Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 17 (Category Theory)

Write the adjunction as the natural bijectionThe unit of an adjunction and counit of an adjunction are the natural transformations whose components areNaturality of the hom-set bijections gives, for $f:FA\to B$ and $g:A\to GB$,For example, postcomposition by $f$ in the first hom-set corresponds to postcomposition by $Gf$ in the second, giving the first formula; naturality in $A$ gives the second. The same naturality says that, for $a:A\to A^{\prime }$ and $b:B\to B^{\prime }$,Thus these components do define the asserted natural transformations $1_{\mathcal{C}}\to GF$ and $FG\to 1_{\mathcal{D}}$.

The two formulas for the adjunction transposes give both triangle identities for an adjunction directly. Apply the inverse bijection to the transpose of $1_{FA}$:Apply the forward bijection to the inverse transpose of $1_{GB}$:These are equalities of components of natural transformations, soNo cancellation assumption on either functor is needed: both identities follow from the inverse hom-set bijections.

Define maps between the hom-sets byUsing naturality of the proposed counit of an adjunction and then the first triangle identity for an adjunction,Using naturality of the proposed unit of an adjunction and then the other triangle identity for an adjunction,Hence $\Phi$ and $\Psi$ are inverse bijections. They are natural in both variables: for $a:A^{\prime }\to A$ and $b:B\to B^{\prime }$, the unit of an adjunction identity givesThis constructs an adjunction $F⊣G$. Taking $f=1_{FA}$ and $g=1_{GB}$ recovers the prescribed unit of an adjunction and counit of an adjunction. Conversely, part (i) shows that any adjunction with those components must have exactly these transpose formulas. The adjunction exists and is unique with the prescribed unit and counit.

Let $\hat{\mathcal{O}}=[\mathcal{O}(S{)}^{\mathrm{op}},\mathbf{Set}]$, with categorical presheaf restrictions $X(U)\to X(V)$ when $V\subseteq U$. The five functors areIn $\Delta A$ all restriction maps are identities. In $\Lambda A$ and $\nabla A$, all nonidentity restrictions are the unique maps permitted by the displayed sets. These definitions are valid also when $A$ is empty: a restriction out of $(\nabla A)(S)$ still has singleton target. Maps of sets act at the displayed $A$-components, and natural transformations act on $\Pi$ and $\Gamma$ by evaluation.

A natural transformation $\Lambda A\to X$ is determined by an arbitrary map $A\to X(\varnothing )$, since all other source components are empty. A natural transformation $X\to \Delta A$ is determined by its empty-open component $X(\varnothing )\to A$: the component at $U$ must first restrict $X(U)\to X(\varnothing )$. Similarly, a natural transformation $\Delta A\to X$ is determined by $A\to X(S)$, followed by the restriction $X(S)\to X(U)$. Finally, a natural transformation $X\to \nabla A$ is determined by $X(S)\to A$, since all proper-open targets are singletons. These descriptions give the hom-set bijectionsThus the five adjoints for constant presheaves on open sets areThe values at the empty open set are genuine presheaf values here; no sheaf condition or sheafification is being imposed.

Let $C\mathcal{A}={\pi }_0\mathcal{A}$ be the set of connected components of a category: two objects of a category are identified when joined by a finite zigzag of morphisms. Let $DA$ be the discrete category on $A$, and let $IA$ be the indiscrete category on $A$, with exactly one morphism for every ordered pair of objects.

A functor $\mathcal{A}\to DA$ must give the same object value along every arrow, hence along every zigzag. Conversely a function ${\pi }_0\mathcal{A}\to A$ uniquely defines such a functor. A functor $DA\to \mathcal{A}$ is just a choice of an object of a category for each element of $A$. A functor $\mathcal{A}\to IA$ is likewise determined by an arbitrary object function, since its arrow images are forced and always compose correctly. ThereforeThis is the adjoint chain for the objects of a category:

Use the preservation obstructions to extending an adjoint chain: a functor having a further left adjoint must preserve categorical limits, while one having a further right adjoint must preserve colimits.

At the left end of the presheaf category chain, $\Lambda$ does not preserve the terminal object. The terminal categorical presheaf has singleton value everywhere, whereas $(\Lambda 1)(S)=\varnothing$ because $S\ne \varnothing$. Thus $\Lambda$ cannot itself be a right adjoint. At the right end, $\nabla$ does not preserve the initial object: $(\nabla \varnothing )(\varnothing )=\{\ast \}$ since the empty open set is proper. Thus $\nabla$ cannot itself be a left adjoint.

