Paths Have a Parameter, Endpoints, and an Image
A path records a continuous motion through a space over a fixed parameter interval. The parameter begins at \(0\) and ends at \(1\), but the path need not move at a constant speed, travel in a straight line, or visit each point of its image only once. To describe a path accurately, keep its map, its two endpoint values, and its image distinct.
Throughout, \(I=[0,1]\) has the standard subspace topology inherited from \(\mathbb{R}\). When the target is \(\mathbb{R}\) or \(\mathbb{R}^2\), those spaces have their standard topologies. For \(\mathbb{R}^2\), the standard topology is the one from the Euclidean metric \(\|(x_1,y_1)-(x_2,y_2)\|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
The map \(\gamma\) contains more information than its image: it specifies which point is visited at each parameter value. Two different maps can have the same image, and a map can visit a point more than once. The endpoints are found by substituting \(t=0\) and \(t=1\); they do not, by themselves, determine the image.
Continuity of Paths in the Plane
For a path into the plane, it is useful to write \(\gamma(t)=(u(t),v(t))\), where \(u\) and \(v\) are its coordinate functions. The following criterion lets us check continuity using real-valued functions.
Proof. Suppose first that \(\gamma\) is continuous, and fix \(t_0\in I\). For every \(t\in I\), \(|u(t)-u(t_0)|\leq\|\gamma(t)-\gamma(t_0)\|\) and \(|v(t)-v(t_0)|\leq\|\gamma(t)-\gamma(t_0)\|\). Given \(\varepsilon>0\), continuity of \(\gamma\) gives a \(\delta>0\) such that \(t\in I\) and \(|t-t_0|<\delta\) imply \(\|\gamma(t)-\gamma(t_0)\|<\varepsilon\). The two inequalities then show that both coordinate differences are less than \(\varepsilon\), proving continuity of \(u\) and \(v\) at \(t_0\).
Conversely, suppose \(u\) and \(v\) are continuous, and fix \(t_0\in I\) and \(\varepsilon>0\). By continuity, there are \(\delta_u>0\) and \(\delta_v>0\) such that \(|t-t_0|<\delta_u\) implies \(|u(t)-u(t_0)|<\varepsilon/2\), and \(|t-t_0|<\delta_v\) implies \(|v(t)-v(t_0)|<\varepsilon/2\), with \(t\in I\) in both cases. Set \(\delta=\min(\delta_u,\delta_v)\). If \(t\in I\) and \(|t-t_0|<\delta\), then \(\|\gamma(t)-\gamma(t_0)\|=\sqrt{|u(t)-u(t_0)|^2+|v(t)-v(t_0)|^2}<\sqrt{\varepsilon^2/4+\varepsilon^2/4}=\varepsilon/\sqrt{2}<\varepsilon\). Thus \(\gamma\) is continuous at \(t_0\). Since \(t_0\) was arbitrary, \(\gamma\) is continuous on \(I\). \(\square\)
In particular, if both coordinates are polynomials or familiar continuous trigonometric functions, the coordinate criterion establishes continuity of the plane path. Endpoint evaluation and image calculation are separate tasks: continuity makes the map a path, while the range of its coordinate expressions determines which points it visits.
Worked Example: An Affine Path in the Real Line
Let \(\gamma:I\to\mathbb{R}\) be given by \(\gamma(t)=3-5t\). Find its initial point, terminal point, and image, and verify that it is a path.
The function \(t\mapsto3-5t\) is continuous, so \(\gamma\) is a path. Its initial point is \(\gamma(0)=3-5(0)=3\), and its terminal point is \(\gamma(1)=3-5(1)=3-5=-2\).
To find the image, let \(s=3-5t\) with \(0\leq t\leq1\). Multiplying the parameter inequalities by \(-5\) reverses their directions: \(-5\leq-5t\leq0\). Adding \(3\) gives \(-2\leq s\leq3\). Conversely, if \(-2\leq s\leq3\), define \(t=(3-s)/5\). The bounds on \(s\) give \(0\leq3-s\leq5\), hence \(0\leq t\leq1\), and \(3-5t=s\). Therefore every \(s\) in this interval is attained, and no other value is attained.
Conclusion and check: \(\gamma(I)=[-2,3]\). Substituting \(t=0\) and \(t=1\) into the formula gives \(3\) and \(-2\), respectively, both of which belong to the computed image.
Worked Example: A Straight Path in the Plane
Define \(\gamma:I\to\mathbb{R}^2\) by \(\gamma(t)=(4t-1,\,2-4t)\). Find its endpoints and describe its image.
The coordinate functions \(4t-1\) and \(2-4t\) are continuous, so the coordinate criterion proves that \(\gamma\) is continuous. Its initial point is \(\gamma(0)=(-1,2)\), and its terminal point is \(\gamma(1)=(3,-2)\).
Write \(\gamma(t)=(x,y)\). The coordinate formulas give \(x=4t-1\) and \(y=2-4t\), so \(x+y=(4t-1)+(2-4t)=1\). As \(t\) ranges over \(I\), \(x=4t-1\) ranges over \([-1,3]\): from \(0\leq t\leq1\), we get \(-1\leq4t-1\leq3\). Conversely, for any \(x\in[-1,3]\), \(t=(x+1)/4\) lies in \([0,1]\), and the corresponding second coordinate is \(1-x\). Thus the image is exactly \(\{(x,1-x):-1\leq x\leq3\}\).
