universal cocone for $GF$ therefore extends to a universal cocone for $G$. In particular,whenever the latter exists, and existence of all colimits of shape $\mathcal{C}$ implies existence of all colimits of shape $\mathcal{D}$.

For the factorization of an arbitrary functor $H:\mathcal{C}\to \mathcal{E}$, letthe set of connected components of a category, where components are defined by finite zigzags. Precomposition by $v:e\to e^{\prime }$ gives a functor $(e^{\prime }\downarrow H)\to (e\downarrow H)$ and therefore a map $P(v):P(e^{\prime })\to P(e)$. Identity and composition laws make $P$ a categorical presheaf. This is a set-valued construction when the relevant comma categories are small or have only a set of components, as in the usual small-category convention for this factorization.

Take $\mathcal{D}$ to be the category of elements of this presheaf. Its objects are pairs $(e,S)$ with $S\in P(e)$; a morphism $(e,S)\to (e^{\prime },S^{\prime })$ is an arrow $v:e\to e^{\prime }$ satisfying $P(v)(S^{\prime })=S$. Define $G:\mathcal{D}\to \mathcal{E}$ by forgetting $S$. Given $v:e\to e^{\prime }$ and an object $(e^{\prime },S^{\prime })$, the unique lift into that object has source $(e,P(v)(S^{\prime }))$ and underlying arrow $v$. This proves that $G$ is a discrete fibration.

Define $F(c)=(Hc,[(c,1_{Hc})])$. For $t:c\to c^{\prime }$, take $F(t)=Ht$. The arrow $t$ in $(Hc\downarrow H)$ joins $(c,1_{Hc})$ to $(c^{\prime },Ht)$, so the required component equation holds. It follows that $F$ is a functor and $GF=H$.

Finally, an object of $((e,S)\downarrow F)$ is a pair $(c,v:e\to Hc)$ whose component in $(e\downarrow H)$ is $S$. Its morphisms are exactly the morphisms of that comma category between objects in $S$. Consequently $((e,S)\downarrow F)$ is precisely the full connected component $S$, and is nonempty and connected. Thus $F$ is final, and every such functor factors as a final functor followed by a discrete fibration.

At a smooth point of an irreducible curve, the local ring ${\mathcal{O}}_{X,p}$ is a discrete valuation ring. A local parameter is an element $t$ generating its maximal ideal; its order at $p$ is one. Embed the projective target as a closed subvariety of ${\mathbb{P}}^N$. Write the rational map generically as $[f_0:\cdots :f_N]$ with $f_i\in K(X)$, and let $m$ be the smallest finite valuation of the nonzero coordinates at $p$. Multiplying by $t^{-m}$ makes every coordinate regular at $p$ and at least one a unit. These functions are regular on some common neighborhood of $p$ and have no common zero after shrinking, so define a morphism to ${\mathbb{P}}^N$. The defining homogeneous equations of the target vanish generically and hence identically on this neighborhood. Thus the rational map extends to the projective target at $p$.

For a nonprojective target, $x\mapsto 1/x$ from ${\mathbb{A}}^1$ rationally to ${\mathbb{A}}^1$ cannot extend at zero: a regular extension would give $xf(x)=1$ there, impossible. In dimension two the morphism $(x,y)\mapsto [x:y]$ on ${\mathbb{C}}^2\setminus \{0\}$ cannot extend to zero, because its constant values along the two punctured coordinate axes are different. This shows why both projectivity of the target and smooth one-dimensional source were used.

For the covariant version of the Yoneda embedding, put $Y(A)=\mathcal{C}(A,-)$. A morphism $u:A\to B$ induces the natural transformation $Y(u):Y(B)\to Y(A)$ whose component sends $v:B\to X$ to $vu$. Thus $Y$ is a functor from the opposite category ${\mathcal{C}}^{\mathrm{op}}$ to the functor category $[\mathcal{C},\mathbf{Set}]$. The locally small category hypothesis makes each value a set.

