Images and Preimages Ask Different Questions
A continuous function cannot pull apart a connected set when we look at its image: continuous images of connected spaces are connected. It is tempting to reverse this statement and expect the preimage of a connected set to be connected as well. That reverse claim is false. We will calculate a preimage under the squaring map, then compare it with examples where the preimage is connected and where it is disconnected in a different way.
All subsets of a topological space below have the subspace topology. A subset \(A\) is connected if it cannot be written as the union of two disjoint, nonempty sets that are open in \(A\). Such a pair of sets is called a separation of \(A\). A set that is not connected is disconnected. In particular, “open in \(A\)” means open in the subspace topology, not necessarily open in the whole space.
To say that a subset is connected always refers to its own subspace topology. Thus, if \(C\subseteq Y\) is connected, the claim that fails is that \(f^{-1}(C)\), with its subspace topology from \(X\), must be connected. Continuity alone does not guarantee it.
Why Continuous Images Preserve Connectedness
Proof. Consider the restriction \(f|_A:A\to f(A)\). It is continuous: if \(U\) is open in \(f(A)\), then \(U=f(A)\cap V\) for some open set \(V\subseteq Y\), and \((f|_A)^{-1}(U)=A\cap f^{-1}(V)\), which is open in \(A\). Suppose, for a contradiction, that \(f(A)\) has a separation \(U,V\). Then \(U\) and \(V\) are disjoint, nonempty, open in \(f(A)\), and have union \(f(A)\). Their preimages under \(f|_A\) are open in \(A\), disjoint, and have union \(A\). Each preimage is nonempty because every point of \(f(A)\) is the image of some point of \(A\). They would therefore separate \(A\), contradicting its connectedness. Hence \(f(A)\) is connected. \(\square\)
The restriction in this proof matters. Sets open in the subspace \(f(A)\) need not be open in \(Y\), so continuity of \(f:X\to Y\) does not by itself say their preimages are open in \(X\). The restriction \(f|_A\) has codomain \(f(A)\), and continuity of that restriction gives exactly the relative openness needed in \(A\).
The Squaring Map and the Positive Real Axis
Worked Example: Complex Squaring
Give \(\mathbb{C}\) its usual Euclidean topology and define \(f:\mathbb{C}\to\mathbb{C}\) by \(f(z)=z^2\). Let \(C=(0,\infty)\), the strictly positive real axis, viewed as a subset of \(\mathbb{C}\). We will find \(f^{-1}(C)\) and prove it is disconnected.
First, \(f\) is continuous. For a fixed \(w\in\mathbb{C}\), if \(|z-w|<1\), then \(|z+w|\leq |z-w|+2|w|<1+2|w|\). Since \(z^2-w^2=(z-w)(z+w)\), choosing \(\delta=\min\{1,\varepsilon/(2|w|+1)\}\) gives \(|z-w|<\delta\) and hence \(|z^2-w^2|<\varepsilon\). This verifies continuity at every \(w\).
Write \(z=x+iy\), where \(x,y\in\mathbb{R}\). Direct multiplication gives \(z^2=(x^2-y^2)+2xy\,i\). For this to be a strictly positive real number, its imaginary part must be zero and its real part must be strictly positive. Thus \(2xy=0\) and \(x^2-y^2>0\). If \(x=0\), the latter inequality becomes \(-y^2>0\), which is impossible. If \(y=0\), it becomes \(x^2>0\), which holds exactly when \(x\ne0\). Therefore \(f^{-1}(C)=\mathbb{R}\setminus\{0\}\), where \(\mathbb{R}\) here denotes the real axis inside \(\mathbb{C}\).
To verify disconnectedness, write this preimage as \(P\cup Q\), where \(P=(-\infty,0)\) and \(Q=(0,\infty)\), both lying on the real axis. They are disjoint and nonempty, and their union is the preimage. They are open in the preimage: \(P\) is the intersection of the preimage with the open half-plane \(\{z\in\mathbb{C}:\operatorname{Re}z<0\}\), and \(Q\) is its intersection with the open half-plane \(\{z\in\mathbb{C}:\operatorname{Re}z>0\}\). So \(P,Q\) give a separation. The target \(C=(0,\infty)\) is an interval in \(\mathbb{R}\), hence connected by the connectedness of intervals.
Check: \(f(1)=1\) and \(f(-1)=1\), so both real rays do map into \(C\). In contrast, \(f(i)=i^2=-1\notin C\), and \(f(0)=0\notin C\). These values agree with the strict inequality defining \(C\) and with the computed preimage.
This example gives the central distinction: the connected set \(C\) is the image of the disconnected preimage, but the image theorem only says that the image of a connected set is connected. It does not say that every preimage of a connected set is connected. In fact, both \(\mathbb{C}\) and \(C\) are connected here, yet \(f^{-1}(C)\) is disconnected.
