Selecting Coordinates From a Product
A product space records a point by all of its coordinates. A projection selects some of those coordinates and discards the rest. In “Product Topology,” the coordinate map \(\pi_i\) was defined for each individual coordinate \(i\). The same idea applies when we retain any chosen collection of coordinates, and the finite-coordinate basis description lets us determine how such projections act on open sets.
A projection may fail to reach every point in its target if the discarded coordinates cannot be filled in. Its fibers describe all the original tuples that give the same retained coordinates. We will make these statements precise, then prove that projections are open in the product topology. Openness does not mean that projections preserve every set-theoretic property: in particular, a projection can send a closed set to a set that is not closed.
The coordinates that are discarded form the complementary index set \(K=I\setminus J\). We may regard a tuple in \(X\) as a pair consisting of its \(J\)-coordinates and its \(K\)-coordinates. This is a coordinate-by-coordinate identification of \(X\) with \(X_J\times X_K\), including when one of the index sets is empty. As usual, a product over an empty index set is a one-point space.
Range and Fibers
To determine whether a point \(y\in X_J\) is in the range of \(p_J\), we ask whether its coordinates can be completed by coordinates in \(X_K\). There are no further restrictions: every tuple in \(X\) is specified by its coordinates. The next result records both the range and the fiber of a projection.
Proof. First suppose that \(X_K\) is nonempty, and choose \(z\in X_K\). For each \(y\in X_J\), combine the coordinates of \(y\) on \(J\) with those of \(z\) on \(K\). Since \(J\) and \(K\) are disjoint and together equal \(I\), this gives a tuple \(x\in X\). By construction, \(p_J(x)=y\). Thus every \(y\in X_J\) is in the range.
Now fix \(y\in X_J\). Any \(x\in p_J^{-1}(\{y\})\) must satisfy \(x_j=y_j\) for every \(j\in J\), by the definition of \(p_J\). Conversely, every tuple with those coordinates projects to \(y\). Its remaining coordinates can be any member of \(X_K\). This proves the fiber description.
If \(X_K=\varnothing\), there is no tuple of coordinates on \(K\). Therefore there is no tuple in \(X\), so \(X=\varnothing\) and the range is empty. For the final statement, when \(X_J\neq\varnothing\), an empty range is not all of \(X_J\), whereas a nonempty \(X_K\) gives range \(X_J\) by the first part. Hence surjectivity is equivalent to \(X_K\neq\varnothing\). \(\square\)
The nonempty-target condition in the final sentence handles an edge case: a function with empty target is surjective when its domain is also empty. The range description covers that case without needing a separate convention about surjectivity. The theorem also emphasizes that the fiber over \(y\), when nonempty, has the same freedom in the discarded coordinates regardless of which \(y\) is chosen.
Worked Example: Projecting a Sphere onto a Coordinate Plane
Let \(S=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\}\), and project onto the first two coordinates. We claim that $$ p_{\{1,2\}}[S]=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}. $$ If \((x,y)\) is the projection of a point in \(S\), then there is a real \(z\) with \(x^2+y^2+z^2=1\). Since \(z^2\geq0\), this implies \(x^2+y^2\leq1\).
Conversely, suppose \(x^2+y^2\leq1\). The number \(1-x^2-y^2\) is nonnegative, so choose \(z=\sqrt{1-x^2-y^2}\). Then \(x^2+y^2+z^2=x^2+y^2+(1-x^2-y^2)=1\), and \((x,y,z)\in S\). Its projection is \((x,y)\). Thus the projection is the closed unit disk, including its boundary. Each point on the boundary has \(z=0\); each point strictly inside has two possible third coordinates, the positive and negative square roots.
Projections of Basic Open Sets
The product topology is generated by conditions on finitely many coordinates. When we project a basic open set, conditions on retained coordinates remain visible in the target. Conditions on discarded coordinates instead ask whether at least one choice of those coordinates is possible. This gives a direct way to calculate the image.
Proof. By the finite-coordinate basis theorem for the product topology, every open set \(O\subseteq X\) is a union of basic sets \(B(F,(U_i))\), where \(F\subseteq I\) is finite and each \(U_i\subseteq X_i\) is open. Because images preserve unions, \(p_J[O]\) is the union of the images of these basic sets. It is enough to show that each such image is open.
Fix one basic set \(B=B(F,(U_i))\). For each discarded coordinate \(i\in K=I\setminus J\), define \(V_i=U_i\) if \(i\in F\), and \(V_i=X_i\) if \(i\notin F\). A tuple of discarded coordinates can accompany a point of \(B\) exactly when it belongs to \(\prod_{i\in K}V_i\). If this product is empty, then \(B\) has no points and \(p_J[B]=\varnothing\), which is open.
