Paths After Applying a Continuous Map
A path in a space is a continuous map from the interval \(I=[0,1]\). If a continuous map is applied to every point of a path, the result is again a path: its endpoints become the images of the original endpoints. This simple operation gives a useful way to prove that an image is path-connected without constructing paths in the image from scratch.
Throughout, \(I=[0,1]\) has its subspace topology from \(\mathbb{R}\). A subset of a topological space has the subspace topology: its open sets are intersections of that subset with open sets of the larger space. When we speak of the image \(f(X)\) as a space, it has the subspace topology inherited from the codomain \(Y\).
Continuity here is topological: a map is continuous if the preimage of every open set is open. We first record why composing continuous maps gives a continuous map.
Proof. Let \(W\) be any open subset of \(Z\). Since \(g\) is continuous, \(g^{-1}(W)\) is open in \(Y\). Since \(f\) is continuous, \(f^{-1}(g^{-1}(W))\) is open in \(X\). The definition of composition gives \((g\circ f)^{-1}(W)=f^{-1}(g^{-1}(W))\). Thus the preimage under \(g\circ f\) of every open set \(W\) is open, so \(g\circ f\) is continuous. \(\square\)
The Continuous-Image Theorem
Proof. If \(X=\varnothing\), then \(f(X)=\varnothing\), which is path-connected under the stated convention. Now suppose \(X\ne\varnothing\), and choose any two points \(y_0,y_1\in f(X)\). By the definition of image, there are \(x_0,x_1\in X\) such that \(f(x_0)=y_0\) and \(f(x_1)=y_1\). Since \(X\) is path-connected, there is a continuous path \(\alpha:I\to X\) with \(\alpha(0)=x_0\) and \(\alpha(1)=x_1\). The composition \(\beta=f\circ\alpha:I\to Y\) is continuous by the proposition, and its values lie in \(f(X)\).
To check continuity when the codomain is specifically \(f(X)\), let \(O\) be open in \(f(X)\). By the subspace topology, \(O=V\cap f(X)\) for some open \(V\subseteq Y\). Since every value of \(\beta\) lies in \(f(X)\), \(\beta^{-1}(O)=\beta^{-1}(V)\), which is open in \(I\) because \(\beta:I\to Y\) is continuous. Hence \(\beta:I\to f(X)\) is continuous. Finally, \(\beta(0)=f(x_0)=y_0\) and \(\beta(1)=f(x_1)=y_1\). We have constructed a path in \(f(X)\) between arbitrary points \(y_0,y_1\), proving that \(f(X)\) is path-connected. \(\square\)
The theorem concerns the image, not necessarily all of \(Y\). It requires neither that \(f\) be one-to-one nor that it be onto \(Y\). A continuous map may identify different points, or its image may be a proper subset of its codomain; the paths in \(X\) still give paths between points of the image.
Worked Example: A Linear Map from the Plane
Give \(\mathbb{R}^2\) and \(\mathbb{R}\) their standard topologies, and define \(f:\mathbb{R}^2\to\mathbb{R}\) by \(f(x,y)=x+y\). The map is continuous because it is a sum of coordinate functions, which are continuous in the standard Euclidean topologies. The plane is path-connected: for \(p,q\in\mathbb{R}^2\), the formula \(\alpha(t)=(1-t)p+tq\) defines a continuous path with endpoints \(p\) and \(q\). The theorem therefore implies that \(f(\mathbb{R}^2)\) is path-connected.
In fact, \(f(\mathbb{R}^2)=\mathbb{R}\): for any \(r\in\mathbb{R}\), \(f(r,0)=r+0=r\). To see concretely how paths are carried across, take \(p=(1,-2)\) and \(q=(3,4)\). The line segment between them is \(\alpha(t)=(1+2t,-2+6t)\), so
At \(t=0\) this gives \(-1=f(p)\); at \(t=1\) it gives \(7=f(q)\); and at \(t=\tfrac12\) it gives \(-1+8(\tfrac12)=3\). The image path lies in \(f(\mathbb{R}^2)=\mathbb{R}\). As a check, the formula \(f\circ\alpha\) is exactly the linear path \((1-t)(-1)+t(7)=-1+8t\) between the two image endpoints.
