Continuity When the Values Lie in a Subspace
A function may take values in a subset \(A\) of a space \(Y\), while it is convenient to regard those values as points of \(Y\). These are two different codomains, so the corresponding definitions of continuity appear to involve different open sets. The subspace topology makes the difference precise: an open set in \(A\) is obtained by intersecting \(A\) with an open set in \(Y\). This gives an exact test for continuity into \(A\), using the map into \(Y\).
Let \((Y,\mathcal{T}_Y)\) be a topological space and \(A\subseteq Y\). The subspace topology on \(A\) is \(\mathcal{T}_A=\{U\cap A:U\in\mathcal{T}_Y\}\). Thus \(A\) is itself a topological space, whether or not \(A\) is open or closed in \(Y\). The inclusion map \(\iota:A\to Y\) is defined by \(\iota(a)=a\).
The Continuity Criterion
Proof. For any open set \(U\in\mathcal{T}_Y\), the definition of the inclusion gives \((\iota\circ g)^{-1}(U)=g^{-1}(U\cap A)\). First suppose \(g\) is continuous. Since \(U\cap A\) is open in \(A\), its preimage \(g^{-1}(U\cap A)\) is open in \(X\). The displayed identity shows that \((\iota\circ g)^{-1}(U)\) is open for every open \(U\) in \(Y\), so \(\iota\circ g\) is continuous.
Conversely, suppose \(\iota\circ g\) is continuous, and let \(V\) be any open set in \(A\). By the definition of the subspace topology, there is an open \(U\in\mathcal{T}_Y\) such that \(V=U\cap A\). Therefore \(g^{-1}(V)=g^{-1}(U\cap A)=(\iota\circ g)^{-1}(U)\), which is open in \(X\) by the assumed continuity. This holds for every open \(V\) in \(A\), so \(g\) is continuous. The argument also covers \(A=\varnothing\): a map into \(A\) can exist only when \(X=\varnothing\), and then all required preimages are open. \(\square\)
The key equality of preimages explains why the statement works. When testing continuity into \(A\), we test preimages of sets \(U\cap A\), not necessarily preimages of sets that are open in \(Y\) and contained in \(A\). Since every open set of \(A\) has the form \(U\cap A\), the two tests match exactly.
Worked Example: Squaring into the Nonnegative Half-Line
Let \(X=\mathbb{R}\) and \(Y=\mathbb{R}\), both with their standard topologies, and let \(A=[0,\infty)\) have the subspace topology from \(Y\). Define \(g:X\to A\) by \(g(x)=x^2\).
First, \(g\) has the stated codomain because \(x^2\geq 0\) for every real \(x\). Its composite with the inclusion is the real-valued function \(F(x)=x^2\). The polynomial function \(F:\mathbb{R}\to\mathbb{R}\) is continuous in the standard topology, so the theorem gives continuity of \(g\) into \(A\).
We can also check one relative open set directly. Let \(V=[0,4)\subseteq A\). This set is open in \(A\), because \(V=(-1,4)\cap A\) and \((-1,4)\) is open in \(\mathbb{R}\). Its preimage is \(g^{-1}(V)=\{x\in\mathbb{R}:0\leq x^2<4\}=(-2,2)\), which is open in \(X\). The strict inequality \(x^2<4\) excludes both endpoints \(-2\) and \(2\), as required.
Check: for \(x=1\), \(g(1)=1\in[0,4)\), and \(1\in(-2,2)\). For \(x=2\), \(g(2)=4\notin[0,4)\), and \(2\notin(-2,2)\). These endpoint tests agree with the computed preimage.
Worked Example: A Map Between Finite Spaces
Let \(X=\{p,q\}\) have the topology \(\mathcal{T}_X=\{\varnothing,\{p\},X\}\). Let \(Y=\{a,b,c\}\) have the topology \(\mathcal{T}_Y=\{\varnothing,\{a\},\{a,b\},Y\}\), and put \(A=\{a,b\}\) with the subspace topology. Intersecting the open sets of \(Y\) with \(A\) gives \(\mathcal{T}_A=\{\varnothing,\{a\},A\}\). Define \(g:X\to A\) by \(g(p)=a\) and \(g(q)=b\).
To test \(g\), compute the preimages of all open sets of \(A\): \(g^{-1}(\varnothing)=\varnothing\), \(g^{-1}(\{a\})=\{p\}\), and \(g^{-1}(A)=X\). Each belongs to \(\mathcal{T}_X\), so \(g\) is continuous.
The composite \(\iota\circ g:X\to Y\) has the same values, now viewed in \(Y\). The preimages of the open sets \(\varnothing,\{a\},\{a,b\},Y\) are, respectively, \(\varnothing,\{p\},X,X\); all are open in \(X\). Hence the composite is continuous as well.
