PlanetMath: properties of direct product

Before proving this, let us observe first that if $f:C\to C$ is a morphism such that \begin{equation} \pi_i=\pi_i\circ f \end{equation}then $f=1_C$ necessarily, since the universal property of product $C$ , $f$ is the unique morphism such that (1) holds, but then $\pi_1=\pi_1\circ 1_C$ as well, and this forces $f=1_C$ .

Proof. Let $$C=\prod_{i\in I} C_i, \quad D=\prod_{j\in J} C_j, \quad\mbox{and}\quad E=\prod_{k\in K} C_k.$$ We break down the proof into two cases:

Suppose one of $J$ and $K$ is the empty set, say, $J=\varnothing$ . Then $D$ is a terminal object, and $K=I$ , so that $E=C$ . In other words, we want to show that $$C\cong D\times C.$$ First, notice that we have morphisms $1_C:C\to C$ and $e_C:C\to D$ (where $e_C$ is unique since $D$ is terminal). If $A$ is any object with morphisms $f:A\to C$ and $e_A: A\to D$ . Any $g:A\to C$ with $f=1_C\circ g$ and $e_A = e_C\circ g$ must result in $f=g$ . This shows that $C$ may be viewed as the product of $D$ and $C$ , or $C\cong D\times C$ .
Now, suppose neither $J$ nor $K$ is empty. We have projection morphisms $f_i:C\to C_i$ for all $i\in I$ , $g_j:D\to C_j$ for all $j\in J$ , and $h_k:E\to C_k$ for all $k\in K$ . Write $F=D\times E$ with projections $p_1:F\to D$ and $p_2: F\to E$ .
For every $i\in I$ , define morphisms $x_i:F\to C_i$ as follows: if $i\in J$ , then $x_i=g_i\circ p_1$ . Otherwise, $x_i=h_i\circ p_2$ . Since $J$ and $K$ are disjoint, this $I$ -indexed set of morphisms is well-defined. By the universality of the product $C$ , we get a unique morphism $x:F\to C$ such that $x_i=f_i\circ x$ .

Next, from the universal properties of the products $D$ and $E$ , we have two unique morphisms $y:C\to D$ and $z:C\to E$ such that $f_j = g_j\circ y$ and $f_k=h_k\circ z$ for any $j\in J$ and $k\in K$ . From the morphisms $y:C\to D$ and $z:C\to E$ and the universality of the product $F$ , we have another unique morphism $f:C\to F$ such that $y=p_1\circ f$ and $z=p_2\circ f$ .

Then $p_1\circ (f \circ x) = (p_1\circ f)\circ x=y\circ x$ . Since $g_j\circ y \circ x=f_j\circ x=x_j=g_j\circ p_1$ for any $j\in J$ , we have $y\circ x=p_1$ , so that $p_1\circ (f\circ x)=p_1$ . Similarly, $p_2\circ (f\circ x)=p_2$ . This shows that $f\circ x=1_F$ . Also, $f_i\circ ( x\circ f) = (f_i\circ x)\circ f = x_i \circ f$ . Now, if $i\in J$ , then $x_i\circ f = g_i\circ p_1 \circ f = g_i \circ y = f_i$ , and if $i\in K$ , then $x_i\circ f = h_i\circ p_2 \circ f = h_i \circ z = f_i$ . As a result, $f_i\circ (x\circ f)=f_i$ for all $i\in I$ , which implies $x\circ f = 1_C$ . This shows that $C\cong F=D\times E$ .

This completes the proof. $\qedsymbol$

This shows that it does not matter whether we say $A\times B$ as the product of $A$ and $B$ , or the product of $B$ and $A$ .

Remarks. All of the properties can be dualized, so that coproducts are unique up to isomorphism, and commutativity and associativity laws hold as well.