Worked Example: The Square Map and a Proper Image
Give \(\mathbb{R}\) its standard topology and define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x^2\). This map is continuous, and \(\mathbb{R}\) is path-connected: between \(a\) and \(b\), the path \(\alpha(t)=(1-t)a+tb\) is continuous and has those endpoints. Thus \(f(\mathbb{R})\) is path-connected.
Its image is exactly \([0,\infty)\). Every square is nonnegative, so \(f(\mathbb{R})\subseteq[0,\infty)\). Conversely, if \(y\ge0\), then \(x=\sqrt{y}\in\mathbb{R}\) satisfies \(f(x)=(\sqrt{y})^2=y\). Hence \([0,\infty)\subseteq f(\mathbb{R})\), proving equality. This is a proper subset of the codomain \(\mathbb{R}\), so the theorem asserts path-connectedness of the image, not that the image must be the whole codomain.
For a numerical path check, take \(a=-2\) and \(b=1\). Then \(\alpha(t)=-2+3t\), and the path in the image is \((f\circ\alpha)(t)=(-2+3t)^2=4-12t+9t^2\). Its endpoints are \(4=f(-2)\) and \(1=f(1)\); at \(t=\tfrac12\), its value is \((-\tfrac12)^2=\tfrac14\), which is nonnegative. Substitution into the expanded expression gives \(4-12(\tfrac12)+9(\tfrac12)^2=4-6+\tfrac94=\tfrac14\), confirming the calculation.
A Parametrized Circle
The theorem also applies when the image is a familiar subspace of a higher-dimensional space. Give \(I\) its subspace topology from \(\mathbb{R}\), and give \(\mathbb{R}^2\) its standard topology. The interval \(I\) is path-connected: for \(s,t\in I\), the map \(u\mapsto(1-u)s+ut\), for \(u\in I\), is continuous, takes values between \(s\) and \(t\), and has endpoints \(s,t\).
Worked Example: A Continuous Parametrization of the Circle
Define \(\gamma:I\to\mathbb{R}^2\) by \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\). The trigonometric coordinate functions are continuous, so \(\gamma\) is continuous. Its image is \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\): the identity \(\cos^2\theta+\sin^2\theta=1\) shows that every value of \(\gamma\) lies on \(S^1\), and the standard parametrization of the unit circle shows that every point of \(S^1\) is attained for some \(t\in[0,1]\). The theorem now gives that \(S^1\), with its subspace topology from \(\mathbb{R}^2\), is path-connected.
For specific values, \(\gamma(0)=(1,0)\), since \(\cos 0=1\) and \(\sin 0=0\). Also, \(\gamma(\tfrac14)=(0,1)\), since \(2\pi(\tfrac14)=\tfrac\pi2\), \(\cos(\tfrac\pi2)=0\), and \(\sin(\tfrac\pi2)=1\). Similarly, \(\gamma(\tfrac12)=(-1,0)\), \(\gamma(\tfrac34)=(0,-1)\), and \(\gamma(1)=(1,0)\), using the corresponding values at \(\pi\), \(3\pi/2\), and \(2\pi\). These checks confirm both that the parametrization stays on the circle and that it returns to its starting point.
For an explicit path between any two points of \(S^1\), choose parameters \(s,t\in I\) that represent them. The map \(u\mapsto\gamma((1-u)s+ut)\) is continuous by composition; its inner parameter belongs to \(I\), since it lies between \(s\) and \(t\). At \(u=0\) its value is \(\gamma(s)\), and at \(u=1\) its value is \(\gamma(t)\). This also checks directly that the paths supplied by the theorem have the required endpoints.
Projection from a Path-Connected Subspace
Worked Example: Projecting a Parabola onto the Line
Let \(X=\{(x,y)\in\mathbb{R}^2:y=x^2\}\) have the subspace topology from \(\mathbb{R}^2\), and define \(\pi:X\to\mathbb{R}\) by \(\pi(x,y)=x\). The projection is continuous: it is the restriction to \(X\) of the continuous first-coordinate projection \(\mathbb{R}^2\to\mathbb{R}\). We verify that \(X\) is path-connected before applying the theorem.
