Proof.

• $\Rightarrow$ Since each of (i) - (iii) is a special case of Definition 6.1.
• $\Leftarrow$ (i) deals with the empty diagram.

  Given $D:J\to C$ with $J$ finite and non-empty, by repeated use of (ii) we can find $A$ with morphisms $D(j)\to A$ for all $j$. Then by repeated use of (ii) we can find $A\to B$ coequalising

  
  for each $\alpha \in \mathrm{mor}J$. □

Proof. By Proposition 4.4(i), enough to show $C$ has all small coproducts.

Given a set-indexeud family $(A_j|j\in J)$ of objects, the finite coproducts ${\sum }_{j\in F}{A}_j$, for $F\subseteq J$ finite, form the vertices of a diagram of shape $P_{f}J=\{F\subseteq J|F\text{finite}\}$ whose edges are coprojections. $P_{f}J$ is directed, and a colimit for this diagram has the universal property of a coproduct ${\sum }_{j\in J}{A}_j$. □

Proof.

• $\Rightarrow$ Let $D:I\to J$ be a diagram with $I$ finite. We have a diagram $E:I^{op}\times J\to Set$ defined by $E(i,j)=J(D(i),j)$.

  For each $i$, $({\mathrm{colim}}_{J}E)(i)$ is a singleton since every $D(i)\to j$ is identified with $1_{D(i)}$ in the colimit, so $\underset{I}{\lim }{\mathrm{colim}}_{J}E$ is a singleton.

  But elements of $\underset{I}{\lim }E(\bullet ,j)$ are cones under $D$ with apex $j$, so if ${\mathrm{colim}}_J\underset{I}{\lim }E$ is nonempty there must be such a cone for some $j$.

• $\Leftarrow$ Suppose given $D:I\times J\to Set$ where $I$ is finite and $J$ is filtered. In general, the colimitof $E:J\to Set$ is the quotient of ${\coprod }_{j\in \mathrm{ob}J}E(j)$ by the smallest equivalence relation identifying $x\in E(j)$ with $D(\alpha )(x)\in E(j^{\prime })$ for all $\alpha :j\to j^{\prime }$ in $J$. For filtered $J$, this identifies $x\in E(j)$ with $x^{\prime }\in E(j^{\prime })$ if and only if there exists $j\stackrel{\alpha }{\to }{j}^{″}\stackrel{{\alpha }^{\prime }}{\leftarrow }{j}^{\prime }$ with $E(\alpha )(x)=E({\alpha }^{\prime })(x^{\prime })$, and moreover if $j=j^{\prime }$ we may assume $\alpha ={\alpha }^{\prime }$.

  Now, given an element $x$ of $\underset{I}{\lim }{\mathrm{colim}}_{J}D$, we can write it as $(x_i|i\in \mathrm{ob}I)$ where $x_i\in {\mathrm{colim}}_{J}D(i,\bullet )$ is an equivalence class of elements $x_{ij}\in D(i,j)$. If $\alpha :i\to i^{\prime }$ in $I$, then $D(\alpha ,j)(x_{i}j)$ and $x_{i^{\prime }{j}^{\prime }}$ represent the same element of ${\mathrm{colim}}_{J}D(i^{\prime },\bullet )$ so by repeated use of Lemma 6.2(ii) we can choose representatives in $D(i,j_0)$ for some fixed $j_0$, and by repeated use of Lemma 6.2(iii) we can assume that these representatives define an element of $\underset{I}{\lim }D(\bullet ,j_0)$. This defines an element of ${\mathrm{colim}}_J\underset{I}{\lim }D$ mapping to the given element of $\underset{I}{\lim }{\mathrm{colim}}_{J}D$.

  The proof of injectivity is similar: if two elements $x,y$ of ${\mathrm{colim}}_J\underset{I}{\lim }D$ have the same image in $\underset{I}{\lim }{\mathrm{colim}}_{J}D$ we can choose representatives $x_j,y_j$ in $\underset{I}{\lim }D(\bullet ,j)$ and then find $j\to j^{\prime }$ so that each of the components $x_{ij}$ and $y_{ij}$ map to the same element of $D(i,j^{\prime })$ under $j\to j^{\prime }$. So $x=y$ in ${\mathrm{colim}}_J\underset{I}{\lim }D$. □

Proof.

