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A product space ${\prod }_i{X}_i$ is the Cartesian product equipped with the product topology, the coarsest topology making every coordinate projection map continuous.

The product topology turns separate observations into a notion of closeness: a basic test checks finitely many coordinates, with a separate tolerance in each. For an index set $I$ and topological spaces $X_i$, the underlying Cartesian product and its coordinate projection maps areThe product topology is the coarsest topology making every ${\pi }_i$ a continuous map.^[ref]

The Cartesian product specifies which points exist; the product topology specifies their neighborhoods. Finite observations motivate its finite-coordinate basis for the product topology, while convergence in a product space and continuity into a product space explain why it is useful. The compactness result for this construction is Tychonoff's theorem.

A projection map from a Cartesian product selects specified coordinates. For the coordinate indexed by $j$,For example, $p_1(a,b)=a$ forgets the second component of an ordered pair.

The product topology makes each coordinate projection map a continuous map. In the two-factor case $X\times Y\to X$, surjectivity depends on whether the omitted factor $Y$ has an element, subject to the empty-set cases for Cartesian projections. For an arbitrary infinite family, simultaneous existence of the other coordinates can also involve the axiom of choice.

A projection is an idempotent self-map $p:X\to X$: applying it twice has the same effect as applying it once, $p\circ p=p$. Equivalently, it fixes every point of its image of a function. The map $p(x,y)=(x,0)$ on the real plane is an example.

A coordinate projection map $X\times Y\to X$ has different source and target, so it is not literally an idempotent self-map. After choosing $y_0\in Y$ and identifying $X$ with $X\times \{y_0\}$, its lifted form $(x,y)\mapsto (x,y_0)$ is a projection. A linear projection additionally respects vector addition and scalar multiplication.

For metric spaces $X,Y$ with $X$ compact, the projection with a compact factor is closed: if $F\subset X\times Y$ is a closed set and $y_n\in {\pi }_Y(F)$ converges to $y$, choose $x_n$ with $(x_n,y_n)\in F$. Sequential compactness supplies $x_{n_k}\to x$, so $(x_{n_k},y_{n_k})\to (x,y)\in F$, proving $y\in {\pi }_Y(F)$. The projection map is also continuous for the sum metric. Without compactness, the closed set $\{(x,y)\in {\mathbb{R}}^2:xy=1\}$ projects to the nonclosed set $\mathbb{R}\setminus \{0\}$.

The statement that every surjective function has a right inverse is equivalent to the axiom of choice.

Given a surjective function $f:X\to Y$, each fiber of a function $f^{-1}(\{y\})$ is nonempty. The axiom of choice selects $g(y)$ from that fiber for every $y\in Y$, so $f(g(y))=y$ and $g$ is a right inverse. When $Y$ is the empty set, the unique empty function supplies $g$ without a choice.

Conversely, take an indexed family $(X_i{)}_{i\in I}$ of nonempty sets. The projection map from the disjoint union $D=\{(i,x):i\in I,x\in X_i\}$ to $I$ is surjective. A right inverse assigns $i\mapsto (i,x_i)$, selecting one $x_i\in X_i$ for every index. This is a choice function, so the asserted splitting of all surjections gives the axiom of choice.

Here is a short proof that uses only the row space. For a vector subspace $W\subseteq {\mathbb{F}}^n$, let $d_j$ be the dimension of its projection map onto the first $j$ coordinates, with $d_0=0$. In any reduced row echelon form with row space $W$, $d_j$ counts the pivot columns among those first $j$ columns: later-pivot rows project to zero, while earlier-pivot rows remain linearly independent. Thus the positions where $d_j-d_{j-1}=1$ determine all pivot positions from $W$ alone.

If those positions are $p_1,\dots ,p_r$, the map sending $w\in W$ to $(w_{p_1},\dots ,w_{p_r})$ is an isomorphism $W\to {\mathbb{F}}^r$: the reduced nonzero rows map to the standard basis. Its $i$th row is therefore the unique vector of $W$ with pivot-coordinate values equal to the $i$th standard basis vector. Any other reduced row echelon form with that row space has identical nonzero rows in identical order. At a fixed matrix size, the remaining rows must be zero. Together with existence by Gaussian elimination, this proves uniqueness.

Given functions $f_i:Y\to X_i$, there is a unique function $g:Y\to X_1\times X_2$ whose coordinate projection maps are $f_1,f_2$. It is the displayed pointwise pairing. Equality of ordered pairs is equality of both coordinates, which proves uniqueness and makes the Cartesian product a categorical product in the Category of sets.