The Inclusion Map
A subset carries its own topology, and that topology need not consist of sets that are open in the ambient space. The inclusion map provides a basic example of how a map can be continuous even when its domain has a different collection of open sets from its codomain. Its continuity follows directly from the definition: compute the inverse image of an open set in the codomain and check that it is open in the domain.
Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). Give \(A\) the subspace topology \(\mathcal{T}_A=\{A\cap U:U\in\mathcal{T}\}\). In particular, a subset of \(A\) is open in \(A\) when it is the intersection of \(A\) with an open subset of \(X\).
The map \(i_A\) does not change the point: it regards a point of \(A\) as a point of \(X\). Continuity is a condition on inverse images of open sets in the codomain. For the inclusion, this inverse image has a particularly direct form.
Proof. A point \(a\in A\) belongs to \(i_A^{-1}(U)\) exactly when \(i_A(a)\in U\). Since \(i_A(a)=a\), this is exactly the condition \(a\in U\). Thus \(i_A^{-1}(U)\) consists of the points that belong both to \(A\) and to \(U\), so \(i_A^{-1}(U)=A\cap U\). \(\square\)
Why the Inclusion Is Continuous
Proof. Assume \(U\) is open in \(X\). By the inverse-image formula, \(i_A^{-1}(U)=A\cap U\). The subspace topology on \(A\) consists of the sets \(A\cap V\) with \(V\) open in \(X\); taking \(V=U\) shows that \(A\cap U\) is open in \(A\). Therefore the inverse image under \(i_A\) of every open set in \(X\) is open in \(A\). By the definition of continuity, \(i_A\) is continuous. This argument also covers \(A=\varnothing\): every such inverse image is empty, which is open in the empty space. \(\square\)
The key point is that the domain's open sets are relative to \(A\). We do not need \(A\cap U\) to be open as a subset of \(X\). The theorem uses exactly the inverse-image test for continuity and the defining form of open sets in the subspace topology.
Worked Example: An Interval Included in the Real Line
Let \(A=[0,2]\) with the subspace topology inherited from \(\mathbb{R}\), and let \(i_A:A\to\mathbb{R}\) be inclusion. Take the ambient open set \(U=(-1,1)\cup(3/2,3)\). Compute its inverse image.
By the inverse-image formula, \(i_A^{-1}(U)=A\cap U\). The intersection of \([0,2]\) with \((-1,1)\) is \([0,1)\): the endpoint \(0\) belongs to both sets, while \(1\) does not belong to \((-1,1)\). The intersection of \([0,2]\) with \((3/2,3)\) is \((3/2,2]\): \(3/2\) is excluded, while \(2\) is included. Hence
This set is open in \(A\), since it is \(A\cap U\) for the ambient open set \(U\). As a check on the endpoint decisions, \(0\in A\cap U\), \(1\notin A\cap U\), \(3/2\notin A\cap U\), and \(2\in A\cap U\). These agree with the displayed intervals.
The Subspace Topology as the Exact Condition
The continuity proof also tells us what happens if \(A\) is given some topology other than its subspace topology. Write \(\mathcal{T}_A\) for that topology on the set \(A\), and keep \(X\) with topology \(\mathcal{T}\). The inverse-image formula remains true because it depends only on the underlying function. Continuity then asks whether all sets \(A\cap U\), for \(U\) open in \(X\), are open in the chosen topology on \(A\).
Proof. Suppose first that \(i_A\) is continuous. For every \(U\in\mathcal{T}\), the set \(U\) is open in \(X\), so \(i_A^{-1}(U)\) is open in \((A,\mathcal{T}_A)\). The inverse-image formula gives \(i_A^{-1}(U)=A\cap U\). Thus every set in the subspace topology belongs to \(\mathcal{T}_A\).
Conversely, suppose every set \(A\cap U\), with \(U\in\mathcal{T}\), belongs to \(\mathcal{T}_A\). For each open \(U\) in \(X\), \(i_A^{-1}(U)=A\cap U\) is therefore open in \((A,\mathcal{T}_A)\). This is precisely the definition of continuity, so \(i_A\) is continuous. Both implications hold for \(A=\varnothing\) as well: the subspace topology then consists only of \(\varnothing\), and the inclusion has empty inverse images. \(\square\)
In this sense, the subspace topology is the coarsest topology on \(A\) that makes the inclusion continuous: any topology on \(A\) for which the inclusion is continuous must contain every subspace-open set. “Coarsest” here means that its collection of open sets is contained in the collection for any other such topology. The subspace topology itself works by the preceding theorem, so it has this property.
