Continuity of Projections

Why Projections Are Continuous

A projection discards some coordinates of a point in a product space. In “Projections From Product Spaces,” we saw that projections are open maps: they send open subsets of the product to open subsets of the target. Continuity asks a different question. It concerns inverse images: if an open set in the target specifies conditions on retained coordinates, what conditions does its inverse image impose on the original product?

The answer follows from the basis for the product topology. A basic open set in a product restricts only finitely many coordinates. Pulling such a set back through a projection imposes those same restrictions on the corresponding coordinates of the original product, so its inverse image is open. This gives continuity even when the product has infinitely many factors or some factors are empty.

Definition: Let $X=\prod_{i\in I}X_i$ have the product topology, and let $J\subseteq I$. The projection onto the coordinates in $J$ is $p_J:X\to X_J=\prod_{j\in J}X_j$, defined by $p_J(x)=(x_j)_{j\in J}$. A map $f:X\to Y$ between topological spaces is continuous if the inverse image $f^{-1}(V)$ is open in $X$ for every open set $V\subseteq Y$.

The continuity criterion through open sets was established earlier in this course. We will use it to prove the projection result. For a basic open set in $X_J$, only finitely many retained coordinates are restricted. Its inverse image under $p_J$ is the basic open set in $X$ imposing those very restrictions. Coordinates outside that finite collection are unrestricted.

The Continuity Theorem

Theorem: Let $X=\prod_{i\in I}X_i$ have the product topology, let $J\subseteq I$, and give $X_J=\prod_{j\in J}X_j$ its product topology. Then the projection $p_J:X\to X_J$ is continuous.

Proof. By the basis description of the product topology, every open set $V\subseteq X_J$ is a union of basic open sets. A basic open set has the form $$ B(G,(U_j))=\{y\in X_J:y_j\in U_j\text{ for every }j\in G\}, $$ where $G\subseteq J$ is finite and each $U_j$ is open in $X_j$. For such a set, the definition of $p_J$ gives $$ p_J^{-1}\bigl(B(G,(U_j))\bigr) =\{x\in X:x_j\in U_j\text{ for every }j\in G\}. $$ Since $G$ is also a finite subset of $I$, the set on the right is basic open in $X$.

Now write $V$ as a union of basic open sets $B_\lambda$. Inverse images preserve unions, so $$ p_J^{-1}(V)=p_J^{-1}\left(\bigcup_\lambda B_\lambda\right) =\bigcup_\lambda p_J^{-1}(B_\lambda). $$ Each set in this union is open in $X$, and a union of open sets is open. Thus $p_J^{-1}(V)$ is open for every open $V\subseteq X_J$. By the continuity criterion, $p_J$ is continuous. $\square$

This argument covers two edge cases without any change. If $J=\varnothing$, then $X_J$ is the one-point product, and its only open sets are the empty set and the whole space; their inverse images are open. If the target product is empty, the domain product must also be empty for the projection to be defined as a map to that target. In that case the inverse-image condition still holds, since the only subset of the empty domain is open.

Worked Example: Projecting an Open Rectangle’s Preimage

Give $\mathbb{R}^3$ and $\mathbb{R}^2$ their product topologies, and let $p_{\{1,3\}}(x,y,z)=(x,z)$. Consider the open rectangle $V=(-2,1)\times(3,5)$ in the target. A point $(x,y,z)$ belongs to $p_{\{1,3\}}^{-1}(V)$ exactly when $x\in(-2,1)$ and $z\in(3,5)$. There is no condition on $y$. Therefore $$ p_{\{1,3\}}^{-1}(V)=(-2,1)\times\mathbb{R}\times(3,5). $$ This is open in $\mathbb{R}^3$, as expected. Notice that continuity is verified by taking an open set in the target and finding its inverse image, not by calculating the image of an open set in the domain.

Worked Example: A Basic Open Set in an Infinite Product

Let $X=\prod_{n=1}^{\infty}\mathbb{R}$, with the product topology, and let $J$ be the set of even positive integers. In $X_J$, consider the basic open set that requires the second coordinate to lie in $(0,2)$ and the sixth coordinate to lie in $(-1,1)$, with no other restrictions. Its inverse image under $p_J$ is $$ \{x\in X:x_2\in(0,2),\ x_6\in(-1,1)\}. $$ This is open in $X$: it restricts just two coordinates, even though the product has infinitely many factors. The coordinates not mentioned in the target basic set, including every odd coordinate, remain unrestricted in the inverse image. The calculation illustrates why finite-coordinate basic sets are the essential feature of the product topology.

Continuity of Maps Into a Product

The same basis argument yields a useful test for continuity into a product. Instead of checking every open set in the product directly, one can check each coordinate function. This is particularly effective when a map assigns infinitely many coordinates: every basic open set in the target constrains only finitely many of them.

