What Must Be Checked for an Open Map?
To decide whether a map is open, the definition seems to ask us to inspect every open subset of its domain. A basis reduces this potentially large task: if each basic open set has an open image, then every open set does. The reason is that every open set is a union of basis elements, and direct images preserve unions. The basis used for the test belongs to the domain; the image must be open in the codomain.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, and let \(f:X\to Y\) be a function. A family \(\mathcal{B}\subseteq\mathcal{T}_X\) is a basis for \(X\) if every open set in \(X\) is a union of members of \(\mathcal{B}\). Equivalently, every point of an open set \(U\) lies in some \(B\in\mathcal{B}\) with \(B\subseteq U\). The empty open set is the union of no basis elements.
This definition concerns images, not preimages. Continuity is tested using preimages of open sets in \(Y\); openness is tested using images of open sets in \(X\). Neither condition is part of the definition of the other.
The Basis Test
Proof. Suppose first that \(f\) is open. Every \(B\in\mathcal{B}\) is open in \(X\), so the definition of an open map gives that \(f(B)\) is open in \(Y\).
Conversely, suppose \(f(B)\) is open in \(Y\) for every \(B\in\mathcal{B}\). Let \(U\) be any open subset of \(X\). Since \(\mathcal{B}\) is a basis, there is a family \((B_i)_{i\in J}\) of basis elements such that \(U=\bigcup_{i\in J}B_i\). For any function, direct images preserve unions: a point \(y\) belongs to \(f(\bigcup_{i\in J}B_i)\) exactly when \(y=f(x)\) for some \(x\in B_i\) for some index \(i\); this is exactly the condition that \(y\in\bigcup_{i\in J}f(B_i)\). Therefore $$ f(U)=f\left(\bigcup_{i\in J}B_i\right)=\bigcup_{i\in J}f(B_i). $$ Each \(f(B_i)\) is open in \(Y\) by assumption, and an arbitrary union of open sets is open. Thus \(f(U)\) is open. This also covers \(U=\varnothing\), whose image is \(\varnothing\), an open set. Since \(U\) was arbitrary, \(f\) is open. \(\square\)
The theorem requires a basis for the domain. There is no requirement to choose a basis for \(Y\): the test is whether each set \(f(B)\) is open in the given topology \(\mathcal{T}_Y\). In practice, the theorem has a useful contrapositive: if one basis element has an image that is not open in \(Y\), then the map is not open.
Worked Example: An Affine Map of the Real Line
Give \(\mathbb{R}\) its standard topology and define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=3x+1\). Use the basis of open intervals to show that \(f\) is open.
The sets \((a,b)\), for real numbers \(a<b\), form a basis for the standard topology on \(\mathbb{R}\). Fix such an interval. If \(x\in(a,b)\), then \(a<x<b\). Multiplying by the positive number \(3\) and adding \(1\) gives \(3a+1<3x+1<3b+1\). Hence \(f(x)\in(3a+1,3b+1)\).
For the reverse inclusion, let \(y\in(3a+1,3b+1)\), and set \(x=(y-1)/3\). Dividing the inequalities by \(3>0\) gives \(a<x<b\), so \(x\in(a,b)\), and \(f(x)=3((y-1)/3)+1=y\). Thus the image is exactly $$ f((a,b))=(3a+1,3b+1). $$ This is an open interval in \(\mathbb{R}\), so every basis element has an open image. By the basis test, \(f\) is open.
Check: for the particular basis element \((2,5)\), the formula gives \(f((2,5))=(7,16)\). Indeed, \(3(2)+1=7\) and \(3(5)+1=16\); the endpoints are excluded because the original interval excludes \(2\) and \(5\).
Worked Example: The Squaring Map Is Not Open
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), with the standard topology on both copies of \(\mathbb{R}\). Show that \(f\) is not open by testing a basis element.
Use the open interval \((-1,1)\), which is a basis element. For every \(x\in(-1,1)\), \(0\leq x^2<1\), so \(f((-1,1))\subseteq[0,1)\). Conversely, if \(t\in[0,1)\), then \(\sqrt{t}\) is defined and \(0\leq\sqrt{t}<1\), so \(\sqrt{t}\in(-1,1)\) and \(f(\sqrt{t})=t\). Therefore $$ f((-1,1))=[0,1). $$ This image is not open in \(\mathbb{R}\): it contains \(0\), but every open neighborhood of \(0\) contains negative numbers and so cannot be contained in \([0,1)\). The image of this one basis element is not open, and the basis test shows that \(f\) is not an open map.
