Why Coordinates Determine a Product-Valued Map
A point of a product space is a tuple: it has one coordinate in each factor. Consequently, a function whose outputs lie in a product can be examined one coordinate at a time. This gives two related ways to describe such a function. Starting with the function, we can extract its coordinate functions; starting with coordinate functions, we can assemble a product-valued function. The key fact is that the assembly is always unique.
Let \((X,\mathcal{T})\) be a topological space, let \(J\) be an index set, and, for each \(i\in J\), let \((X_i,\mathcal{T}_i)\) be a topological space. We give \(\prod_{i\in J}X_i\) the product topology. Its underlying set consists of tuples \((x_i)_{i\in J}\) with \(x_i\in X_i\) for every \(i\). The projection \(\pi_i:\prod_{j\in J}X_j\to X_i\) sends a tuple to its \(i\)-th coordinate. The arguments about coordinates below concern functions of sets; by themselves, they make no assertion about continuity.
The order of operations matters in this definition: first evaluate \(f\), obtaining a tuple, and then project to its \(i\)-th entry. Thus \(f_i(x)=\pi_i(f(x))\). The full tuple \(f(x)\) contains all these entries, not just one of them.
The Coordinate Factorization Theorem
Proof. First suppose \(f:X\to\prod_{i\in J}X_i\) is given. For any \(x\in X\), \(f(x)\) is a tuple in the product, and its \(i\)-th coordinate is \(\pi_i(f(x))=f_i(x)\). A tuple in a product is specified by all its coordinates, so \(f(x)=(f_i(x))_{i\in J}\). In particular, any two functions into the product with the same coordinate maps have the same value at every \(x\), and hence are equal.
Now suppose functions \(f_i:X\to X_i\) are given for every \(i\in J\). For each \(x\in X\), the assignment \(i\mapsto f_i(x)\) has a value in \(X_i\) for every \(i\in J\), so it defines a tuple in \(\prod_{i\in J}X_i\). Define \(f(x)=(f_i(x))_{i\in J}\). For every \(i\in J\), \(\pi_i(f(x))=f_i(x)\), so the coordinate maps of \(f\) are the prescribed functions. If \(g:X\to\prod_{i\in J}X_i\) has those same coordinate maps, then for every \(x\in X\) all coordinates of \(g(x)\) equal the corresponding coordinates of \(f(x)\). Thus \(g(x)=f(x)\) for every \(x\), and \(g=f\). If \(X\) is empty, the pointwise definition has no inputs to evaluate and still defines the unique function from \(X\); if \(J\) is empty, the product has just the empty tuple and the same construction gives the unique function into it. \(\square\)
The theorem is a statement about functions and tuples. It does not require any topological hypothesis on the sets, and it does not require the functions to be continuous. When the sets are topological spaces, the same set-theoretic conclusion applies to the underlying functions.
Proof. If \(f=g\), composing both sides with \(\pi_i\) gives \(\pi_i\circ f=\pi_i\circ g\) for every \(i\). Conversely, if those coordinate maps agree, then for each \(x\in X\), the tuples \(f(x)\) and \(g(x)\) have equal \(i\)-th coordinates for every \(i\in J\). Tuples with all coordinates equal are equal, so \(f(x)=g(x)\) for every \(x\), and therefore \(f=g\). When \(J\) is empty, the product has one point, so any two functions from \(X\) into it are equal; the condition on all coordinates is vacuous and the conclusion still holds. \(\square\)
Building Maps into Products
Worked Example: Constructing a Map into a Pair
Give \(\mathbb{R}\) its standard topology and give \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\) the product topology. Suppose the two coordinate functions are \(f_1:\mathbb{R}\to\mathbb{R}\), \(f_1(x)=2x+1\), and \(f_2:\mathbb{R}\to\mathbb{R}\), \(f_2(x)=x^2\). Construct the unique function \(f:\mathbb{R}\to\mathbb{R}^2\) with those coordinate maps.
By the theorem, the value at \(x\) must be the ordered pair of coordinate values: $$ f(x)=(f_1(x),f_2(x))=(2x+1,x^2). $$ For example, at \(x=-1\), the first coordinate is \(2(-1)+1=-2+1=-1\), and the second is \((-1)^2=1\). Thus \(f(-1)=(-1,1)\). At \(x=2\), the first coordinate is \(2(2)+1=4+1=5\), and the second is \(2^2=4\). Thus \(f(2)=(5,4)\).
Check: applying the projections gives \(\pi_1(f(x))=\pi_1(2x+1,x^2)=2x+1=f_1(x)\) and \(\pi_2(f(x))=\pi_2(2x+1,x^2)=x^2=f_2(x)\). In particular, \(\pi_1(f(2))=5\) and \(\pi_2(f(2))=4\), as required.
