What Makes a Map Open?
Continuity asks whether preimages of open sets are open. Openness asks a different question: do images of open sets remain open? The answer depends on the domain and codomain topologies, not just on the formula for the function. For instance, the image of an interval under squaring can be open in the half-line \([0,\infty)\) with its subspace topology, but not open in \(\mathbb{R}\).
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. A subset of either space is called open with respect to that space’s topology. In particular, for \(A\subseteq[0,\infty)\), openness in \([0,\infty)\) means that \(A=V\cap[0,\infty)\) for some open set \(V\subseteq\mathbb{R}\), where \([0,\infty)\) has the subspace topology inherited from the standard topology on \(\mathbb{R}\).
Being an open map is a property of a function together with the topologies on its domain and codomain. It is not the same condition as continuity: continuity tests preimages, while openness tests images. We will decide openness by considering arbitrary open sets, often by finding a suitable open neighborhood around each point in their images.
A Neighborhood Test for Open Maps
Proof. Suppose first that \(f\) is open. Given an open \(U\subseteq X\) and \(x\in U\), the set \(f(U)\) is open in \(Y\) by definition, and it contains \(f(x)\). Set \(V=f(U)\). Then \(V\) is open and \(f(x)\in V\subseteq f(U)\).
Conversely, suppose the stated neighborhood condition holds. Let \(U\subseteq X\) be open. For every \(y\in f(U)\), there is some \(x\in U\) with \(f(x)=y\). The assumed condition gives an open set \(V_y\subseteq Y\) with \(y\in V_y\subseteq f(U)\). Thus \(f(U)\) is the union of the open sets \(V_y\), as \(y\) ranges over \(f(U)\), and is open. If \(f(U)=\varnothing\), it is open by the topology axioms. Therefore \(f\) is open. \(\square\)
The test is useful because it reduces a set-wide question to local checks: for each point of an image, find an open neighborhood that stays inside that image. The neighborhood must be open in the codomain. Merely finding an interval around a point is not enough if that interval is not open in the codomain topology being used.
Squaring: The Codomain Changes the Answer
Worked Example: Squaring into the Half-Line
Consider \(q:\mathbb{R}\to[0,\infty)\), \(q(x)=x^2\), with the standard topology on \(\mathbb{R}\) and the subspace topology on \([0,\infty)\). We prove that \(q\) is open.
Let \(U\subseteq\mathbb{R}\) be open, and choose any \(y\in q(U)\). There is an \(x\in U\) such that \(y=x^2\). If \(x\ne0\), openness of \(U\) gives \(\delta>0\) with \((x-\delta,x+\delta)\subseteq U\). We can decrease \(\delta\), if needed, so that this interval does not contain \(0\). Squaring maps this interval onto the open interval whose endpoints are the squares of its endpoints, in increasing order. That interval contains \(x^2\) and lies in \(q(U)\), so it is an open neighborhood of \(y\) in \([0,\infty)\).
If \(x=0\), choose \(\delta>0\) with \((-\delta,\delta)\subseteq U\). Its image is \([0,\delta^2)\): zero is attained at \(0\), and every value \(t\) with \(0<t<\delta^2\) is attained at \(\sqrt{t}\in(-\delta,\delta)\). This image is open in \([0,\infty)\), since \([0,\delta^2)=(-1,\delta^2)\cap[0,\infty)\), and \((-1,\delta^2)\) is open in \(\mathbb{R}\). Thus in both cases \(y\) has an open neighborhood in the codomain contained in \(q(U)\). The pointwise test proves that \(q\) is open.
For a direct image calculation, take \(U=(-2,1)\). Its points have squares from zero up to, but not including, \(4\). Every \(t\in[0,4)\) occurs: \(t=0\) is the square of \(0\), and for \(0<t<4\), \(\sqrt{t}\in(0,2)\), while the negative square root works when \(\sqrt{t}\geq1\). More simply, for \(0<t<1\) the positive root is in \((0,1)\), and for \(1\leq t<4\) the negative root is in \((-2,-1]\). Hence \(q(U)=[0,4)\). This is open in the half-line because \([0,4)=(-1,4)\cap[0,\infty)\).
Check: \(0\) belongs to \(U\), so its square \(0\) must belong to \(q(U)\); the displayed image includes it. The value \(4\) is not attained because the only real numbers with square \(4\) are \(-2\) and \(2\), neither of which belongs to \((-2,1)\).
Here is the interval calculation behind the local argument, including the cases where an endpoint is zero. For an open interval \((a,b)\) with \(a<b\), if \(0<a<b\), its square image is \((a^2,b^2)\). If \(a=0<b\), its image is \((0,b^2)\), not \([0,b^2)\), because \(0\notin(0,b)\). If \(a<b<0\), its image is \((b^2,a^2)\). If \(a<b=0\), its image is \((0,a^2)\), again because \(0\) is excluded. If \(a<0<b\), its image is \([0,M)\), where \(M=\max(a^2,b^2)\): zero is attained, but the larger endpoint square is not. Each of these images is open in \([0,\infty)\), including the intervals of the form \((0,c)\), since \((0,c)=(-1,c)\cap[0,\infty)\) for \(c>0\).