Check: the open set \(\{b\}\) in the subspace \(A\) is not an available continuity test, because \(\{b\}\notin\mathcal{T}_A\). Its preimage \(\{q\}\) is not open in \(X\), but this does not contradict continuity of \(g\). The topology of the codomain determines which preimages must be tested.
A Range Criterion for Maps into a Subset
Proof. The range assumption \(f(X)\subseteq A\) makes \(g(x)=f(x)\) a well-defined map into \(A\). For every \(x\in X\), \((\iota\circ g)(x)=\iota(f(x))=f(x)\), so \(\iota\circ g=f\). Applying the theorem to \(g\) proves that \(g\) is continuous if and only if \(f\) is continuous into \(Y\). \(\square\)
Worked Example: A Bounded Function into an Open Interval
Let \(X=\mathbb{R}\) and \(Y=\mathbb{R}\), both with the standard topology, and let \(A=(-1,1)\) have the subspace topology. Define \(g(x)=\frac{x}{1+|x|}\).
For every \(x\), \(1+|x|>|x|\), so \(\left|\frac{x}{1+|x|}\right|<1\). Thus the formula does define a map into \(A\). The real-valued function \(F(x)=\frac{x}{1+|x|}\) is continuous: the absolute-value function is continuous, \(1+|x|\) is never zero, and sums, quotients with nonzero denominator, and products of continuous real functions are continuous. The range criterion therefore proves that \(g:X\to A\) is continuous.
For an explicit relative-open-set check, take \(V=(-\frac12,\frac12)\subseteq A\). Because this is also open in \(\mathbb{R}\), it is open in \(A\). Membership in its preimage is determined by \(\left|\frac{x}{1+|x|}\right|<\frac12\). Since \(1+|x|>0\), this is equivalent to \(2|x|<1+|x|\), then to \(|x|<1\). Hence \(g^{-1}(V)=(-1,1)\), open in \(X\).
Check: at \(x=2\), \(g(2)=\frac{2}{3}\), which is not in \((-\frac12,\frac12)\), and \(2\notin(-1,1)\). At \(x=\frac12\), \(g(\frac12)=\frac{1/2}{3/2}=\frac13\), which is in \((-\frac12,\frac12)\), and \(\frac12\in(-1,1)\). Both tests match the preimage calculation.
Why the Subspace Topology Is Essential
The equivalence depends on using the subspace topology on \(A\). If \(A\) is assigned some other topology, the open sets used to test continuity into \(A\) may no longer be the intersections \(U\cap A\). Then continuity of the composite need not detect continuity into that differently topologized set.
Common pitfall. In the first example, \(V=[0,4)\) is open in \(A=[0,\infty)\), but it is not open in \(\mathbb{R}\). A mistaken test might say that continuity into \(A\) cannot be checked because \(V\) is not ambient-open. The correction is to express it as \(V=U\cap A\), with \(U=(-1,4)\) open in \(\mathbb{R}\), and use the identity \(g^{-1}(V)=(\iota\circ g)^{-1}(U)\). The ambient set \(U\) need not equal \(V\); its intersection with \(A\) is what matters.
Here is a concrete failure if the topology assumption is dropped. Give \(Y=\{a,b,c\}\) the indiscrete topology \(\{\varnothing,Y\}\), so the subspace topology on \(A=\{a,b\}\) is also indiscrete. Now instead assign \(A\) the discrete topology \(\{\varnothing,\{a\},\{b\},A\}\). Let \(X=\{p,q\}\) have topology \(\{\varnothing,\{p\},X\}\), and define \(g(p)=a\), \(g(q)=b\). As a map into the discrete \(A\), \(g\) is not continuous: the open set \(\{b\}\) has preimage \(\{q\}\), which is not open in \(X\). But the composite into the indiscrete \(Y\) is continuous, because the only open sets of \(Y\) have preimages \(\varnothing\) and \(X\), both open. This does not contradict the theorem: the discrete topology on \(A\) is not the subspace topology inherited from \(Y\).
The criterion is useful whenever the values of a function are constrained to a subset: first verify that the map lands in that subset, then view the same assignment in the ambient space. With the inherited topology, the ambient continuity test is exactly equivalent to continuity into the subset, not merely a sufficient condition.
Check Your Understanding
For each question, use the definition of the subspace topology and the preimage identity from the theorem.
- Let \(A=[0,1]\subseteq\mathbb{R}\). Write the relative-open-set expression for \(V=[0,\frac12)\subseteq A\).
- Suppose \(f:X\to Y\) is continuous and \(f(X)\subseteq A\). What topology must \(A\) carry for the range criterion to apply?
- In the finite-space example, why does the fact that \(g^{-1}(\{b\})=\{q\}\) is not open fail to disprove continuity into \(A\)?
- For \(g(x)=\frac{x}{1+|x|}\), solve \(\left|\frac{x}{1+|x|}\right|<\frac13\) and identify the resulting preimage.
- In the common-pitfall example, which open set of the assigned discrete topology on \(A\) has a nonopen preimage?