Two Definitions of Infinite Cartesian Product

Question

In my one of lecture notes, there are two definitions of infinite Cartesian product, and it reads that we can construct a unique bijection between them.

One way to define an infinite Cartesian product $\prod_{i \in I}X_i$ of a family of sets $\{X_i\}_{i \in I}$ is as follows:

$\prod_{i \in I}X_i=\{f:I\rightarrow \bigcup_{i\in I} X_i : f(i)\in X_i\}$

On the other hand, another way to define an infinite Cartesian product of a family of sets is as follows:

We could have defined $\prod_{i \in I}X_i$ as a set together with a family of maps $\{p_j : \prod_{i \in I}X_i\rightarrow X_j\}_{j\in I}$ having the universal property.

Now, in the lecture note, the universal property is as follows:

Given any set $Z$ and any family of maps $\{g_i : Z\rightarrow X_i\}_{i\in I}$, there exists a unique function $g:Z\rightarrow \prod_{i \in I}X_i$ such that $g_i=p_i\circ g$ for all $i\in I$.

But, I think in order to prove that there is a unique bijection between the two different infinite Cartesian products as defined above, we need another assumption that a family of maps $\{p_j : \prod_{i \in I}X_i\rightarrow X_j\}_{j\in I}$ mentioned at the first quote has to be a family of canonical projections such that $p_j(x)=x(j)$ where $x$ is a function from $I$ to $\bigcup_{i\in I}X_i$ such that $x(i)\in X_i$.

So, will we need this new assumption, or not? If not, could somebody give a proof for it?

Answer 1

Here is how the universal property should be used.

First, the universal property should be stated abstractly in the following manner. I'll use a different abstract letter for the universal property, since by using the letter $Z$ in the comments I think that I just created confusion.

A set $Q$ and a family of maps $\{q_i : Q \to X_i\}_{i \in I}$ satisfies the universal property if for any other set $Z$ and any other family of maps $\{g_i : Z \to X_i\}_{i \in I}$ there exists a unique map $h : Z \to Q$ such that $q_i \circ h = g_i$ for all $i \in I$.

Next, one should prove that the universal property does indeed hold for the set $Q=\Pi_i X_i$ and the family of maps $q_i = p_i$ where $p_i(f)=f(i)$ is the standard projection function onto $X_i$. To do this, suppose that $Z$ is any set and $\{g_i : Z \to X_i\}_{i \in I}$ is any family of maps. Define the map $h : Z \to \Pi_i X_i$ by $h(z) = f$ where $f(i) = g_i(z)$, check that indeed $p_i \circ h = g_i$. Also check uniqueness: if $h' : Z \to \Pi_i X_i$ satisfies $\pi_i \circ h' = g_i$ then $h'(z)$ must satisfy $h'(z)(i) = g_i(z)$, which determines $h=h'$ uniquely.

Now there is an abstract argument which is the same for almost any situation where you have a universal property. In this particular situation the abstract argument goes like this:

• Suppose that $Q$ is a set and $\{q_i : Q \to X_i\}_{i \in I}$ is a family of maps which satisfies the universal property.
• Suppose that $Q'$ is another set and $\{q'_i : Q' \to X_i\}_{i \in I}$ is another family of maps which satisfies the universal property.
• By the universal property for $Q'$, there exists a unique map $h : Q \to Q'$ such that $q'_i \circ h = q_i$ for all $i \in I$.
• Also by the universal property for $Q$, there exists a unique map $h' : Q' \to Q$ such that $q_i \circ h' = q'_i$ for all $i \in I$.
• We now have two maps $Q \to Q$, namely $h' \circ h$ and $Id_Q$, such that $q_i \circ (h' \circ h)$ and $q_i \circ Id_Q$ are both equal to $q_i$. It follows, by the universality property for $Q$, that $h' \circ h = Id_Q$.
• We also have two maps $Q' \to Q$, namely $h \circ h'$ and $Id_{Q'}$, such that $q'_i \circ (h \circ h')$ and $q'_i \circ Id_{Q'}$ are both equal to $q'_i$. It follows, by the universality property for $Q'$, that $h \circ h' = Id_{Q'}$.

We conclude that $h$ and $h'$ are inverse functions of each other and so each is a bijection. (I've always been impressed by how this argument uses four different applications of the universal property).

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So, how does this apply to your question regarding two different cartesian products according to the second definition, and the possible need for a new assumption?

The trouble with your question is that you have not stated the second quote (the definition of a set with a family of maps satisfying the universal property) in the correct level of abstraction. That is why, in my discussion above, I have separated out the abstract statement of what a set with a family of maps satisfying the universal property means from the proof that the given set $\Pi_i X_i$ from the first definition with the given standard projections functions $p_i : \Pi_i X_i \to X_i$ does indeed satisfy the universal property (your first definition of $\Pi_i X_i$ is correct).

So, for instance, you could take $Q = \Pi_i X_i$, and you could take $q_i : \Pi_i X_i \to X_i$ to be some totally weird collection of functions, not the standard projection functions, as long as it satisfies the universal property. For example, pick a bijection $\sigma : \Pi_i X_i \to \Pi_i X_i$ and define $q_i = p_i \circ \sigma$. This choice of $Q$ and of maps $q_i$ does indeed satisfy the universal property, as you can check. And so there does indeed exist a unique bijection $h : \Pi_i X_i \to \Pi_i X_i$ that satisfies $p_i \circ h = q_i$. And in fact we know that bijection has to be, namely $\sigma$.

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I hope this helps. Getting the right level of abstraction straight when trying to understand universal properties can be a head banger.