Worked Example: The Same Formula with a Different Codomain
Now regard \(q\) as the function \(q:\mathbb{R}\to\mathbb{R}\), still given by \(q(x)=x^2\), with the standard topology on both copies of \(\mathbb{R}\). The open set \(U=(-1,1)\) has image \([0,1)\). This image is not open in \(\mathbb{R}\): every open interval containing \(0\) includes negative numbers, and no negative number lies in \([0,1)\). Thus this version of \(q\) is not an open map.
Check: zero is attained at \(x=0\), and every \(t\) with \(0<t<1\) is attained at \(x=\sqrt t\in(-1,1)\). The value \(1\) is not attained, since its only real square roots are \(-1\) and \(1\), both outside \(U\). So the image is exactly \([0,1)\), which is open in the half-line but not in the real line.
Other Standard Maps
Worked Example: The Cubing Map
Let \(c:\mathbb{R}\to\mathbb{R}\) be \(c(x)=x^3\), with the standard topology on both spaces. We show directly that it is open. Take an open set \(U\subseteq\mathbb{R}\) and \(x\in U\). Choose \(\delta>0\) such that \((x-\delta,x+\delta)\subseteq U\). The function \(t\mapsto t^3\) is strictly increasing: if \(s<t\), then \(t^3-s^3=(t-s)(t^2+ts+s^2)>0\). The second factor is positive for \(s<t\), since it equals \((t+s/2)^2+3s^2/4\) and can be zero only when \(s=t=0\), which contradicts \(s<t\). By continuity and the intermediate value theorem, the image of the interval is exactly \(((x-\delta)^3,(x+\delta)^3)\). This is an open neighborhood of \(x^3\) contained in \(c(U)\). The pointwise test proves that \(c\) is open.
For example, \(c((-1,2))=(-1,8)\): the endpoints map to \((-1)^3=-1\) and \(2^3=8\), and strict increase with continuity gives every value strictly between them, but neither endpoint is attained. The image is open in \(\mathbb{R}\), as required.
Check: \(c(0)=0\), and \(-1<0<8\), so the computed image contains the value of an interior point. The endpoint values are excluded because \(-1\) and \(2\) are not in the domain interval.
Worked Example: Identity Maps and Constant Maps
For any topological space \((X,\mathcal{T})\), the identity function \(\operatorname{id}_X:X\to X\) is open. If \(U\in\mathcal{T}\), then \(\operatorname{id}_X(U)=U\), which is open by the assumption \(U\in\mathcal{T}\). This argument applies to every space, including the empty space.
By contrast, the constant function \(k:\mathbb{R}\to\mathbb{R}\), \(k(x)=0\), is not open for the standard topology on \(\mathbb{R}\). The open set \((-1,1)\) has image \(\{0\}\). The singleton \(\{0\}\) is not open in \(\mathbb{R}\), because every open interval containing \(0\) contains points other than \(0\). Hence \(k\) fails the definition of an open map.
Check: every \(x\in(-1,1)\) maps to \(0\), so the constant-map image really is \(\{0\}\); it is not an interval around \(0\) and is not open in the standard real topology.
Composition and a Common Pitfall
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 \(Z\). The composition satisfies \((g\circ f)(U)=g(f(U))\), so the image of every open \(U\) is open in \(Z\). Therefore \(g\circ f\) is open. \(\square\)
This result gives a way to construct open maps from simpler ones, but both openness assumptions matter to this proof: the first makes the intermediate image open, and the second makes its image open in the final codomain.
Common pitfall. A frequent error is to calculate \(q((-1,1))=(0,1)\) for \(q(x)=x^2\). This drops the value \(0\), even though \(0\in(-1,1)\) and \(q(0)=0\). The correct image is \([0,1)\). The correction matters: \((0,1)\) is open in \([0,\infty)\), while the actual image \([0,1)\) is also open there, but neither set is the correct image calculation. In the standard topology on \(\mathbb{R}\), \([0,1)\) is not open. Always check whether endpoints are attained by points in the domain set, and state which topology is being used to judge openness.
One useful consequence of the definition is that if \(f:X\to Y\) is open, then \(f(X)\) is open in \(Y\), because \(X\) itself is open in every topological space. The converse does not follow from the definition: knowing that the full image is open says nothing by itself about the images of other open subsets. To establish openness, test every open subset, or use the pointwise test to provide neighborhoods for all its image points.
Check Your Understanding
For each question, keep the domain and codomain topologies explicit, and check whether endpoints are attained.
- For \(q:\mathbb{R}\to[0,\infty)\), compute \(q((-3,2))\) and decide whether it is open in the codomain.
- Is the squaring map \(q:\mathbb{R}\to\mathbb{R}\) open? Give an open set that tests your answer.
- Use the pointwise test to explain why the cubing map \(\mathbb{R}\to\mathbb{R}\) is open.
- Prove directly that the identity map \(\operatorname{id}_X:X\to X\) is open for an arbitrary topological space \(X\).
- For the constant map \(k:\mathbb{R}\to\mathbb{R}\), find an open set whose image is not open.