For the category of small categories chain, take two functors from the terminal category $1$ to the two-object indiscrete category $I\{0,1\}$, selecting its distinct objects. Their equalizer is the empty category. But their images under $C={\pi }_0$ are the same function between singleton sets, whose equalizer is a singleton. Hence $C$ does not preserve equalizers and has no further left adjoint. Finally, $I1⨿I1$ is the two-object discrete category, while $I(1⨿1)$ has arrows between its distinct objects. The canonical comparison is not an isomorphism of categories, so $I$ fails to preserve this coproduct and has no further right adjoint.

Let $\mathcal{C}$ be a locally small category, let $h_A=\mathcal{C}(-,A)$ be a representable presheaf, and let $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$. The Yoneda lemma gives a bijection, natural in $A$ and $X$,For $x\in X(A)$, define a natural transformation ${\alpha }^x:h_A\to X$ byFor $v:C\to B$, the functor composition law gives $X(v){\alpha }_B^x(f)=X(fv)(x)={\alpha }_C^x(fv)$, proving naturality. Conversely, naturality of any $\alpha :h_A\to X$ along $f:B\to A$ givesThus $\alpha$ is determined by its displayed element, and ${\alpha }_A^x(1_A)=x$ proves that the constructions are inverse.

A natural transformation $\theta :X\to Y$ sends the element to ${\theta }_A(x)$, which corresponds to $\theta {\alpha }^x$. A morphism $u:A^{\prime }\to A$ induces $h_u:h_{A^{\prime }}\to h_A$; precomposition by $h_u$ sends the element to $X(u)(x)$. This proves naturality in both variables and completes the Yoneda lemma proof.

The Yoneda embedding sends $A$ to $h_A=\mathcal{C}(-,A)$ and $u:A\to B$ to postcomposition $h_u:h_A\to h_B$. Apply the Yoneda lemma with $X=h_B$:The inverse sends $u$ precisely to $h_u$, because its component at $C$ maps $f:C\to A$ to $uf$. Hence the induced maps on all hom-sets are bijections. The Yoneda embedding is full and faithful.

An isomorphism $f:A\to B$ gives a natural isomorphism $h_f:h_A\to h_B$ by postcomposition, with inverse $h_{f^{-1}}$.

Conversely suppose $\alpha :h_A\to h_B$ is a natural isomorphism, with inverse $\beta$. Since the Yoneda embedding is a full and faithful functor, there are unique $f:A\to B$ and $g:B\to A$ with $h_f=\alpha$ and $h_g=\beta$. Their composites satisfyFaithfulness gives $gf=1_A$ and $fg=1_B$. ThusThe naturality requirement ensures that all incoming morphisms are respected; unrelated componentwise bijections alone would not justify the conclusion.

Here a universal element means a pair $(A,u)$, with $u\in X(A)$, such that for every $B$ and every $x\in X(B)$ there is a unique morphism $f:B\to A$ satisfying $X(f)(u)=x$. Equivalently,By the Yoneda lemma these component maps constitute a natural transformation $h_A\to X$. If $u$ is universal, every component is bijective, so this is a natural isomorphism and $X$ is a representable presheaf. Conversely a natural isomorphism $h_A\to X$ takes $1_A$ to an element $u$ with exactly this property. In the category of elements of the contravariant functor $X$, $(A,u)$ is a terminal object: every $(B,x)$ has a unique arrow to it.

First construct every finite product in a category by iterating binary products in a category; the empty product is the terminal object. Let $D:J\to \mathcal{C}$ be a diagram in a category with finitely many objects and arrows. Form the finite products in a categoryDefine $a,b:P\rightrightarrows Q$ by their componentsTake their equalizer $e:L\to P$. Its components $p_j={\pi }_{j}e$ satisfy $D(u)p_i=p_j$, so they form a cone over a diagram.

For any other categorical cone $x_j:X\to D(j)$, the product in a category property supplies a unique $x:X\to P$ with ${\pi }_{j}x=x_j$. The cone equations imply $ax=bx$, so the equalizer property supplies a unique $\bar{x}:X\to L$ with $e\bar{x}=x$. These are exactly the required equations $p_j\bar{x}=x_j$, with uniqueness. Thus $L$ is a categorical limit. For empty $J$, both products are terminal and this construction still gives a terminal limit. All finite limits exist.

