Continuity into a Subspace
A function may take values in a subset of a space, but continuity depends on the topology assigned to that subset. For a subset \(A\) of a topological space \(Y\), the subspace topology declares a set open in \(A\) when it has the form \(A\cap V\) for some open set \(V\) in \(Y\). This makes it possible to test continuity of a map into \(A\) by considering the same values in the ambient space \(Y\).
Throughout, let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, and let \(A\subseteq Y\) carry the subspace topology. A map \(f:X\to Y\) is continuous if the preimage of every open subset of \(Y\) is open in \(X\). Write \(i:A\to Y\) for the inclusion map, \(i(a)=a\).
Proof. For any open set \(V\subseteq Y\), the set \(i^{-1}(V)\) is \(A\cap V\), which is open in \(A\) by the definition of the subspace topology. If \(g\) is continuous, then \((i\circ g)^{-1}(V)=g^{-1}(i^{-1}(V))\) is open in \(X\), so \(i\circ g\) is continuous.
Conversely, suppose \(i\circ g\) is continuous, and let \(O\) be open in \(A\). By the subspace topology, \(O=A\cap V\) for some open \(V\subseteq Y\). Since \(g(x)\in A\) for every \(x\in X\), we have \(g^{-1}(O)=g^{-1}(A\cap V)=(i\circ g)^{-1}(V)\). This is open in \(X\) by the assumed continuity of \(i\circ g\). Thus \(g\) is continuous. \(\square\)
The fact that \(g\) takes values in \(A\) is part of the statement: it is what makes \(g:X\to A\) a map and gives the preimage identity in the proof. The theorem says that once the target is \(A\), testing continuity there is exactly equivalent to testing the same function as a \(Y\)-valued map.
Proof. The image condition ensures that \(\widetilde f\) is well-defined, and \(f=i\circ\widetilde f\). The theorem therefore says precisely that \(\widetilde f\) is continuous if and only if \(f\) is continuous. \(\square\)
Maps into Intervals
For an interval \(I=[0,1]\), use the subspace topology inherited from \(\mathbb{R}\) with its standard topology. A useful practical consequence is that a map with values in \(I\) can be tested as a real-valued function: prove it is real-valued and continuous, and separately check that its values lie in \(I\). The range check is essential; continuity alone does not make a function \(I\)-valued.
Worked Example: A Map into the Unit Interval
Define \(g:\mathbb{R}\to I\) by \(g(x)=\dfrac{x^2}{1+x^2}\), where \(\mathbb{R}\) has its standard topology and \(I\) has the subspace topology. Verify its range and continuity, and compute \(g(2)\).
For every real \(x\), \(x^2\geq 0\), so \(g(x)\geq 0\). Also \(1+x^2>0\), and \(x^2<1+x^2\), so \(g(x)<1\). Hence \(0\leq g(x)<1\), which verifies \(g(x)\in[0,1]\) for every \(x\). As a real-valued function, \(g\) is a quotient of polynomials with denominator \(1+x^2\), which never vanishes; the standard continuity rules for real functions therefore show that it is continuous into \(\mathbb{R}\). By the corestriction criterion, \(g:\mathbb{R}\to I\) is continuous.
Substituting \(x=2\) gives \(g(2)=\dfrac{2^2}{1+2^2}=\dfrac{4}{5}\). This value lies in \(I\), since \(0\leq \dfrac45<1\). As a further range check, \(g(0)=\dfrac{0^2}{1+0^2}=0\), and for every finite \(x\) the strict inequality \(g(x)<1\) shows that the endpoint \(1\) is not attained.
Maps into the Circle
Let \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\), with the subspace topology inherited from the Euclidean topology on \(\mathbb{R}^2\). To define a map into \(S^1\), it is enough to give two continuous coordinate functions and verify that their values satisfy the equation \(u^2+v^2=1\). The continuity criterion then turns continuity as an \(\mathbb{R}^2\)-valued map into continuity as an \(S^1\)-valued map.
Worked Example: A Parametrization of the Circle
Let \(I=[0,1]\) have its subspace topology from \(\mathbb{R}\), and define \(g:I\to\mathbb{R}^2\) by \(g(t)=(\cos(2\pi t),\sin(2\pi t))\). Show that \(g\) defines a continuous map \(I\to S^1\), and evaluate it at \(t=0\), \(t=\dfrac14\), and \(t=1\).
The coordinate functions are continuous, so \(g:I\to\mathbb{R}^2\) is continuous. For every \(t\in I\), the identity \(\cos^2(2\pi t)+\sin^2(2\pi t)=1\) gives \(g(t)\in S^1\). Thus the same formula, regarded as a map \(I\to S^1\), is continuous by the corestriction criterion.
At \(t=0\), \(2\pi t=0\), so \(g(0)=(\cos 0,\sin 0)=(1,0)\). At \(t=\dfrac14\), \(2\pi t=\dfrac{\pi}{2}\), so \(g(\dfrac14)=(\cos(\dfrac{\pi}{2}),\sin(\dfrac{\pi}{2}))=(0,1)\). At \(t=1\), \(2\pi t=2\pi\), so \(g(1)=(\cos(2\pi),\sin(2\pi))=(1,0)\).
