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Here is a precise base change of a pullback square statement. Suppose $A=B{\times }_{D}C$ is a pullback in a category and $t:D^{\prime }\to D$ is a morphism. Assume the pullbacks in a categoryexist, where $A\to D$ is the common composite. The induced square with vertices $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ is a pullback in a category; equivalently,Indeed, compatible morphisms $T\to B^{\prime }$ and $T\to C^{\prime }$ give morphisms $b:T\to B$, $c:T\to C$, and a common $v:T\to D^{\prime }$, with the composites of $b,c$ to $D$ both equal to $tv$. The original pullback in a category gives a unique morphism $h:T\to A$ with projections $b,c$. Its composite to $D$ is $tv$, so the defining pullback in a category for $A^{\prime }$ gives a unique morphism $T\to A^{\prime }$ above $h,v$. Its maps to $B^{\prime }$ and $C^{\prime }$ are the prescribed ones by uniqueness in those pullbacks in a category. Conversely, these projections recover $b,c,v$, so every possible factorization is the same. This proves the universal property without requiring arbitrary categorical limits in the ambient category.

Keep the notation of the preceding part. For $u:T\to D$ and $w:T\to B$ satisfying $fu=gw$, we have $efu=egw=dcw$. The outside pullback in a category gives a unique $z:T\to A$ with $az=u$ and $cbz=cw$. Moreover $gbz=faz=fu=gw$. The morphisms $bz,w:T\to B$ thus have the same composites with $g$ and $c$. Uniqueness in the right pullback in a category gives $bz=w$. Any factorization through the left square also factors through the outside rectangle, so it is unique there. The left square is a pullback. Together these two arguments prove the pullback pasting lemma.

Name the upper objects $A,B,C$ and the lower objects $D,E,F$, so the morphisms are $b:A\to B$, $c:B\to C$, $a:A\to D$, $g:B\to E$, $d:C\to F$, $f:D\to E$, and $e:E\to F$. The commutative diagram gives $gb=fa$ and $dc=eg$.

Take $u:T\to D$ and $v:T\to C$ with $efu=dv$. The right pullback in a category gives a unique $w:T\to B$ with $gw=fu$ and $cw=v$. The left pullback in a category then gives a unique $z:T\to A$ with $az=u$ and $bz=w$. It follows that $cbz=v$, giving the required factorization through the outside rectangle. Any other factorization has the same middle morphism by uniqueness in the right pullback in a category, and therefore is the same $z$ by uniqueness in the left pullback in a category. The outside rectangle is a pullback.

Let $1$ be a terminal object. The pullback in a category of $X\to 1\leftarrow Y$ has precisely the universal property of the binary product in a category $X\times Y$: every pair of morphisms to $X,Y$ is compatible over $1$. Repeated binary products in a category, together with $1$ as the empty product, give all finite products in a category.

For $f,g:X\rightrightarrows Y$, form the pullback in a category of the categorical diagonal ${\Delta }_Y:Y\to Y\times Y$ along $⟨f,g⟩:X\to Y\times Y$. Its projection $e:E\to X$ satisfies $fe=ge$. If $h:T\to X$ satisfies $fh=gh$, the pair $h,fh$ gives a unique morphism into this pullback in a category. Conversely any such morphism must project to that pair. Hence $e$ is the equalizer of $f,g$. The preceding construction now gives all finite limits.

For a functor $D:\mathcal{J}\to \mathcal{C}$ with finite indexing category, form the finite products in a categoryDefine $s,t:P\rightrightarrows Q$ by requiring the coordinate indexed by $u:i\to j$ to be $D(u){\pi }_i$ for $s$ and ${\pi }_j$ for $t$. Let $e:L\to P$ be their equalizer, and put ${\lambda }_j={\pi }_{j}e$. The equality $se=te$ says exactly that $D(u){\lambda }_i={\lambda }_j$ for every morphism $u$, so these morphisms form a categorical cone.

For any other categorical cone $(x_j:X\to D(j))$, the product in a category gives a unique $x:X\to P$ with ${\pi }_{j}x=x_j$. The categorical cone equations give $sx=tx$, so the equalizer gives a unique $\bar{x}:X\to L$ with $e\bar{x}=x$. This is exactly the universal property of a categorical limit. Empty indexing products are terminal objects, so the construction includes the empty diagram in a category. Thus finite products and equalizers give all finite limits.