Use the explicit construction of small limits from products and equalizers from part (i), here with finite indexing sets. If $F$ preserves finite products in a category and equalizers, then $F(P)$ and $F(Q)$, with their image projections, are the products of the objects $FD(j)$ and of the target objects indexed by arrows of $J$.

The image maps $F(a),F(b)$ have componentsAlso $F(e):F(L)\to F(P)$ is their equalizer. Therefore the same universal property proof makes $F(L)$, with projections $F(p_j)$, a categorical limit of $FD$. Any other chosen limit of $D$ is uniquely isomorphic to this construction, and the functor carries that isomorphism to an isomorphism. Consequently $F$ preserves every finite limit. Preservation of finite products includes the empty product, so the terminal object is also preserved.

Given $A,B$, their product in a category is the pullback of the unique arrows $A\to 1\leftarrow B$, where $1$ is a terminal object. Having constructed this product in a category, define the equalizer of $f,g:A\rightrightarrows B$ as the pullbackwhere ${\delta }_B$ is the categorical diagonal. A map $x:X\to A$ factors through this pullback exactly when $fx=gx$: the accompanying arrow to $B$ must equal this common composite. The factorization is unique, proving the equalizer property.

Thus the hypotheses give a terminal object, binary products in a category and all equalizers. Part (i) now gives all finite limits.

Let $D:J\to [\mathcal{C},\mathcal{S}]$ be a small diagram in a category. Since $\mathcal{S}$ is a complete category, choose at each $C\in \mathcal{C}$ a categorical limitFor $f:C\to C^{\prime }$, the arrows $D_j(f)p_{j,C}$ form a cone over a diagram at $C^{\prime }$, because the diagram arrows $D_j\to D_k$ are natural transformations. There is therefore a unique arrow $L(f):L(C)\to L(C^{\prime })$ withThe universal property immediately gives $L(1_C)=1_{L(C)}$ and $L(gf)=L(g)L(f)$: both sides have the same composites with every limit projection. Thus $L$ is a functor, and the displayed equations make the $p_j$ natural transformations.

Given any categorical cone $q_j:X\to D_j$ in the functor category, its components induce unique $q_C:X(C)\to L(C)$. For $f:C\to C^{\prime }$, compare $L(f)q_C$ and $q_{C^{\prime }}X(f)$ after each $p_{j,C^{\prime }}$; naturality of $q_j$ makes their composites equal. Hence $q$ is a natural transformation, and uniqueness is componentwise. This proves that $L$ is a categorical limit in $[\mathcal{C},\mathcal{S}]$.

The functor category is complete, and every evaluation functor preserves limits, since evaluating the constructed limit at $C$ gives exactly the chosen limit in $\mathcal{S}$. This is the pointwise limits in a functor category construction; it also covers the empty diagram and its terminal object.

Let $a:X\to Y$ be a natural transformation of categorical presheaves. If every $a_C:X(C)\to Y(C)$ is a surjective function, and $r,s:Y\rightrightarrows Z$ satisfy $ra=sa$, then $r_C{a}_C=s_C{a}_C$ at every $C$. Surjectivity gives $r_C=s_C$ for all $C$, hence $r=s$. Thus $a$ is an epimorphism.

Conversely suppose $y\in Y(C)$ is outside the image of some $a_C$. Construct the pushout witness for failure of pointwise surjectivity explicitly: at each $D$ takeRestrictions send $[(z,k)]$ to $[(Y(f)z,k)]$; naturality of $a$ ensures that every generating identification is respected. Thus $P$ is a categorical presheaf, and the two inclusions $j_0,j_1:Y\to P$ are natural transformations with $j_{0}a=j_{1}a$.

All identifications concern the same element $z$ in its two copies, so an element outside the image is never identified with its other copy. Consequently $j_{0,C}(y)\ne j_{1,C}(y)$, proving $a$ is not an epimorphism. ThereforeThis is the pointwise epimorphism in a functor category criterion. It requires neither a componentwise choice of preimages nor a natural section.

For $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$, let $\int X$ be its category of elements. Its objects are $(A,a)$ with $a\in X(A)$; an arrow $(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $X(f)b=a$. This is a small category, because $\mathcal{C}$ is small and all values of $X$ are sets.

Send $(A,a)$ to the representable presheaf $h_A$, and send an arrow $f$ to postcomposition $h_f$. There is a cocone to $X$ whose map at $(A,a)$ isThe functor laws give its naturality and the required cocone compatibility.