• (i) This is just like Example 5.14(a): Given a filtered diagram $D:J\to C$ and a colimit for $UD$ with apex $L$, then $L^n$ is the colimit of $U{D}^n$ for all $n$, so each $n$-ary operation on the $D(j)$’s induces an $n$-ary operation on $L$, and $L$ also inherits all the equations defining $C$, so there’s a unique lifting of the colimit cone under $UD$ to a colimit cone for $D$.
• (ii) Follows from (i) and Theorem 6.5, since $U$ also creates finite limits (and reflects isomorphisms). □

Proof.

• (i) $\Rightarrow$ (ii) By Lemma 4.10, $(1\downarrow F)$ has finite limits so ${(1\downarrow F)}^{op}$ is filtered.
• (ii) $\Rightarrow$ (iii) Consider the diagram ${(1\downarrow F)}^{op}\stackrel{U}{\to }{C}^{op}\stackrel{Y}{\to }[C,Set]$ where $U$ is the forgetful functor and $Y$ is the Yoneda embedding. A cone under this diagram (with apex $G$, say) yields a family of morphisms $C(A,\bullet )\stackrel{{\lambda }_{(A,x)}}{\to }G$ for each $x\in FA$, subject to compatibility conditions which say that $(Gf)\mathrm{\Phi }({\lambda }_{(A,x)})=\mathrm{\Phi }({\lambda }_{(B,y)})$ for every $f:(A,x)\to (B,y)$ in $(1\downarrow F)$, i.e. such that $x\mapsto \mathrm{\Phi }({\lambda }_{(A,x)})$ is a natural transformation $F\to G$. So the cone $(C(A,\bullet )\stackrel{\mathrm{\Psi }(x)}{\to }F|(A,x)\in \mathrm{ob}(1\downarrow F))$ has the universal property of a colimit for the diagram.
• (iii) $\Rightarrow$ (i) Functors of the form $C(A,\bullet )$ preserve any limits which exist, so this follows from Theorem 6.5 plus the fact that colimits in $[C,Set]$ are computed pointwise. □

Proof. Given a model $M:T^{op}\to Set$, we have $M[n]\cong M{[1]}^n$ for all $n$. Also, any morphism $M{[1]}^n\to M{[1]}^p$ induced by a morphism $[p]\to [n]$ in $T$ is determined by its composites with the projections $M{[1]}^p\to M[1]$, so specifying $M$ on morphisms is determined by its effect on morphisms with domain $[1]$.

So, given a set $A$, specifying a model $M$ with $M[1]=A$ is equivalent to specifying operations ${\alpha }_A:A^n\to A$ for each $\alpha :[1]\to [n]$ in $T$, subject to ${(v_i)}_A(a_1,\dots ,a_n)=a_i$ whenever $v_i:[1]\to [n]$ is the $i$-th coprojection, and

Proof.

• (ii) $\Rightarrow$ (i) Let $T$ be the full subcategory of $C$ on the free algebras $F[n]$, for $n\in \mathbb{N}$. Then $T$ is a Lawvere theory, and for every object $A$ of $C$, the functor $C(\bullet ,A)$ restricted to $T$ preserves finite products, so it’s a model of $T$. This defines a functor $T-Mod(Set)\stackrel{Y}{\leftarrow }{Set}^{𝕋}$; but $T-Mod(Set)\simeq {Set}^{{𝕋}^{\prime }}$ for some finitary monad ${𝕋}^{\prime }$ on $Set$, so we get a functor ${Set}^{𝕋}\stackrel{Y}{\to }{Set}^{{𝕋}^{\prime }}$ which is the identity on underlying sets.

  In this situation, $Y$ is induced by a morphism of monads ${𝕋}^{\prime }\to 𝕋$, i.e. a natural transformation $𝜃:T^{\prime }\to T$ commuting with the units and multiplications. (Clearly, such a $𝜃$ induces a functor ${Set}^{𝕋}\to {Set}^{{𝕋}^{\prime }}$ sending $(A,\alpha )$ to $(A,\alpha {𝜃}_A)$).

  But we know ${𝜃}_{[n]}$ is bijective for all $n$, since elements of the free algebras on $[n]$ are just morphisms $[1]\to [n]$ in $T$. But both functors are finitary, so ${𝜃}_A$ is bijective for all $A$, i.e. it’s an isomorphism of monads. □