Check: At the computed point \(g(\dfrac14)=(0,1)\), the circle equation gives \(0^2+1^2=1\); at \(g(0)=g(1)=(1,0)\), it gives \(1^2+0^2=1\). These evaluations agree with the range verification for every \(t\).
Maps into Spheres
For \(S^2=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\}\), use the subspace topology from Euclidean \(\mathbb{R}^3\). The same method applies: establish continuity of the three coordinate functions, verify the sphere equation, and then use the subspace criterion. A formula with a denominator also requires checking that the denominator is nonzero throughout its domain.
Worked Example: A Map from the Plane into the Sphere
Give \(\mathbb{R}^2\) its Euclidean topology and define \(F:\mathbb{R}^2\to\mathbb{R}^3\) by \(F(u,v)=\left(\dfrac{2u}{1+u^2+v^2},\dfrac{2v}{1+u^2+v^2},\dfrac{u^2+v^2-1}{1+u^2+v^2}\right)\). Show that its values lie in \(S^2\), prove continuity into \(S^2\), and evaluate it at \((0,0)\) and \((\dfrac34,0)\).
Set \(r^2=u^2+v^2\), so \(r^2\geq0\) and the denominator \(1+r^2\) is strictly positive. Each coordinate is a quotient of continuous polynomial functions with nonzero denominator, so \(F:\mathbb{R}^2\to\mathbb{R}^3\) is continuous. To check its range, compute the sum of the squares of its coordinates: \[ \frac{4u^2+4v^2+(r^2-1)^2}{(1+r^2)^2} =\frac{4r^2+(r^4-2r^2+1)}{(1+r^2)^2} =\frac{r^4+2r^2+1}{(1+r^2)^2} =\frac{(1+r^2)^2}{(1+r^2)^2} =1. \] Thus \(F(u,v)\in S^2\) for every \((u,v)\), and the corestriction \(\mathbb{R}^2\to S^2\) is continuous.
At \((0,0)\), \(r^2=0\) and the denominator is \(1+0=1\), so \(F(0,0)=(0,0,-1)\). Its squared coordinate sum is \(0^2+0^2+(-1)^2=1\). At \((\dfrac34,0)\), \(r^2=(\dfrac34)^2+0^2=\dfrac9{16}\), and the denominator is \(1+\dfrac9{16}=\dfrac{25}{16}\). The first coordinate is \(\dfrac{2(\frac34)}{\frac{25}{16}}=\dfrac{3}{2}\cdot\dfrac{16}{25}=\dfrac{24}{25}\); the second is \(0\); the third is \(\dfrac{\frac9{16}-1}{\frac{25}{16}}=\dfrac{-\frac7{16}}{\frac{25}{16}}=-\dfrac7{25}\). Therefore \(F(\dfrac34,0)=(\dfrac{24}{25},0,-\dfrac7{25})\).
Check: The squared coordinate sum of this last value is \((\dfrac{24}{25})^2+0^2+(-\dfrac7{25})^2=\dfrac{576}{625}+\dfrac{49}{625}=\dfrac{625}{625}=1\), as required.
A Common Error About Relative Openness
Continuity into a subspace does not require preimages of subspace-open sets to be open in the ambient target. They must be open in the domain, and a set open in a subspace need not be open in the ambient space. Confusing these two meanings can lead to a false rejection of a continuous map.
Common pitfall. Consider the inclusion \(j:I\to\mathbb{R}\), where \(I=[0,1]\) has the subspace topology. A mistaken calculation takes the open set \(V=(-1,1)\subseteq\mathbb{R}\), obtains \(j^{-1}(V)=[0,1)\), observes that \([0,1)\) is not open in \(\mathbb{R}\), and concludes that \(j\) is not continuous. The error is the final test: \([0,1)\) is open in \(I\), because \([0,1)=I\cap(-1,1)\), with \((-1,1)\) open in \(\mathbb{R}\). More generally, for any open \(V\subseteq\mathbb{R}\), \(j^{-1}(V)=I\cap V\), which is open in \(I\). Thus \(j\) is continuous.
The reverse issue also matters: a formula may be continuous into the ambient space but fail to define a map into the proposed subspace if some value lies outside it. For example, the constant real-valued function \(h(x)=2\) is continuous as a map into \(\mathbb{R}\), but it does not define a map into \(I\), because \(h(x)=2\notin[0,1]\) for every \(x\). The range condition and the continuity test are separate checks.
Check Your Understanding
For each question, distinguish the ambient topology from the subspace topology where needed.
- Let \(A=[-1,1]\subseteq\mathbb{R}\), and define \(g:\mathbb{R}\to A\) by \(g(x)=\dfrac{x}{1+|x|}\). Verify the range condition and explain how continuity into \(A\) can be tested.
- Suppose \(g:X\to A\subseteq Y\). Write the preimage identity that relates \(g^{-1}(O)\) to \((i\circ g)^{-1}(V)\) when \(O=A\cap V\).
- For \(j:[0,1]\to\mathbb{R}\) the inclusion, compute \(j^{-1}((0,2))\) and state whether it is open in \([0,1]\) and in \(\mathbb{R}\).
- Define a continuous map from \(\mathbb{R}\) into \(S^1\) using two coordinate functions, and verify the circle equation for all inputs.
- Why must one check that a proposed ambient-valued map has image contained in \(A\) before calling it a map into the subspace \(A\)?