Choose for each $j\in \mathcal{J}$ a categorical limit $L_j$ of $F(-,j)$, with projections ${\lambda }_{i,j}:L_j\to F(i,j)$. For $v:j\to k$, the morphisms $F(1_i,v){\lambda }_{i,j}$ form a categorical cone over $F(-,k)$, since the two actions of the product functor $F$ commute. Define $L(v):L_j\to L_k$ by the equationsExistence and uniqueness follow from the universal property. The identity morphism of $L_j$ satisfies the equations for $v=1_j$, and $L(w)L(v)$ satisfies those for $wv$, so uniqueness gives $L(1_j)=1_{L_j}$ and $L(wv)=L(w)L(v)$. Hence $L:\mathcal{J}\to \mathcal{D}$ is a functor.

If $L_j^{\prime }$ with projections ${\lambda }_{i,j}^{\prime }$ are other choices, there is a unique isomorphism $u_j:L_j\to L_j^{\prime }$ carrying every projection to the corresponding projection. The same equations show $u_{k}L(v)=L^{\prime }(v)u_j$, because their composites with all ${\lambda }_{i,k}^{\prime }$ agree. Thus the choices give a unique natural isomorphism compatible with the chosen categorical limit categorical cones. Without the compatibility condition, uniqueness of an arbitrary natural isomorphism is not asserted.

For $D:\mathcal{I}\to \mathcal{C}$, define the functor $\mathrm{Cone}(-,D):{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$ bywhere $\Delta X$ is the constant diagram functor; a morphism into $X$ acts by precomposition on every categorical cone leg. A categorical limit is a representation of a functor of the formThe image of $1_L$ is the universal categorical cone $\lambda :\Delta L\to D$. By naturality of the representing isomorphism, the image of $h:X\to L$ is $\lambda \circ \Delta h$. Thus this definition is exactly existence and uniqueness of factorization of every categorical cone through $\lambda$. Conversely that universal property gives the representing natural bijection.

The category of elements $\int X$ has objects $(W,x)$ for $x\in X(W)$, and morphisms $f:(W,x)\to (V,y)$ with $X(f)y=x$. Send $(W,x)$ to the representable presheaf $H_W$, and a morphism $f$ to $H_{\bullet }(f)$. The element $x$ supplies a natural transformation $H_W\to X$ with component $h\mapsto X(h)x$, by the Yoneda lemma. These transformations form a cocone.

Pointwise at $U$, the colimit of this diagram in a category has representatives $(W,x,h:U\to W)$, with relation $(W,X(f)y,h)\sim (V,y,fh)$. These are exactly the relations of the preceding categorical coend. Its proved natural isomorphism with $X(U)$ therefore identifies the cocone with the universal colimit cocone. More directly, every representative equals $(U,X(h)x,1_U)$ through the morphism $h:(U,X(h)x)\to (W,x)$; two representatives with the same image therefore agree. As $\mathcal{C}$ is small, $\int X$ is small, so this is an ordinary small colimit. Consequently

Let $T:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$. A categorical end is an object $E$ with morphisms $e_A:E\to T(A,A)$ satisfying, for every $f:A\to B$,It is universal among such families: any $x_A:X\to T(A,A)$ satisfying the same equations is uniquely $e_{A}h$ for one $h:X\to E$. This is a universal dinatural transformation from the constant value $E$, and is written $E={\int }_{A}T(A,A)$.

Dually a categorical coend is $Q$ with morphisms $j_A:T(A,A)\to Q$ satisfyingAny family $x_A:T(A,A)\to X$ satisfying these equations is uniquely $h{j}_A$ for one $h:Q\to X$. Write $Q={\int }^{A}T(A,A)$. For small $\mathcal{C}$, the categorical end can be constructed as the equalizer of ${\prod }_{A}T(A,A)\rightrightarrows {\prod }_{f:A\to B}T(A,B)$ when these objects exist. The categorical coend is the coequalizer of ${\coprod }_{f:A\to B}T(B,A)\rightrightarrows {\coprod }_{A}T(A,A)$ when those objects exist. The two morphisms impose exactly the displayed equations for a dinatural transformation.