Check: \(0\) is included in the image because \(f(0)=0\), while \(1\) is excluded because \(x^2=1\) requires \(x=1\) or \(x=-1\), neither of which belongs to \((-1,1)\).
A Finite-Space Test
The same argument applies when the spaces are not metric spaces and their open sets are not intervals. For a finite example, it is important to specify both topologies: openness of an image is determined by the codomain topology.
Worked Example: A Map Between Finite Spaces
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}_X=\{\varnothing,\{a\},\{a,b\},X\}\), and let \(Y=\{0,1\}\) have topology \(\mathcal{T}_Y=\{\varnothing,\{1\},Y\}\). Define \(f:X\to Y\) by \(f(a)=1\), \(f(b)=0\), and \(f(c)=0\). Determine whether \(f\) is open.
Take \(\mathcal{B}=\{\{a\},\{a,b\},X\}\). Each member is open in \(X\). Every open set in \(\mathcal{T}_X\) is either empty or one of these three sets, so \(\mathcal{B}\) is a basis. Compute the image of each basis element: $$ f(\{a\})=\{1\},\qquad f(\{a,b\})=\{0,1\}=Y,\qquad f(X)=\{0,1\}=Y. $$ All three images are open in \(Y\), because \(\{1\}\) and \(Y\) belong to \(\mathcal{T}_Y\). Thus \(f\) is open.
Check: the four open subsets of \(X\) have images \(\varnothing,\{1\},Y,Y\), respectively. Each is indeed open in \(Y\), confirming the conclusion directly.
Why the Basis Test Works—and Where It Can Fail
The proof uses two precise facts: every open subset of the domain is a union of basis elements, and a function carries a union to the union of the images. It does not claim that images of open sets are always open. The squaring example shows why the test must actually be carried out: the image of a basic open interval can have a boundary point that prevents it from being open.
Common mistake. A tempting but incorrect calculation is \(f((-1,1))=(-1,1)\) for \(f(x)=x^2\), on the grounds that the input interval has endpoints \(-1\) and \(1\). This overlooks that squares cannot be negative. The membership calculation gives \(x^2\geq0\) and \(x^2<1\) for every \(x\in(-1,1)\); conversely every \(t\in[0,1)\) is the square of \(\sqrt{t}\in(-1,1)\). The correct image is \([0,1)\), not \((-1,1)\), and it is not open. Checking the image set itself, including whether its endpoints are included, is essential.
One useful consequence concerns composition. If \(f:X\to Y\) and \(g:Y\to Z\) are both open maps, then their composition is open.
Proof. Let \(U\) be open in \(X\). Since \(f\) is open, \(f(U)\) is open in \(Y\). Since \(g\) is open, \(g(f(U))\) is open in \(Z\). By the definition of composition, \(g(f(U))=(g\circ f)(U)\). Thus the image of every open \(U\subseteq X\) under \(g\circ f\) is open in \(Z\), so \(g\circ f\) is open. \(\square\)
The basis test is useful because it replaces a check over all open sets by one over a chosen family of generators. It gives an exact equivalence, not just a sufficient condition: if a map is open, every basis element has open image; if every basis element has open image, the map is open. A failure on even one basis element settles the question negatively.
Check Your Understanding
For each question, state the topologies involved and identify whether the test uses images or preimages.
- In the standard topology on \(\mathbb{R}\), use the interval basis to test the map \(f(x)=x+4\).
- For \(f(x)=|x|\) from \(\mathbb{R}\) to \(\mathbb{R}\), find the image of \((-1,1)\) and decide whether this alone proves that \(f\) is not open.
- In the basis test, why is it enough to check a basis for the domain rather than a basis for the codomain?
- Let \(X\) have the discrete topology and \(Y\) any topological space. Must every map \(f:X\to Y\) be open? Justify your answer using the basis test.
- Suppose every basis element of \(X\) has an open image under \(f:X\to Y\). Explain why an arbitrary union of those basis elements also has an open image.