Take arbitrary \((a,a^2),(b,b^2)\in X\). Define \(\alpha(t)=((1-t)a+tb,((1-t)a+tb)^2)\). Its coordinates are continuous functions of \(t\), and its second coordinate is the square of its first, so \(\alpha(t)\in X\) for all \(t\in I\). Its endpoints are \(\alpha(0)=(a,a^2)\) and \(\alpha(1)=(b,b^2)\). Thus \(X\) is path-connected. The theorem implies that \(\pi(X)\) is path-connected; moreover, \(\pi(X)=\mathbb{R}\), because every \(r\in\mathbb{R}\) is the projection of \((r,r^2)\in X\).
For example, connect \((-1,1)\) to \((2,4)\) using \(\alpha(t)=(-1+3t,(-1+3t)^2)\). The projected path is \((\pi\circ\alpha)(t)=-1+3t\), with endpoints \(-1\) and \(2\). At \(t=\tfrac13\), the point on the parabola is \((0,0)\), and its projection is \(0=-1+3(\tfrac13)\). At \(t=\tfrac23\), the point is \((1,1)\), and its projection is \(1=-1+3(\tfrac23)\). The projected formula therefore agrees with the projection at these intermediate points as well as at both endpoints.
Common Pitfall: Testing Continuity with the Wrong Preimage
To use the theorem, the map must actually be continuous. A common error is to choose an open set whose preimage happens to be open and treat that as evidence of discontinuity. Continuity requires that every open set have open preimage; to disprove continuity, one must find an open set whose preimage is not open.
For a concrete check, give \(\{0,1\}\) the discrete topology, in which every subset is open, and define \(g:\mathbb{R}\to\{0,1\}\) by \(g(x)=0\) when \(x<0\) and \(g(x)=1\) when \(x\ge0\). The open set \(\{0\}\) has preimage \((-\infty,0)\), which is open in \(\mathbb{R}\); that preimage does not show that \(g\) is discontinuous. Instead, \(\{1\}\) is open and \(g^{-1}(\{1\})=[0,\infty)\), which is not open in \(\mathbb{R}\): every open interval around \(0\) contains negative numbers. This correctly proves that \(g\) is not continuous.
The relative topology matters just as much on the interval. Give \(I=[0,1]\) its subspace topology and define \(h:I\to\{0,1\}\), with the discrete topology on the codomain, by \(h(t)=0\) for \(t<\tfrac12\) and \(h(t)=1\) for \(t\ge\tfrac12\). The preimage of \(\{0\}\) is \([0,\tfrac12)=(-1,\tfrac12)\cap I\), so it is open in \(I\). But the preimage of the open set \(\{1\}\) is \([\tfrac12,1]\), which is not open in \(I\): any relatively open neighborhood of \(\tfrac12\) contains points of \(I\) less than \(\tfrac12\). Thus \(h\) is not continuous. In both cases the corrected preimage proves discontinuity, and the theorem cannot be applied to the map in question.
The essential mechanism is specific and dependable: start with two points of the image, choose preimages, join those preimages by a path in the domain, and compose that path with the continuous map. The endpoints of the new path are the chosen image points, and the subspace topology ensures it is a path in the image itself.
Check Your Understanding
For each question, identify the spaces and topologies involved before applying the theorem.
- If \(f:X\to Y\) is continuous and \(X\) is path-connected, which space is guaranteed to be path-connected: \(Y\) or \(f(X)\)? What topology is used on it?
- Give a path in \(\mathbb{R}\) from \(a\) to \(b\), and verify its endpoints.
- For \(f(x)=x^2\) on \(\mathbb{R}\), determine \(f(\mathbb{R})\) and explain why the theorem applies.
- In the discrete topology on \(\{0,1\}\), why does the preimage of \(\{1\}\) under \(g(x)=0\) for \(x<0\) and \(g(x)=1\) for \(x\ge0\) prove that \(g\) is not continuous?
- For the circle parametrization \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\), write a path from \(\gamma(s)\) to \(\gamma(t)\) using a parameter \(u\in I\).