Compute the colimit pointwise as a disjoint union modulo its diagram identifications. At $C$ a representative is a triple $(A,a,g:C\to A)$, and it maps to $X(g)a\in X(C)$. Every $x\in X(C)$ is the image of $(C,x,1_C)$. Moreover $g$ is an arrow $(C,X(g)a)\to (A,a)$ in $\int X$, so its diagram relation identifiesThus every representative is identified with the canonical representative of its image. Two representatives have equal images exactly when their canonical representatives agree. This proves a componentwise bijection; restriction maps respect it. The resulting natural isomorphism is the canonical colimit presentation of a presheaf:

A Cartesian closed category has finite products in a category and, for all $X,Y$, an exponential object $Y^X$ with a natural bijectionEquivalently the functor $-\times X$ has a right adjoint. Its counit of an adjunction is the evaluation map of an exponential object $Y^X\times X\to Y$.

For the presheaf category, define the exponential in a presheaf category byFor $f:D\to C$, its restriction sends $\alpha$ to $\alpha \circ (h_f\times 1_X)$. Composition of these restriction maps follows from composition of the $h_f$, so this is a categorical presheaf. Define evaluation map of an exponential object byFor $f:D\to C$, naturality of $\alpha$ givesThus $e$ is a natural transformation.

Given a natural transformation $\beta :Z\times X\to Y$, define its currying byNaturality of $\beta$ proves that ${\hat{\beta }}_C(z)$ is a natural transformation $h_C\times X\to Y$. For $f:C^{\prime }\to C$, the same formula proves that $\hat{\beta }:Z\to Y^X$ is natural. Evaluating at $g=1_C$ gives $e(\hat{\beta }\times 1_X)=\beta$.

Conversely, if $\gamma :Z\to Y^X$ and $\beta =e(\gamma \times 1_X)$, thenby naturality of $\gamma$. Hence the two constructions are inverse and natural in $Z,Y$. Products are available pointwise, including the terminal singleton presheaf, so the presheaf category is Cartesian closed. The exponential uses natural transformations out of $h_C\times X$; simply exponentiating the individual values generally does not give this object.

Write the Yoneda embedding as $h:\mathcal{C}\to \hat{\mathcal{C}}$. For any finite product in a category, its universal property givesThese bijections are natural, so $h$ preserves finite products. Now use the exponential in a presheaf category, followed by the Yoneda lemma and the exponential object property in $\mathcal{C}$:Every step is natural in $D$, so these component bijections form a natural isomorphismUnder this isomorphism, the image of the evaluation $B^A\times A\to B$ is the presheaf evaluation: both correspond to the same identity of the exponential under the transpose bijection. Thus the Yoneda embedding preserves exponentials, including their evaluation maps.

If $F⊣U$, let $1$ denote a singleton set. The adjunction gives the natural bijectionsThus $U$ is represented by $F1$, proving $A\Rightarrow R$.

If $U\cong \mathcal{C}(R,-)$ is a covariant representable functor, a morphism $R\to {\lim }_j{C}_j$ is exactly a compatible family of morphisms $R\to C_j$. Consequently the comparison mapis a bijection for every existing small categorical limit. This proves $R\Rightarrow L$ by the universal property of a limit. In particular,

Suppose $U(C)\cong \mathcal{C}(R,C)$ naturally and $\mathcal{C}$ has small coproducts in a category. DefineFor a function $v:X\to Y$, define $F(v)$ by sending the $x$-summand by the identity of $R$ to the $v(x)$-summand. The coproduct property gives the identity and composition laws, so $F$ is a functor. There are natural bijectionsThe first chooses a morphism on every summand; the second views that family as a function. Hence this is the left adjoint to a covariant representable functor:For empty $X$, the formula uses the empty coproduct, an initial object, and the same bijection remains valid.

Only $L\Rightarrow A$ remains. Assume $U$ preserves small categorical limits, and choose a small cogenerating family $(Q_i{)}_{i\in I}$ in $\mathcal{C}$. Fix a set $X$. The comma category $(X\downarrow U)$ is a complete category: underlying limits in $\mathcal{C}$, together with limit preservation by $U$, lift compatible maps from $X$. It is locally small, since its arrows form subsets of the hom-sets in $\mathcal{C}$.

