How Connectedness Behaves Under a Map
The previous tutorial constructed connected components as maximal connected subsets and showed that they partition a metric space. A continuous map can change how those pieces fit together: it may send different components to the same set, or send a component to only part of a larger connected piece in the target. The useful first step is to track where each component can go.
We will use the theorem from “Connectedness in Metric Spaces” that continuous images of connected spaces are connected. We will also use the Construction of Components Theorem: a component contains every connected subset of the space that contains any one of its points. These results let us describe the relation between components without repeating the proof that continuous images preserve connectedness.
Proof. The component \(C_X(x)\) is connected. By the theorem on continuous images of connected spaces, \(f[C_X(x)]\) is connected in \(Y\). Also, \(x\in C_X(x)\), so \(f(x)\in f[C_X(x)]\). The Construction of Components Theorem, applied in \(Y\), says that every connected subset containing \(f(x)\) is contained in \(C_Y(f(x))\). Therefore \(f[C_X(x)]\subseteq C_Y(f(x))\). \(\square\)
The conclusion is an inclusion, not necessarily equality. The image of a component is connected and lies in one target component, but it need not fill that component. Moreover, the target component may receive images from several different components of \(X\).
Worked Example: Two Components Join in the Image
Let \(X=[0,1]\cup[2,3]\), with the metric inherited from \(\mathbb{R}\), and define \(f:X\to\mathbb{R}\) by $$ f(x)= \begin{cases} x, & x\in[0,1],\\ x-1, & x\in[2,3]. \end{cases} $$
The two intervals are separated in \(X\), and each is connected. They are the components of \(X\): a connected subset of their union cannot meet both sides of this separation. The function is continuous on each interval, and each interval is open in the subspace \(X\). Since these two open sets cover \(X\), the restrictions combine to give a continuous function on \(X\).
The image of the first component is \(f[[0,1]]=[0,1]\). The image of the second is \(f[[2,3]]=[1,2]\), since subtracting \(1\) sends the endpoints \(2,3\) to \(1,2\). Hence $$ f[X]=[0,1]\cup[1,2]=[0,2]. $$ The target space \(f[X]\) is connected, so it has just one component, namely \([0,2]\). Each component image is a proper subset of that target component, and the two images meet at \(1\). Thus distinct components in the domain can contribute pieces that join in the image.
When Component Structure Is Preserved
A homeomorphism preserves more than connectedness: it preserves which connected subsets are maximal. This follows by applying the component inclusion theorem in both directions. The reverse direction matters because a larger connected set in the target must also be carried back to a connected set in the domain.
Proof. Since \(f\) is continuous, the Continuous Maps Send Components into Components Theorem gives \(f[C_X(x)]\subseteq C_Y(f(x))\).
For the opposite inclusion, \(f^{-1}:Y\to X\) is continuous, and \(C_Y(f(x))\) is connected. Therefore \(f^{-1}[C_Y(f(x))]\) is connected. It contains \(x\), because \(f(x)\in C_Y(f(x))\). The Construction of Components Theorem in \(X\) now gives \(f^{-1}[C_Y(f(x))]\subseteq C_X(x)\). Applying \(f\) to this inclusion yields \(C_Y(f(x))\subseteq f[C_X(x)]\). The two inclusions prove equality.
Every component of \(X\) has the form \(C_X(x)\) for some \(x\), so its image is a component of \(Y\). Conversely, if \(D\) is a component of \(Y\), choose \(y\in D\) and write \(y=f(x)\), which is possible because \(f\) is onto. Then the equality just proved gives \(D=f[C_X(x)]\). Finally, distinct components of \(X\) have distinct images: if two images were equal, injectivity of \(f\) would make the original components equal. Thus the correspondence is one-to-one and onto. \(\square\)
This result applies to any homeomorphism, including a change of coordinates that alters distances but preserves the open sets. It also applies to a continuous map that is a homeomorphism onto its image, when the image is given the subspace metric. In that case, components correspond between the domain and the image subspace; the image subspace need not be a component of some larger ambient space.
Worked Example: A Homeomorphism Preserves Two Components
Let \(X=[-2,-1]\cup[1,2]\), and define \(f:X\to Y\) by \(f(x)=x^3\), where \(Y=[-8,-1]\cup[1,8]\). The function is continuous and strictly increasing on \(X\), and its inverse on \(Y\) is \(f^{-1}(y)=\sqrt[3]{y}\), which is continuous. Thus \(f\) is a homeomorphism from \(X\) onto \(Y\).