Suppose instead that \(\prod_{i\in K}V_i\) is nonempty. A tuple in \(X_J\) is in \(p_J[B]\) exactly when its coordinates satisfy the basic-set conditions in \(F\cap J\). Indeed, those conditions are necessary; and, given a tuple satisfying them, any member of \(\prod_{i\in K}V_i\) supplies the discarded coordinates needed to form a point of \(B\). Therefore $$ p_J[B]=\{y\in X_J:y_j\in U_j\text{ for every }j\in F\cap J\}. $$ This is a basic open set in the product topology on \(X_J\); if \(F\cap J\) is empty, it is all of \(X_J\). In either case \(p_J[B]\) is open. Since \(p_J[O]\) is a union of such open images, it is open. \(\square\)
This proof also identifies why a basic-set image can sometimes be empty: a condition on a discarded coordinate may have no solutions, or an unrestricted discarded factor may itself be empty. When the discarded coordinates can be chosen, their restrictions disappear from the description of the image, while the retained-coordinate restrictions remain. No assumption that all factors are nonempty is needed for the theorem.
Worked Example: Projecting an Open Box
In \(\mathbb{R}^3\) with its product topology, let \(B=(-2,1)\times(3,5)\times(-1,4)\), and project onto the first and third coordinates. A point \((x,z)\) belongs to the image exactly when there is a \(y\in(3,5)\) such that \((x,y,z)\in B\). Since \((3,5)\) is nonempty, such a \(y\) always exists whenever \(x\in(-2,1)\) and \(z\in(-1,4)\). Thus $$ p_{\{1,3\}}[B]=(-2,1)\times(-1,4). $$ The condition on the discarded second coordinate is needed for the original box to have points, but it does not restrict the projected coordinates further.
Worked Example: A Closed Set with a Nonclosed Projection
Consider \(H=\{(x,y)\in\mathbb{R}^2:xy=1\}\). This set is closed in \(\mathbb{R}^2\). To check this directly, take \((a,b)\notin H\) and put \(c=|ab-1|>0\). Choose \(\delta=\min\{1,c/[4(|a|+|b|+1)]\}>0\). If \(|x-a|<\delta\) and \(|y-b|<\delta\), then \(|x|\leq |a|+1\), and $$ |xy-ab|\leq |x||y-b|+|b||x-a|<\delta(|a|+|b|+1)\leq c/4. $$ It follows that \(|xy-1|\geq |ab-1|-|xy-ab|>3c/4>0\). The open rectangle around \((a,b)\) therefore misses \(H\), so the complement of \(H\) is open.
Nevertheless, its projection onto the first coordinate is \(p_{\{1\}}[H]=\mathbb{R}\setminus\{0\}\). No point of \(H\) has first coordinate zero. For every \(x\neq0\), taking \(y=1/x\) gives \((x,y)\in H\). The projected set is not closed in \(\mathbb{R}\), since it contains \(1/n\) for every positive integer \(n\), these numbers converge to \(0\), and \(0\) is not in the set. Thus openness of projections does not imply preservation of closed sets.
What Projections Preserve—and What They Do Not
A projection forgets information, so its image describes which retained coordinates occur at least once among the points of a set. In a product \(X_J\times X_K\), this means that \(y\in p_J[A]\) exactly when there is some \(z\in X_K\) such that the combined tuple \((y,z)\) belongs to \(A\). This “there exists a choice of the other coordinates” interpretation is often the most useful way to calculate a projection.
The open-map theorem concerns images of open sets in the product topology. It does not say that every image is open, or that arbitrary sets inherit their original properties after projection. The hyperbola example shows that even a closed set can have a nonclosed image. Nor should a projection be assumed surjective without checking the discarded factors: if they cannot supply a tuple, the domain may be empty. Keeping track of retained coordinates, discarded coordinates, and the existence of completions prevents these common errors.
Check Your Understanding
Use the range, fiber, and open-map descriptions of projections to answer the following questions.
- For \(p_J:\prod_{i\in I}X_i\to\prod_{j\in J}X_j\), what must be possible in the discarded coordinates for a given target point to lie in the range?
- Describe the fiber over \(y\in X_J\) when the complementary product \(X_{I\setminus J}\) is nonempty.
- Why does a basic open set project to either the empty set or a basic open set in the target?
- What is the projection of the unit sphere in \(\mathbb{R}^3\) onto its first two coordinates, and why are all points in that disk attained?
- Why does the fact that projections are open not imply that they send closed sets to closed sets?