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properties of direct product
Let $\mathcal{C}$ be a category. This entry lists some of the basic properties of categorical direct product:
Proposition 1.
(uniqueness of products) A product $(C,\{\pi_{i}\}_{{i\in I}})$ of objects $\{C_{i}\}_{{i\in I}}$, if it exists, is unique up to isomorphism.
Before proving this, let us observe first that if $f:C\to C$ is a morphism such that
$\pi_{i}=\pi_{i}\circ f$  (1) 
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.
If $(D,\{g_{i}\}_{{i\in I}})$ is another product of $\{C_{i}\}_{{i\in I}}$, we get two unique morphisms $x:D\to C$ and $y:C\to D$ such that $\pi_{i}=g_{i}\circ y$ and $g_{i}=\pi_{i}\circ x$ for all $i\in I$. So $\pi_{i}=(\pi_{i}\circ x)\circ y=\pi_{i}\circ(x\circ y)$. From the previous paragraph, we see that $x\circ y=1_{C}$. Similarly, $g_{i}=g_{i}\circ(y\circ x)$, so that $y\circ x=1_{D}$. This shows that $C$ is isomorphic to $D$. ∎
This justifies writing $\prod_{{i\in I}}C_{i}$ (with $\pi_{i}$) as the product of $\{C_{i}\}_{{i\in I}}$. In case $I$ has cardinality $2$, we write $C_{1}\times C_{2}$ as the product of $C_{1}$ and $C_{2}$. Also, when $I=\varnothing$, we set the product as any terminal object $T$ in $\mathcal{C}$.
Proposition 2.
If $I$ is the disjoint union of $J$ and $K$, then
$\prod_{{i\in I}}C_{i}\cong\prod_{{j\in J}}C_{j}\times\prod_{{k\in K}}C_{k},$ 
assuming all products exist.
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 welldefined. 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. ∎
Corollary 1.
(commutativity of products) $A\times B\cong B\times A$, if one (and hence the other) exists.
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 $A$.
Corollary 2.
(associativity of products) $A\times(B\times C)\cong A\times B\times C\cong(A\times B)\times C$, whenever the products are defined.
Remarks. All of the properties can be dualized, so that coproducts are unique up to isomorphism, and commutativity and associativity laws hold as well.
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