Theorem: Let $Z$ be a topological space, let $X=\prod_{i\in I}X_i$ have the product topology, and let $f:Z\to X$. For each $i\in I$, define the coordinate map $f_i:Z\to X_i$ by $f_i=\pi_i\circ f$, where $\pi_i$ is the projection onto the $i$-th coordinate. Then $f$ is continuous if and only if every $f_i$ is continuous.

Proof. Suppose first that $f$ is continuous. Each coordinate projection $\pi_i:X\to X_i$ is a projection of the type just considered, so it is continuous. The composition theorem for continuous maps then shows that $f_i=\pi_i\circ f$ is continuous for each $i$.

Conversely, suppose every $f_i$ is continuous. Take a basic open set $B(F,(U_i))$ in $X$, where $F\subseteq I$ is finite and each $U_i$ is open in $X_i$. A point $z\in Z$ belongs to $f^{-1}(B(F,(U_i)))$ exactly when $f_i(z)\in U_i$ for every $i\in F$. Hence $$ f^{-1}(B(F,(U_i)))=\bigcap_{i\in F} f_i^{-1}(U_i). $$ Each $f_i^{-1}(U_i)$ is open in $Z$, and this is a finite intersection of open sets, so it is open. Every open set in $X$ is a union of basic open sets; its inverse image under $f$ is therefore a union of open sets and is open. The continuity criterion proves that $f$ is continuous. If $I=\varnothing$, then $X$ is a one-point space, every map $Z\to X$ is continuous, and the coordinate condition is vacuous. $\square$

Worked Example: Checking a Map Into the Plane Coordinate by Coordinate

Define $f:\mathbb{R}\to\mathbb{R}^2$ by $f(t)=(t,|t|)$, using the product topology on $\mathbb{R}^2$. Its coordinate maps are $f_1(t)=t$ and $f_2(t)=|t|$. The identity map $f_1$ is continuous. For $f_2$, the triangle inequality gives $$ \bigl||s|-|t|\bigr|\leq |s-t| $$ for all real $s,t$. Thus, for every $\varepsilon>0$, choosing $\delta=\varepsilon$ ensures that $|s-t|<\delta$ implies $\bigl||s|-|t|\bigr|<\varepsilon$. So $f_2$ is continuous. By the coordinatewise theorem, $f$ is continuous.

Now project $f(t)$ onto its second coordinate. The composition $p_{\{2\}}\circ f:\mathbb{R}\to\mathbb{R}$ is continuous by the composition theorem, and it equals $t\mapsto |t|$. This gives the same function by composing continuous maps, while the coordinatewise test establishes continuity of the map into the product in the first place.

Worked Example: A Projection Restricted to a Subspace

Let $A=\{(t,t^2):t\in\mathbb{R}\}\subseteq\mathbb{R}^2$, with the subspace topology, and define $q:A\to\mathbb{R}$ by $q(x,y)=x$. This is the restriction to $A$ of the first-coordinate projection $p_{\{1\}}:\mathbb{R}^2\to\mathbb{R}$. For an open set $U\subseteq\mathbb{R}$, $$ q^{-1}(U)=A\cap (U\times\mathbb{R}). $$ The set $U\times\mathbb{R}$ is open in $\mathbb{R}^2$, and its intersection with $A$ is open in the subspace topology on $A$. Therefore $q$ is continuous. This calculation is also an instance of the earlier theorem that the restriction of a continuous map to a subspace is continuous.

Continuity Is Not the Same as Openness

A common source of confusion is to treat continuity and openness as the same property. Continuity of $p_J$ says that inverse images of open target sets are open in the domain. Openness says that images of open domain sets are open in the target. The preceding tutorial established separately that projections from product spaces are open. The continuity proof here uses inverse images and does not follow merely from the fact that a projection is open.

There is a similar distinction concerning closed sets. Continuity guarantees that inverse images of closed target sets are closed, by the closed-set characterization of continuity. It does not guarantee that images of closed sets are closed. The earlier example of a closed hyperbola whose projection is $\mathbb{R}\setminus\{0\}$ shows why that direction cannot be assumed.

The basis calculation also explains why the product topology is the natural topology for this result. Basic open sets restrict finitely many coordinates, so their inverse images under coordinate projections remain open. A projection can discard arbitrarily many coordinates without making the inverse-image conditions more complicated: only the finitely many restrictions that occur in the chosen basic set need to be carried back.

Check Your Understanding

Use the inverse-image description and the product-basis results to answer the following questions.

What is the inverse image of a basic open set in $X_J$ under $p_J$, and why is it open in $X$?
Why does the proof of continuity still work when the index set $I$ is infinite?
State the coordinatewise criterion for a map into a product to be continuous.
For a projection $p_{\{1,3\}}:\mathbb{R}^3\to\mathbb{R}^2$, what conditions describe the inverse image of $U\times V$, where $U,V\subseteq\mathbb{R}$ are open?
Explain the difference between continuity of a projection and its property of being an open map.

What You'll Learn