Worked Example: A Coordinate Line in the Plane
Give \(X=\mathbb{R}^2\) its Euclidean topology and let \(A=\{(t,0):t\in\mathbb{R}\}\) have the subspace topology. Consider the inclusion \(i_A:A\to\mathbb{R}^2\) and the open ball \(U=B((1,0),2)\). Find \(i_A^{-1}(U)\).
A point \((t,0)\in A\) is in \(U\) exactly when its Euclidean distance from \((1,0)\) is less than \(2\). That distance is \(\sqrt{(t-1)^2+(0-0)^2}=\sqrt{(t-1)^2}=|t-1|\). Thus membership requires \(|t-1|<2\), which is equivalent to \(-2<t-1<2\), and then to \(-1<t<3\). Consequently
The endpoints are excluded because at \(t=-1\) the distance is \(|-1-1|=2\), and at \(t=3\) it is \(|3-1|=2\); the ball uses a strict distance inequality. At \(t=1\), the distance is \(|1-1|=0<2\), so \((1,0)\) is included. The inverse image is open in \(A\) because it is the intersection of \(A\) with the ambient open ball \(U\).
Worked Example: An Open Cap on the Circle
Give \(X=\mathbb{R}^2\) its Euclidean topology and let \(A=S^1=\{(x,y):x^2+y^2=1\}\) have the subspace topology. For the open ball \(U=B((1,0),1)\), compute the inverse image under inclusion.
A point \((x,y)\in S^1\) belongs to \(U\) exactly when \((x-1)^2+y^2<1\). Using \(x^2+y^2=1\), the left side becomes \(x^2-2x+1+y^2=2-2x\). Therefore the ball condition is \(2-2x<1\), or \(x>1/2\). It follows that
The inequality is strict: points of the circle with \(x=1/2\) are excluded because their squared distance from \((1,0)\) is \(2-2(1/2)=1\), not less than \(1\). The point \((1,0)\) is included because its squared distance is \(2-2(1)=0<1\). This cap is open in the subspace \(S^1\) because it is the intersection of the circle with the ambient open ball.
A Common Misreading
Common mistake. One might claim that the inclusion \(i_A:A\to X\) is continuous only when \(A\) is open in \(X\). This confuses openness in the domain with openness in the codomain. Take \(A=[0,1]\subseteq\mathbb{R}\) with the subspace topology and \(U=(-1,2)\), which is open in \(\mathbb{R}\). The inverse image is
The set \([0,1]\) is not open in \(\mathbb{R}\): every open interval about \(0\) contains negative real numbers and so is not contained in \([0,1]\). But \([0,1]\) is open in the domain \(A\), because it equals \(A\cap(-1,2)\). The inverse-image test asks for openness in the domain, not in the ambient space. The inclusion is therefore continuous, regardless of whether \(A\) itself is open in \(X\).
The inverse-image formula makes the role of the subspace topology precise: it is exactly the topology that declares all the sets needed for continuity of the inclusion to be open. In concrete settings, one can test the map by taking an ambient open set, intersecting it with the subset, and checking that the result is open relative to that subset.
Check Your Understanding
For each question, use the inverse-image formula and specify which space's topology is being used to judge openness.
- Let \(A=[-1,1]\subseteq\mathbb{R}\) with its subspace topology. Find the inverse image of \(U=(-2,0)\cup(1/2,3)\) under \(i_A\).
- For \(A=\{(t,0):t\in\mathbb{R}\}\subseteq\mathbb{R}^2\), find the inverse image of \(B((0,0),3)\) under inclusion.
- Suppose a topology \(\mathcal{T}_A\) on \(A\) contains the subspace topology. Prove directly from the open-set definition that \(i_A:(A,\mathcal{T}_A)\to X\) is continuous.
- Can \(A\) be empty in the inclusion theorem? Explain what the inverse image of an open set in \(X\) is in that case.
- Give an example of a subset \(A\) that is not open in \(X\), but whose inclusion into \(X\) is continuous when \(A\) has the subspace topology.