A Path Has a Codomain as Well as Endpoints
A path is a continuous map from the interval \(I=[0,1]\), with its standard topology. When the space in which the path travels is a subset \(A\) of a larger space \(X\), the topology on \(A\) is the subspace topology: its open sets are the sets \(A\cap U\), where \(U\) is open in \(X\). This matters because saying that a map takes values in \(A\) is not by itself enough to say that it is continuous as a map into \(A\).
If a continuous map \(\gamma:I\to X\) begins and ends in \(A\), it is an ambient path with endpoints in \(A\). It is a path in \(A\) only if all its values lie in \(A\). The following result makes the relationship exact: when the image is contained in \(A\), continuity into \(A\) and continuity into \(X\) agree.
Proof. Let \(i:A\to X\) be the inclusion map, \(i(a)=a\). For any open \(U\subseteq X\), \(i^{-1}(U)=A\cap U\), which is open in the subspace topology on \(A\). Thus \(i\) is continuous. If \(\alpha:I\to A\) is continuous, then its composite \(i\circ\alpha:I\to X\) is continuous.
Conversely, suppose \(\gamma:I\to X\) is continuous and \(\gamma(I)\subseteq A\). Regard the same rule as a map \(\alpha:I\to A\). Any open set \(O\subseteq A\) has the form \(A\cap U\) for some open \(U\subseteq X\). Since every value of \(\gamma\) lies in \(A\), \(\alpha^{-1}(O)=\gamma^{-1}(U)\). This set is open in \(I\), so \(\alpha\) is continuous. Requiring the stated endpoint values then makes \(\alpha\) a path from \(a\) to \(b\) in \(A\). \(\square\)
The criterion separates two checks. First, check continuity as a map into the ambient space; then check that every value lies in the proposed subspace. For paths described by coordinates, the second check often reduces to an equation or inequality that must hold for every \(t\in I\).
Testing a Path in an Interval
Worked Example: An Ambient Path That Exits the Interval
Give \(X=\mathbb{R}\) its standard topology and \(A=[0,1]\) the subspace topology. Consider two ambient maps from \(I\) to \(\mathbb{R}\), both with endpoints in \(A\). One stays in \(A\); the other does not.
First take \(\lambda(t)=\frac15+\frac35t\). Because \(0\leq t\leq1\), we have \(0\leq\frac35t\leq\frac35\). Therefore \(\frac15\leq\lambda(t)\leq\frac45\), so \(\lambda(I)\subseteq[0,1]\). Its endpoints are \(\lambda(0)=\frac15\) and \(\lambda(1)=\frac15+\frac35=\frac45\). The function is continuous, so the subspace path criterion shows that \(\lambda\), regarded as a map \(I\to A\), is a path in \(A\).
Now take \(\gamma(t)=\frac15+\frac35t+\frac35\sin(\pi t)\). This is continuous and has endpoints \(\gamma(0)=\frac15+\frac35(0)+\frac35\sin(0)=\frac15\) and \(\gamma(1)=\frac15+\frac35+\frac35\sin(\pi)=\frac45\). But at \(t=\frac12\), \(\sin(\pi/2)=1\), and hence \(\gamma(\frac12)=\frac15+\frac3{10}+\frac35=\frac2{10}+\frac3{10}+\frac6{10}=\frac{11}{10}>1\). Thus \(\gamma(I)\) is not contained in \(A\). It is an ambient path with endpoints in \(A\), but it is not a path in \(A\).
Check: The failure is precisely at an intermediate parameter value: \(\gamma(\frac12)=\frac{11}{10}\notin[0,1]\). In contrast, the bound \(\frac15\leq\lambda(t)\leq\frac45\) verifies the interval condition for every \(t\in I\).
This example also shows why checking only the endpoints cannot establish that a path lies in a subspace. An endpoint condition constrains just two values of the map; the image condition constrains every value.
Equations for Paths on a Circle
Worked Example: A Chord Is Not a Path in the Circle
Let \(X=\mathbb{R}^2\) have its Euclidean topology, and let \(A=S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) have the subspace topology. The ambient line segment from \((1,0)\) to \((-1,0)\) is \(\gamma(t)=(1-2t,0)\). Its endpoints are \(\gamma(0)=(1,0)\) and \(\gamma(1)=(-1,0)\), and both satisfy \(x^2+y^2=1\). However, \(\gamma(\frac12)=(1-2(\frac12),0)=(0,0)\), and \(0^2+0^2=0\ne1\). The image is not contained in \(S^1\), so this ambient path does not define a path in \(S^1\).