We construct a weakly initial set in this comma category. Given $f:X\to U(C)$, intersect all subobjects $m:B\hookrightarrow C$ through which $f$ lifts under $U$. There are only a set of subobjects because $\mathcal{C}$ is well-powered; the family includes $1_C$. Their intersection exists by completeness. Also $U$ preserves monomorphisms, since a map is monic precisely when its self-pullback diagonal is invertible, and $U$ preserves such pullbacks.

Thus all lifts of $f$ are unique and compatible, and preservation of the intersection givesNo proper subobject of $C_0$ supports a lift of $f_0$: its composite into $C$ would belong to the intersected family, forcing it to contain $C_0$. This is a minimal supported subobject.

For each $i$, the mapis injective. Indeed, equal images mean $f_0$ lifts to the equalizer of the two maps, because $U$ preserves that equalizer. Minimality forces the equalizer to be invertible, so the maps agree.

The cogenerator property makes the evaluation mapa monomorphism: any two arrows into $C_0$ that agree after all these projections are equal. The indexing pairs $(i,h)$ inject into the fixed setTheir image is some subset $J\subseteq B_X$. Consequently $C_0$ is isomorphic to a subobject of a product $P_J$ of the $Q_i$ indexed by $J$. There are only a set of such subsets $J$, only a set of subobjects of each $P_J$, and only a set of maps $X\to U(B)$ for each representative subobject $B$.

Collect all these pairs $(B,x:X\to U(B))$. Every $(C,f)$ receives a comma arrow from one of them, via the isomorphic copy of $(C_0,f_0)$ and $m_0$. Hence they form a weakly initial set. The initial-object lemma for complete categories with a weakly initial set, proved explicitly in Question 8, gives an initial object of $(X\downarrow U)$. Its map from $X$ is a universal arrow from an object to a functor. Choosing these universal arrows for all $X$ defines a left adjoint to $U$.

ThusThe products indexed by realized subsets $J$ are important: one cannot assume that $C_0$ has a map to every cogenerator and simply add arbitrary missing coordinates.

The general adjoint functor theorem says that, for a functor $U:\mathcal{C}\to \mathcal{D}$ with $\mathcal{C}$ a complete category and both categories locally small, $U$ has a left adjoint exactly when it preserves small categorical limits and satisfies the solution-set condition. The latter means that for each $D\in \mathcal{D}$ there is a set of arrows $f_i:D\to U(C_i)$ such that every $f:D\to U(C)$ factors asEquivalently, $(D\downarrow U)$ has a weakly initial set.

The limit form of the special adjoint functor theorem says that, if in addition $\mathcal{C}$ is well-powered and has a small cogenerating family, then $U$ has a left adjoint exactly when it preserves small limits. These are the limit/left-adjoint forms; their duals interchange limits with colimits, left with right adjoints, and cogenerators with generators.

For the first route, let $\mathcal{E}$ be a complete category that is locally small and has a weakly initial set $(W_i)$. Put $P={\prod }_i{W}_i$. It is weakly initial: for any $X$, choose an arrow $W_i\to X$ and precompose with the corresponding product projection. The set $\mathcal{E}(P,P)$ indexes a simultaneous equalizerIt exists by completeness, for example as the equalizer of the two maps $P\rightrightarrows {\prod }_{u:P\to P}P$ with components $u$ and $1_P$. Since $E\to P\to X$ exists for each $X$, $E$ is also weakly initial.

Choose $r:P\to E$. Applying the defining equation to $er:P\to P$ gives $ere=e$, whence $re=1_E$ because $e$ is a monomorphism. If $v:E\to E$ is any endomorphism, apply it to $evr:P\to P$ to obtain $evre=e$, hence $ev=e$ and $v=1_E$. Thus $E$ has only its identity endomorphism.

For two arrows $a,b:E\rightrightarrows X$, take their equalizer $j:Y\to E$. Weak initiality of $E$ gives $t:E\to Y$. Since $jt$ is an endomorphism of $E$, $jt=1_E$. The monic arrow $j$ with this right inverse is invertible; therefore $a=b$. Existence of an arrow $E\to X$ and this uniqueness show that $E$ is initial.

For the general adjoint functor theorem, limit preservation makes each comma category $(D\downarrow U)$ complete, and the solution-set condition makes it weakly initially generated by a set. The lemma provides an initial object $(FD,{\eta }_D)$. For $v:D\to D^{\prime }$, the universal property uniquely defines $Fv:FD\to F{D}^{\prime }$ byUniqueness gives the functor laws. The same property gives natural bijectionsso $F⊣U$. Conversely a right adjoint preserves limits, by applying its hom-set bijections to cones, and its unit $(FD,{\eta }_D)$ is already an initial object of each comma category, hence a singleton solution set. This proves necessity as well as sufficiency.