Worked Example: Constructing a Map into a Triple
Give \(\mathbb{R}^2\) and \(\mathbb{R}^3\) their standard product topologies. For \((x,y)\in\mathbb{R}^2\), specify three coordinate functions by \(g_1(x,y)=x+y\), \(g_2(x,y)=xy\), and \(g_3(x,y)=x-y\). The unique map with these coordinates is $$ g(x,y)=(x+y,xy,x-y). $$ At the input \((1,-2)\), the first coordinate is \(1+(-2)=-1\), the second is \(1\cdot(-2)=-2\), and the third is \(1-(-2)=3\). Therefore \(g(1,-2)=(-1,-2,3)\).
Check: the three projections recover the specified values: \(\pi_1(g(1,-2))=-1=1+(-2)\), \(\pi_2(g(1,-2))=-2=1\cdot(-2)\), and \(\pi_3(g(1,-2))=3=1-(-2)\). This confirms the value using each coordinate separately.
Infinite Tuples and Coordinate-by-Coordinate Equality
For an infinite product, an output is still one tuple with one entry for every index. The number of coordinates changes, but the factorization rule does not: prescribe a function for each index, and at each input collect all the prescribed values into a tuple. The construction makes sense only when the \(i\)-th value lies in the \(i\)-th factor.
Worked Example: A Function into an Infinite Product
Give \(I=[0,1]\) the subspace topology from the standard real line. Give \([0,1]^{\mathbb{N}}=\prod_{n\in\mathbb{N}}[0,1]\) the product topology. For each \(n\in\mathbb{N}\), define \(h_n:I\to[0,1]\) by \(h_n(t)=t^n\). These really do take values in \([0,1]\): if \(0\le t\le1\) and \(n\ge1\), then \(0\le t^n\le1\). The coordinate factorization theorem therefore gives $$ h:I\longrightarrow[0,1]^{\mathbb{N}},\qquad h(t)=(t^n)_{n\in\mathbb{N}}. $$ At \(t=\frac12\), the first three coordinates are \((\frac12)^1=\frac12\), \((\frac12)^2=\frac14\), and \((\frac12)^3=\frac18\). Hence $$ h\left(\frac12\right)=\left(\frac12,\frac14,\frac18,\frac1{16},\ldots\right), $$ where the entry at index \(n\) is \((\frac12)^n\).
Check: for every \(n\in\mathbb{N}\), \(\pi_n(h(t))=t^n=h_n(t)\), which verifies the coordinate formula at every index, not just the first three. At the endpoint \(t=0\), since \(n\ge1\), each coordinate is \(0^n=0\), so \(h(0)=(0,0,\ldots)\). At \(t=1\), each coordinate is \(1^n=1\), so \(h(1)=(1,1,\ldots)\). Both endpoint tuples belong to \([0,1]^\mathbb{N}\), as the construction requires.
The infinite example also illustrates why checking only a few coordinates is not enough to prove two arbitrary elements of an infinite product are equal. To establish equality, one must verify equality at every index. In a formula such as \(h(t)=(t^n)_{n\in\mathbb{N}}\), the index \(n\) specifies which coordinate is being described; it is not a claim that a finite initial list of entries determines the rest.
A Common Pitfall: Swapping Coordinates
When constructing a map into a product, the order of its coordinates must match the order of its factors. For instance, let \(u_1(x)=x+1\) and \(u_2(x)=2x\), and construct \(u:\mathbb{R}\to\mathbb{R}^2\). The correct formula is \(u(x)=(x+1,2x)\). A common wrong calculation is to write \(u(3)=(6,4)\), putting the second coordinate first. The correction is \(u_1(3)=3+1=4\) and \(u_2(3)=2\cdot3=6\), so \(u(3)=(4,6)\). Checking the projections exposes the error: \(\pi_1(6,4)=6\ne4=u_1(3)\), whereas \(\pi_1(4,6)=4\) and \(\pi_2(4,6)=6\).
The central practical rule is therefore precise: write down the coordinates in the same indexed order as the factors, and verify the result by applying each projection. The coordinate factorization theorem gives existence and uniqueness of the resulting function, but it does not settle whether that function is continuous. Continuity is an additional property that requires its own hypotheses and argument.
Check Your Understanding
In each question, use the coordinate maps and the defining property of tuples.
- Let \(a(x)=x-2\) and \(b(x)=x^2+1\) for \(x\in\mathbb{R}\). Write the unique map \(\mathbb{R}\to\mathbb{R}^2\) with coordinates \(a,b\), and evaluate it at \(x=3\).
- Suppose \(f,g:X\to Y\times Z\) and \(\pi_1\circ f=\pi_1\circ g\), while \(\pi_2\circ f\ne\pi_2\circ g\). Can \(f=g\)? Explain using a point where the second coordinate maps differ.
- For \(t\in[0,1]\), define \(k_n(t)=t^{n+1}\) for \(n\in\mathbb{N}\). Write the corresponding map into \([0,1]^\mathbb{N}\) and list its first three coordinates at \(t=\frac12\).
- Explain why prescribing one function \(f_i:X\to X_i\) for every index \(i\in J\) is enough to define a function into \(\prod_{i\in J}X_i\).
- If two maps into an infinite product agree on the first ten coordinates, what additional information is needed to conclude that the maps are equal?