The Yoneda lemma says that, for a functor $F:\mathcal{C}\to \mathbf{Set}$, evaluation at the identity morphism gives a natural bijectionGiven $a\in F(A)$, define ${\theta }_X^a(v)=F(v)(a)$. For $w:X\to X^{\prime }$, the functor law gives $F(w){\theta }_X^a(v)=F(wv)(a)={\theta }_{X^{\prime }}^a(wv)$, proving naturality. Conversely, naturality of $\theta$ at $v:A\to X$ gives ${\theta }_X(v)=F(v){\theta }_A(1_A)$. These two constructions are inverse, because $F(1_A)(a)=a$.

The bijection is natural in both variables. A natural transformation $\beta :F\to H$ sends the element $a$ to ${\beta }_A(a)$, matching postcomposition by $\beta$. For $u:A\to B$, precomposing with $Y(u)$ sends $\theta$ to a transformation from $Y(B)$ and its distinguished element is ${\theta }_B(u)=F(u)(a)$. This is exactly the required dependence on $A$.

Taking $F=Y(B)$ gives $\mathrm{Nat}(Y(A),Y(B))\cong \mathcal{C}(B,A)$. Consequently the covariant Yoneda embedding is full and faithful, with the reversal of arrows accounted for by its opposite category domain.

An object of the slice category $\mathcal{C}/A$ is a morphism $x:X\to A$; a map from $x$ to $y:Y\to A$ is $u:X\to Y$ with $yu=x$. The identity morphism $1_A:A\to A$ is a terminal object of this slice category, even when $\mathcal{C}$ itself has no terminal object.

For a cospan $X\stackrel{u}{\to }Z\stackrel{v}{\leftarrow }Y$ in the slice category, form its underlying pullback in a category $P$ in $\mathcal{C}$, with maps $p:P\to X$ and $q:P\to Y$. If $x,y,z$ are the structure maps to $A$, then $xp=zup=zvq=yq$. This common map equips $P$ with its slice structure. An underlying mediating map into $P$ is automatically over $A$, because its composite with $xp$ is the prescribed structure map. Hence the same square is a pullback in a category in $\mathcal{C}/A$.

Part (b) applies to the slice's terminal object and pullbacks in a category. Every slice category has all finite limits when the ambient category has pullbacks in a category. The finite completeness of slice categories result needs no ambient terminal-object hypothesis.

A monomorphism $m:A\to B$ satisfies $mu=mv\Rightarrow u=v$ for every parallel pair into $A$. An epimorphism $e:X\to Y$ satisfies $ue=ve\Rightarrow u=v$ for every parallel pair out of $Y$. A strong monomorphism is a monomorphism with the following lifting property: whenever $e:X\to Y$ is an epimorphism and $mu=ve$, with $u:X\to A$ and $v:Y\to B$, there is a unique $d:Y\to A$ with $de=u$ and $md=v$. A regular monomorphism is a morphism occurring as the equalizer of two parallel maps. A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism.

Suppose $m:A\to B$ equalizes $p,q:B\rightrightarrows D$. If $mu=mv$, the common composite has two factorizations through the equalizer, so its uniqueness property gives $u=v$. Hence every regular monomorphism is indeed a monomorphism.

Now take the lifting square just described. The equations $pmu=qmu$ imply $pve=qve$. Cancel the epimorphism $e$ to obtain $pv=qv$. The equalizer gives a unique $d$ with $md=v$. Since $mde=ve=mu$ and $m$ is a monomorphism, $de=u$. Thus every regular monomorphism is a strong monomorphism; the existence of the lift uses epic cancellation, and its uniqueness uses the equalizer property.

The comma category $(X\downarrow G)$ has objects $(A,x)$ with $A\in \mathcal{C}$ and a function $x:X\to G(A)$. Its morphisms $(A,x)\to (B,y)$ are maps $f:A\to B$ in $\mathcal{C}$ satisfying $G(f)x=y$. The identity morphism and composition in a category are inherited from $\mathcal{C}$; the functor laws make the defining equations stable under both.

For a singleton $1$, its map $1\to G(A)$ is just an element $a\in G(A)$. Thus $(1\downarrow G)$ is the covariant category of elements of $G$, with arrows $f:(A,a)\to (B,b)$ satisfying $G(f)(a)=b$.