For the second route, assume the general adjoint functor theorem and the additional hypotheses of the Special adjoint functor theorem. Let $(Q_i{)}_{i\in I}$ be a small cogenerating family. Fix $D\in \mathcal{D}$ and an arrow $f:D\to UC$. A limit-preserving functor preserves monomorphisms, by the self-pullback criterion. Intersect the set of all subobjects $B\hookrightarrow C$ supporting $f$. Preservation of this categorical limit gives a minimal supporting pair $(C_0,f_0:D\to U{C}_0)$ and a monomorphism $m_0:C_0\to C$ with $U(m_0)f_0=f$.

If $g,h:C_0\rightrightarrows Q_i$ have $Ug{f}_0=Uh{f}_0$, their equalizer supports $f_0$ and must be invertible by minimality. ThusThe evaluation map into the product of all these cogenerator targets is monic, because the family coseparates arrows. Its coordinates inject into the fixed setFor every subset $J\subseteq B_D$, form the product $P_J$ whose coordinate $(i,z)$ has target $Q_i$. Each minimal $C_0$ is isomorphic to a subobject of one such product, using precisely its realized coordinates. There is a set of subsets $J$; the well-powered category condition gives a set of representative subobjects of each product. For each representative $B$, include every arrow $D\to UB$, a set by local smallness of $\mathcal{D}$.

This is a solution-set condition: the minimal pair is isomorphic to one of these representatives, and its inclusion into $C$ supplies the required factorization of $f$. The general adjoint functor theorem now gives the left adjoint. This is the cogenerator bound for comma-category solution sets proof of the special theorem. Necessity follows again because a right adjoint preserves limits. Both alternative proof routes are therefore supplied.

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying the unit and multiplication of a monad lawsAn algebra for a monad is an object $A$ equipped with $a:TA\to A$ such thatA morphism of algebras for a monad $f:(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $fa=bT(f)$. Identities and composition obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^T$ and its forgetful functor $U^T(A,a)=A$.

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with induced monad $T=UF$, the Eilenberg-Moore comparison functor isThe functor $U$ is monadic when it has such a left adjoint and this comparison is an equivalence of categories; a stricter convention requires an isomorphism of categories over $\mathcal{C}$. We will identify which limit-lifting conclusion each convention supports in part (iii).

For $F⊣U$ with unit of an adjunction $\eta$ and counit of an adjunction $\epsilon$, putThe triangle identities for an adjunction giveFor associativity, naturality of $\epsilon$ at ${\epsilon }_{FA}:FUFA\to FA$ saysApply $U$ to obtain ${\mu }_{A}T({\mu }_A)={\mu }_A{\mu }_{TA}$. Thus this is the monad induced by an adjunction.

Conversely, start with a monad $(T,\eta ,\mu )$. The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad laws ensure that these are algebras for a monad and morphisms of algebras for a monad. For an algebra $(B,b)$, there are inverse bijectionsThe inverse really is an algebra map becauseFor a free-algebra map $g$, its algebra equation and the monad unit law giveFor a base map $f$, naturality and the algebra unit law give $bT(f){\eta }_A=b{\eta }_{B}f=f$. The formulas are natural, establishing the free-forgetful Eilenberg-Moore adjunction $F^T⊣U^T$. Its unit is ${\eta }_A$, and its counit at $(B,b)$ is $b$. Hence its induced multiplication is $\mu$, and every monad arises from an adjunction.

First prove the exact lifting statement for the forgetful functor $U^T:{\mathcal{C}}^T\to \mathcal{C}$. Take any diagram in a category of algebras for a monad $(A_j,a_j)$ whose underlying diagram has a chosen categorical limit $p_j:L\to A_j$. The arrows $a_{j}T(p_j):TL\to A_j$ form a categorical cone, since every diagram arrow is a morphism of algebras for a monad. Thus the universal property gives a unique arrow $a:TL\to L$ satisfyingThe projections are jointly monic: two arrows into $L$ with the same composites with every $p_j$ are equal. Therefore the algebra identities can be checked after these projections. For the unit,For multiplication,The algebra law for $a_j$ makes the right-hand sides equal. Thus $(L,a)$ is an algebra for a monad, and each $p_j$ is a morphism of algebras for a monad. Its action is unique with this property.