Changing the Set Changes the Preimage
Worked Example: Including Zero in the Ray
Keep \(f(z)=z^2\), but now take \(C=[0,\infty)\), including zero. The same calculation starts with \(z=x+iy\) and requires \(2xy=0\) and \(x^2-y^2\geq0\). When \(x=0\), the inequality is \(-y^2\geq0\), so \(y=0\). When \(y=0\), the inequality is \(x^2\geq0\), which holds for every real \(x\). Therefore \(f^{-1}(C)\) is the entire real axis, including the origin.
The real axis is connected: it is a copy of \(\mathbb{R}\), which is an interval and is connected. The calculation also checks directly at the boundary point: \(f(0)=0\in C\). Thus including zero changes the preimage from two rays to the whole real axis. The target \(C=[0,\infty)\) is itself an interval, so it is connected as well.
Check: For \(z=2+0i\), \(f(z)=4\in[0,\infty)\); for \(z=0\), \(f(z)=0\in[0,\infty)\); and for \(z=i\), \(f(z)=-1\notin[0,\infty)\). These values agree with the computed real-axis preimage.
Worked Example: A Singleton Target
Define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(x)=x^2\), with the standard topology on both copies of \(\mathbb{R}\), and take \(C=\{1\}\). The map is continuous because it is the real restriction of the continuous squaring map. A singleton is connected: it cannot be expressed as the union of two disjoint nonempty subsets. Solving \(g(x)=1\) gives \(x^2=1\), or \((x-1)(x+1)=0\), so \(g^{-1}(C)=\{-1,1\}\).
This two-point subspace is disconnected. Its subsets \(\{-1\}\) and \(\{1\}\) are disjoint, nonempty, and cover it. They are open in the subspace: \((-3/2,-1/2)\cap\{-1,1\}=\{-1\}\), while \((1/2,3/2)\cap\{-1,1\}=\{1\}\). Thus they form a separation.
Check: \(g(-1)=(-1)^2=1\) and \(g(1)=1^2=1\), so both points belong to the preimage. If \(x\in\mathbb{R}\) is in the preimage, \((x-1)(x+1)=0\) forces \(x=1\) or \(x=-1\), so there are no other points.
What Can Be Reversed?
The image theorem does give a valid statement in the other direction of reasoning, but it is not the false preimage claim. If \(f^{-1}(C)\) is connected and nonempty, then its image under \(f\) is connected by the theorem. That image is exactly \(C\cap f(X)\): a point is in \(f(f^{-1}(C))\) precisely when it belongs to \(C\) and is attained by \(f\). Consequently, if \(f\) is onto \(C\), connectedness of \(f^{-1}(C)\) implies connectedness of \(C\). Without surjectivity, this conclusion applies only to \(C\cap f(X)\); there may be points of \(C\) not attained by \(f\).
For the squaring example, \(f\) is not onto \(C=(0,\infty)\) when its domain is restricted to the real axis: it attains positive values, but that real-domain example still has two preimage rays. In the complex example \(f\) does attain every positive real number, since \(f(\sqrt{t})=t\) for each \(t>0\). The preimage is nevertheless disconnected. Surjectivity therefore does not make preimages of connected sets connected; it only makes the valid implication above work when the preimage is already known to be connected.
Common pitfall. A common incorrect argument says, “\(C\) is connected and \(f\) is continuous, so \(f^{-1}(C)\) is connected.” The image theorem has the opposite direction: it starts with a connected subset of the domain and concludes that its image is connected. For the real squaring map, an incorrect computation might also say \(x^2>0\) exactly when \(x>0\). The correct condition is \(x\ne0\), since both \(2^2=4>0\) and \((-2)^2=4>0\), while \(0^2=0\) is not strictly positive. Hence \(g^{-1}((0,\infty))=(-\infty,0)\cup(0,\infty)\), not just \((0,\infty)\).
Check Your Understanding
For each question, distinguish the topology on a subset from openness in its ambient space.
- For \(f(z)=z^2\) on \(\mathbb{C}\), compute \(f^{-1}((-\infty,0))\) and decide whether it is connected.
- For \(g(x)=x^2\) on \(\mathbb{R}\), compute \(g^{-1}(\{4\})\) and exhibit a separation if it is disconnected.
- Explain why continuity of \(f:X\to Y\) alone does not justify saying that the preimage of a set open in the subspace \(f(A)\) is open in \(X\).
- Suppose \(f^{-1}(C)\) is connected and nonempty. Which set is guaranteed to be connected: \(C\), or \(C\cap f(X)\)?
- For real squaring, compare the preimages of \((0,\infty)\) and \([0,\infty)\). Which one contains the origin, and why?