Many types of universal constructions that appear in a wide variety of mathematical contexts can be realized as categorical limits and colimits. To define a limit we first need the notion of a cone of a diagram. A diagram of type \(J\) in a category \(\mathcal C\) is a simply a functor \(F \colon J \to \mathcal C\). We imagine \(J\) as an indexing for the objects and morphisms in \(\mathcal C\) under consideration. When \(J\) is a finite category it can be visualized as a directed graph.

A cone of \(F\) is a pair \((N, \Psi)\) where \(N\) is an object of \(\mathcal C\) and \(\Psi\) is a collection of \(\mathcal C\)-morphisms \(\Psi_X \colon N \to F(X) \) (one for each object \(X\) in \(J\)) such that \(\Psi_Y = F(f) \circ \Psi_X\) for every \(J\)-morphism \(f \colon X \to Y\). Given two cones \((N, \Psi)\) and \((M, \Phi)\) we call a \(\mathcal C\)-morphism \(u \colon N \to M\) a cone morphism from \((N, \Psi)\) to \((M, \Phi)\) provided that \(u\) respects the cone property, which is to say that for every object \(X\) in \(J\) the morphism \(\Psi_X\) factors through \(u\), i.e., \(\Psi_X = \Phi_X \circ u\). We will write \(u \colon (N, \Psi) \to (M, \Phi)\) to denote that \(u\) is a cone morphism.

A limit cone of the diagram \(F\) is a cone \((L, \Phi)\) of \(F\) such that for every other cone \((N, \Psi)\) of \(F\) there exists a unique cone morphism \(u \colon (N, \Psi) \to (L, \Phi)\). In this sense a limit \(L\) is a universal cone. A co-cone and colimit are the corresponding dual notions, so in particular a co-cone is a cone of a diagram of type \(J\) in the opposite category \(\mathcal C^{op}\).

One observation to make is that a limit of \(F\) is unique up to canonical cone isomorphism. For this reason we typically speak of *the* limit of \(F\). What this means is that if \((L, \Phi)\) and \((L', \Phi')\) are limits of \(F\), then there exists a unique cone isomorphism between them. To see this, first note that the identity morphism \(\text{id}_L \colon L \to L\) is actually a cone morphism \(\text{id}_L \colon (L, \Phi) \to (L, \Phi)\). By the uniqueness property of limits, this is the only such cone morphism. Moreover, since \(L\) and \(L'\) are both limits, there exist unique cone morphisms \(u \colon (L', \Phi') \to (L, \Phi)\) and \(u' \colon (L, \Phi) \to (L', \Phi')\). If we can show that \(u \circ u' \colon L \to L\) and \(u' \circ u \colon L' \to L'\) are cone morphisms, then it must follow that \(u \circ u' = \text{id}_L\) and \(u' \circ u = \text{id}_{L'}\), which proves that \(u\) is in fact a unique cone isomorphism. To this end, observe that for any object \(X\) in \(J\) it follows that \(\Phi'_X = \Phi_X \circ u\) and \(\Phi_X = \Phi'_X \circ u'\), so \(\Phi_X = \Phi'_X \circ u'= (\Phi_X \circ u) \circ u' = \Phi_X \circ (u \circ u')\) and similarly \(\Phi'_X = \Phi'_X \circ (u' \circ u)\). So \(u \circ u'\) and \(u' \circ u\) are cone morphisms, as desired.

The product and co-product of objects in a category \(\mathcal C\) are special cases of limits and co-limits. When \(J\) consists precisely of two distinct objects \(X\) and \(Y\) with no non-identity morphisms, and \(F \colon J \to \mathcal C\) is a functor, the product of \(F(X)\) and \(F(Y)\) is the limit \((F(X) \times F(Y), \pi)\) of \(F\) and the co-product of these two objects is the co-limit \((F(X) \amalg F(Y), \iota)\).

For a specific example products and co-products consider the category \(\mathcal C\) of sets and let \(J\) be as above. If \(F \colon J \to \mathcal C\) is any diagram, we claim that the Cartesian product \(F(X) \times F(Y)\) together with the projection maps is the limit of \(F\). To see this, first set \(A:=F(X)\) and \(B:=F(Y)\) and let \(\pi_A\) and \(\pi_B\) denote the respective projections from \(A \times B\). We must show that if \(N\) is any other set with maps \(f \colon N \to A\) and \(g \colon N \to B\), then there exists a unique map \(u \colon N \to A \times B\) such that \(f = \pi_A \circ u\) and \(g = \pi_B \circ u\). First, the map \(f \times g \colon N \to A \times B\) taking an element \(x \in N\) to the tuple \((f(x), g(x))\) clearly satisfies \(\pi_A \circ (f \times g) = f\) and \(\pi_B \circ (f \times g) = g\), so \(f \times g\) is a cone morphism. Now suppose \(u \colon (N, \{ f, g \}) \to (A \times B, \pi)\) is a cone morphism. Then \(\pi_A \circ u = f\) and \(\pi_B \circ u = g\). If \(u(x)=(a,b)\) for an element \(x \in N\), then \(a = (\pi_A \circ u) (x)= f(x)\) and \(b = (\pi_B \circ u) (x) = g(x)\), so \(uh(x)= f \times g (x)\). This proves that \(u = f \times g\), so \(f \times g\) is the unique cone morphism from \(N\) to \(A \times B\).

Next we claim that the disjoint union \(A \amalg B\) together with the injections \(\iota_A \colon A \to A \amalg B\) and \(\iota_B \colon B \to A \amalg B\) is the co-product of \(A\) and \(B\). To see this, we must show that if \(N\) is any other set and there are maps \(f \colon A \to N\) and \(g \colon B \to N\), then \(f\) factors uniquely through \(\iota_A\) and \(g\) factors uniquely through \(\iota_B\). More specifically this means we must show there is a unique map \(u \colon A \amalg B \to N\) such that \(f = u \circ \iota_A\) and \(g = u \circ \iota_B\). Define a map \(f \amalg g \colon A \amalg B \to N\) by \((f \amalg g) (x) = f(x)\) if \(x \in A\) and \((f \amalg g) (x) = g(x)\) if \(x \in B\). If \(u \colon (A \amalg B, \iota) \to (N, \{ f, g \} )\) is a cone morphism, then \(f(a) = (u \circ \iota_A) (a)\) for all \(a \in A\) and \(g(b) = (u \circ \iota_B) (b)\) for all \(b \in B\). So in particular, if \(a \in A\) (viewed as a subset of \(A \amalg B\)) we have \(u(a) = f(a) = (f \amalg g) (a)\) and if \(b \in B\) (viewed as a subset of \(A \amalg B\)) then \(u(b) = g(b) = (f \amalg g) (b)\), so \(u = f \amalg g\). This proves that \(f \amalg g\) is a unique cone morphism, as desired.