A different ambient path between the same endpoints is \(\eta(t)=(\cos(\pi t),\sin(\pi t))\). Its endpoints are \(\eta(0)=(\cos 0,\sin 0)=(1,0)\) and \(\eta(1)=(\cos\pi,\sin\pi)=(-1,0)\). For every \(t\in I\), \((\cos(\pi t))^2+(\sin(\pi t))^2=1\). So \(\eta(t)\in S^1\) for every \(t\); the trigonometric coordinate functions are continuous, and the subspace path criterion gives a path in \(S^1\).
Check: At the midpoint, \(\eta(\frac12)=(\cos(\pi/2),\sin(\pi/2))=(0,1)\), and \(0^2+1^2=1\). The chord fails the circle equation at its midpoint; the second path satisfies it at every parameter value by the identity \(\cos^2 u+\sin^2 u=1\).
For a subset described by an equation, a reliable test is to substitute the path coordinates into that equation. For \(S^1\), the required condition is \(x(t)^2+y(t)^2=1\) for all \(t\), not just at \(t=0\) and \(t=1\).
A Detour in the Punctured Plane
Worked Example: Avoiding the Missing Point
Give \(X=\mathbb{R}^2\) its Euclidean topology and \(A=\mathbb{R}^2\setminus\{(0,0)\}\) the subspace topology. The straight-line path \(\sigma(t)=(1-2t,0)\) has endpoints \((1,0)\) and \((-1,0)\), both in \(A\). At \(t=\frac12\), it equals \((0,0)\), which is excluded from \(A\). Thus \(\sigma\) is not a path in the punctured plane.
Consider instead \(\tau(t)=(1-2t,t(1-t))\). Its endpoints are \(\tau(0)=(1,0)\) and \(\tau(1)=(-1,0)\), since \(0(1-0)=0\) and \(1(1-1)=0\). Both coordinate functions are continuous. To check the image condition, suppose \(\tau(t)=(0,0)\). Its first coordinate would force \(1-2t=0\), so \(t=\frac12\). But then its second coordinate is \(\frac12(1-\frac12)=\frac14\ne0\), a contradiction. Thus \(\tau(t)\ne(0,0)\) for every \(t\in I\), so \(\tau(I)\subseteq A\). By the subspace path criterion, \(\tau\) is a path in \(A\).
Check: At \(t=\frac12\), \(\tau(\frac12)=(0,\frac14)\), which is not the removed point. More generally, the argument shows that there is no parameter where both coordinates vanish, so the path avoids the puncture throughout.
A Common Mistake and a Practical Test
Common mistake: “The path \(\gamma(t)=(1-2t,0)\) is a path in \(S^1\) because its endpoints \((1,0)\) and \((-1,0)\) lie in \(S^1\).” The endpoint calculation is correct, but it does not establish that the image lies in \(S^1\). Substituting \(t=\frac12\) gives \(\gamma(\frac12)=(0,0)\), and \(0^2+0^2=0\ne1\). The correction is to check the defining condition for membership in the subspace for every parameter value, or to find one parameter value where it fails. Here one failing value rules out the path.
The ambient continuity test and the image test answer different questions. A continuous formula into \(X\) may leave \(A\); a formula whose values lie in \(A\) may be treated as a map into \(A\), and the subspace path criterion guarantees its continuity there. The key checks can be summarized as follows.
| Question | What to verify | What it establishes |
|---|---|---|
| Is the formula continuous? | Continuity as a map \(I\to X\) | Ambient continuity |
| Does it stay in \(A\)? | \(\gamma(t)\in A\) for every \(t\in I\) | The formula can be regarded as a map \(I\to A\) |
| Is it a path from \(a\) to \(b\) in \(A\)? | Both checks, plus \(\gamma(0)=a\) and \(\gamma(1)=b\) | A path in the subspace |
In an interval, the image condition is an inequality; on a circle, it is an equation; in a punctured plane, it is avoidance of one point. These examples illustrate the same principle without relying on the shape suggested by a drawing: a path in a subspace is a continuous map whose entire image belongs to that subspace.
Check Your Understanding
For each question, distinguish endpoint membership from the condition on the whole image.
- In \(A=[-1,1]\subseteq\mathbb{R}\), test whether \(\gamma(t)=2t-1\) is a path in \(A\), and give its endpoints.
- For \(S^1\subseteq\mathbb{R}^2\), does \(\alpha(t)=(\cos(2\pi t),\sin(2\pi t))\) define a path in the circle? Verify the image condition.
- In \(\mathbb{R}^2\setminus\{(0,0)\}\), explain why the straight segment from \((1,1)\) to \((-1,-1)\) fails to be a path in the punctured plane.
- State the subspace path criterion and identify where the subspace topology enters its proof.
- Give a continuous ambient map with endpoints in \(A=[0,1]\) that is not a path in \(A\), and verify one parameter value at which it leaves \(A\).