Check: the formula gives \((-1,2)\) when \(x=-1\) and \((3,-2)\) when \(x=3\). Both satisfy \(x+y=1\), as does every point in the displayed image. The converse parameter calculation shows that the description includes every point on this portion of the line, not merely points satisfying a necessary equation.
A Curved Path and Its Trace
For curved paths, it is often helpful to identify a familiar equation satisfied by every image point and then use the parameter range to establish that every point in the proposed trace is actually reached.
Worked Example: An Upper Semicircle
Let \(\gamma:I\to\mathbb{R}^2\) be defined by \(\gamma(t)=(\cos(\pi t),\sin(\pi t))\). Find its endpoints and image.
The coordinate functions are continuous, so the coordinate criterion shows that \(\gamma\) is a path. At \(t=0\), \(\gamma(0)=(\cos0,\sin0)=(1,0)\). At \(t=1\), \(\gamma(1)=(\cos\pi,\sin\pi)=(-1,0)\).
Set \(\theta=\pi t\). Because \(0\leq t\leq1\), the angle ranges over \(0\leq\theta\leq\pi\). For every such \(\theta\), the identity \(\cos^2\theta+\sin^2\theta=1\) gives \(x^2+y^2=1\), and \(\sin\theta\geq0\) gives \(y\geq0\). So the image is contained in \(\{(x,y):x^2+y^2=1,\ y\geq0\}\).
Conversely, take any point \((x,y)\) satisfying \(x^2+y^2=1\) and \(y\geq0\). The standard parametrization of the unit circle gives an angle \(\theta\in[0,\pi]\) with \(x=\cos\theta\) and \(y=\sin\theta\). Set \(t=\theta/\pi\); then \(t\in[0,1]\) and \(\gamma(t)=(x,y)\). Hence the image is exactly the upper semicircle \(\{(x,y):x^2+y^2=1,\ y\geq0\}\).
Check: the endpoints \((1,0)\) and \((-1,0)\) both satisfy \(x^2+y^2=1\) and \(y\geq0\). They agree with the direct evaluations at \(t=0\) and \(t=1\).
Why the Full Image Matters
Endpoint values identify where a path starts and finishes, but say little about what happens between those parameter values. A useful general fact about the image is that it is compact. Here a space is compact if every open cover has a finite subcover. The interval \(I\) is compact by the Heine–Borel theorem, which states that a closed and bounded interval in \(\mathbb{R}\) is compact.
Proof. Let \(\{V_\alpha\}\) be an open cover of \(\gamma(I)\) in its subspace topology. By the definition of the subspace topology, for each \(\alpha\) there is an open set \(U_\alpha\) in \(X\) such that \(V_\alpha=U_\alpha\cap\gamma(I)\). Continuity of \(\gamma\) implies that each preimage \(\gamma^{-1}(U_\alpha)\) is open in \(I\). These preimages cover \(I\): for any \(t\in I\), the point \(\gamma(t)\) belongs to some \(V_\alpha\), and hence to \(U_\alpha\), so \(t\in\gamma^{-1}(U_\alpha)\). Compactness of \(I\) gives finitely many indices \(\alpha_1,\ldots,\alpha_n\) whose preimages cover \(I\). For each \(t\in I\), membership in one of these preimages means \(\gamma(t)\in V_{\alpha_i}\) for some \(i\). Therefore \(V_{\alpha_1},\ldots,V_{\alpha_n}\) cover \(\gamma(I)\), proving that the image is compact. \(\square\)
This result is a useful check on computations: a path’s trace cannot be an arbitrary subset of its target. The theorem does not say that endpoint values determine the trace, nor that every compact set can be the image of a path.
Common pitfall: using endpoints as the image. Consider \(\eta:I\to\mathbb{R}\) defined by \(\eta(t)=(2t-1)^2\). Its initial and terminal points are both \(1\), since \(\eta(0)=(-1)^2=1\) and \(\eta(1)=1^2=1\). A wrong calculation would conclude that its image is \(\{1\}\), using only those endpoint values. But \(0\leq t\leq1\) implies \(-1\leq2t-1\leq1\), so \(0\leq(2t-1)^2\leq1\). Every value \(s\in[0,1]\) is attained: choose \(t=(1+\sqrt{s})/2\), which lies in \([0,1]\), and then \(\eta(t)=(\sqrt{s})^2=s\). Thus \(\eta(I)=[0,1]\), not \(\{1\}\). In particular, \(\eta(1/2)=0\). The correction is to calculate the range over the entire parameter interval, rather than infer it from the two endpoint values.
When describing any explicit path, state the map first, evaluate it at \(0\) and \(1\), and then justify the full image. For a plane path, coordinate continuity supplies a direct continuity check; equations and parameter bounds help establish the trace. Keeping these tasks separate avoids confusing a path with its endpoints or with the set of points it visits.
Check Your Understanding
For each map below, distinguish its endpoint values from its full image.
- For \(\alpha(t)=2+3t\) from \(I\) to \(\mathbb{R}\), find the initial point, terminal point, and image.
- For \(\beta(t)=(t,1-t)\) from \(I\) to \(\mathbb{R}^2\), describe its image as a subset of the plane.
- Verify using the coordinate criterion that \(\delta(t)=(t^2,3t)\) is a path into \(\mathbb{R}^2\).
- For \(\rho(t)=(\cos(2\pi t),\sin(2\pi t))\), find the two endpoint values and its full image.
- Explain why knowing only \(\gamma(0)\) and \(\gamma(1)\) does not, in general, determine \(\gamma(I)\), using a specific formula from this tutorial.