The components of \(X\) are \([-2,-1]\) and \([1,2]\), while the components of \(Y\) are \([-8,-1]\) and \([1,8]\). Direct calculation gives \(f[-2,-1]=[-8,-1]\) and \(f[1,2]=[1,8]\). The components correspond exactly, as the theorem predicts. The gaps between the intervals are part of the respective spaces’ component structure: neither homeomorphism nor its inverse can turn the two components into one.
Connected Images and Intermediate Values
A particularly useful consequence concerns real-valued functions. Earlier in the course, we established that a subset of \(\mathbb{R}\) is connected if and only if it is an interval. Combining that characterization with the theorem on continuous images gives an intermediate value principle even when the domain is not an interval.
Proof. Since \(X\) is connected and \(f\) is continuous, \(f[X]\) is connected in \(\mathbb{R}\). By the theorem that connected subsets of \(\mathbb{R}\) are intervals, \(f[X]\) contains every real number between any two of its elements. Both \(f(x_1)\) and \(f(x_2)\) belong to \(f[X]\), so \(r\in f[X]\). By the definition of image, there is \(c\in X\) with \(f(c)=r\). \(\square\)
When \(f(x_1)=f(x_2)\), the only number lying between those two values is their common value, and either \(x_1\) or \(x_2\) supplies the required point. Thus the argument covers this case as well as the case of distinct endpoint values.
Worked Example: Intermediate Values on a Circle
Consider the unit circle \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\), and the function \(g:S^1\to\mathbb{R}\) given by \(g(u,v)=u\). The circle is connected: the continuous map \(t\mapsto(\cos t,\sin t)\) from the interval \([0,2\pi]\) has image \(S^1\), so the theorem on continuous images of connected spaces applies. The coordinate function \(g\) is continuous.
At the points \((-1,0)\) and \((1,0)\), the values of \(g\) are \(-1\) and \(1\), respectively. The intermediate value theorem for connected domains therefore says that every \(r\in[-1,1]\) is \(g(u,v)\) for some point \((u,v)\in S^1\). In this example, one can verify the conclusion directly: for each such \(r\), the point \((r,\sqrt{1-r^2})\) lies on \(S^1\), because \(r^2+(\sqrt{1-r^2})^2=1\), and its first coordinate is \(r\). The connected-image argument gives the interval of values without requiring this explicit formula.
Worked Example: A Continuous Function on a Disconnected Domain
Let \(X=[0,1]\cup[3,4]\) and define \(h:X\to\mathbb{R}\) by \(h(x)=x\). The domain is disconnected, with components \([0,1]\) and \([3,4]\), and the function is continuous. Its image is \(h[X]=[0,1]\cup[3,4]\), which is not an interval: for example, \(2\) lies between \(1\) and \(3\) but does not belong to the image.
There is no contradiction with the Intermediate Value Property on a Connected Domain, because its domain hypothesis fails. Continuity alone does not ensure that every value between two function values is attained. In particular, choosing \(x_1=1\) and \(x_2=3\) gives function values \(1\) and \(3\), while \(h\) never takes the value \(2\) on \(X\).
What the Image Can and Cannot Tell Us
Connected images are useful because they transfer a global property of the domain into a restriction on the range. For a continuous real-valued function on a connected space, the range cannot have a gap. For a general metric-space target, the same principle says the image lies in one connected component when the domain itself is connected.
The converse claims require care. A connected image does not show that the domain is connected: in the first worked example, a disconnected domain maps onto the connected interval \([0,2]\). Nor does the image of a component have to be a whole component, as that same example demonstrates. Equality is guaranteed for homeomorphisms because both the map and its inverse preserve connectedness; continuity in only the forward direction gives the component inclusion, not the reverse inclusion.
A reliable way to use these results is to identify the set being mapped, check continuity on that set, and then apply the connected-image theorem. If the goal is to identify a component rather than merely prove connectedness, also check maximality or use a homeomorphism to transfer the component structure. Keeping these conclusions distinct prevents connectedness of an image from being mistaken for connectedness of its domain or for equality with a target component.
Check Your Understanding
Use the component and connected-image results to answer the following questions.
- Why must the image of a component under a continuous map lie inside one component of the target?
- In the first worked example, which two component images meet, and what is their union?
- Where is continuity of the inverse used in the proof that a homeomorphism maps components onto components?
- Why does a continuous real-valued function on a connected domain attain every value between two values in its range?
- What hypothesis fails when a continuous function on a disconnected domain misses a value between two of its function values?