Continuity Is a Condition on Preimages
A function between topological spaces is continuous when the topology of its domain is compatible with the open sets of its codomain in a particular way: pulling back any open set in the codomain must produce an open set in the domain. The word “preimage” matters. Continuity does not require open sets to have open images.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, and let \(f:X\to Y\) be a function. For \(V\subseteq Y\), its preimage is \(f^{-1}(V)=\{x\in X:f(x)\in V\}\). The preimage is a subset of the domain, even when \(f\) is not one-to-one or onto.
The definition quantifies over all open sets \(V\) in the codomain, including \(\varnothing\) and \(Y\). Their preimages are \(\varnothing\) and \(X\), which are open by the topology axioms. For a particular function, the substantive checks concern the other open sets in the codomain. The topologies on both spaces must be kept in view: the same set-theoretic function can be continuous for one choice of topologies and fail to be continuous for another.
Identity and Constant Maps
The identity map on a set \(X\) is \(\operatorname{id}_X:X\to X\), defined by \(\operatorname{id}_X(x)=x\). A constant map has the form \(f:X\to Y\), \(f(x)=y_0\) for every \(x\in X\), where \(y_0\in Y\) is fixed. These maps make the preimage test especially explicit.
Proof. For the identity map, let \(V\in\mathcal{T}\). By the definition of the identity, \(\operatorname{id}_X^{-1}(V)=V\), which is open in \((X,\mathcal{T})\). Thus \(\operatorname{id}_X\) is continuous. Now let \(f:X\to Y\) be constant with value \(y_0\), and let \(V\) be any open subset of \(Y\). If \(y_0\in V\), every \(x\in X\) satisfies \(f(x)=y_0\in V\), so \(f^{-1}(V)=X\). If \(y_0\notin V\), no \(x\in X\) satisfies \(f(x)\in V\), so \(f^{-1}(V)=\varnothing\). Both \(X\) and \(\varnothing\) are open in the domain, so the constant map is continuous. If \(X=\varnothing\), the preimage is always \(\varnothing\); the same argument applies. \(\square\)
Worked Example: The Identity on One Topological Space
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\). Check the continuity of \(\operatorname{id}_X:(X,\mathcal{T})\to(X,\mathcal{T})\).
Test each open set in the codomain. The preimages are \(\operatorname{id}_X^{-1}(\varnothing)=\varnothing\), \(\operatorname{id}_X^{-1}(\{a\})=\{a\}\), \(\operatorname{id}_X^{-1}(\{a,b\})=\{a,b\}\), and \(\operatorname{id}_X^{-1}(X)=X\). Each is in \(\mathcal{T}\), so the identity is continuous. This example also illustrates why the topologies belong in the statement of a continuity test: here, domain and codomain use exactly the same collection of open sets.
Check: for an arbitrary \(V\subseteq X\), the identity fixes every point, so \(x\in\operatorname{id}_X^{-1}(V)\) exactly when \(x\in V\). Hence the preimage equals \(V\), as the general proof predicts.
Worked Example: A Constant Map with a Discrete Codomain
Give \(X=\mathbb{R}\) the standard topology and give \(Y=\{p,q\}\) the discrete topology, in which every subset is open. Define \(f:X\to Y\) by \(f(x)=p\) for every \(x\in\mathbb{R}\). Check every codomain open set.
The four open sets of \(Y\) have preimages \(f^{-1}(\varnothing)=\varnothing\), \(f^{-1}(\{p\})=\mathbb{R}\), \(f^{-1}(\{q\})=\varnothing\), and \(f^{-1}(Y)=\mathbb{R}\). The empty set and the whole real line are open in the standard topology on \(\mathbb{R}\). Thus every open set in the codomain has an open preimage, and \(f\) is continuous.
Check: the value \(p\) lies in \(\{p\}\), so that set pulls back to the entire domain; the value \(p\) does not lie in \(\{q\}\), so that set pulls back to the empty set. These are precisely the two possibilities in the proof for a constant map.
When the Same Set Has Different Topologies
Suppose the domain and codomain have the same underlying set \(X\), but use topologies \(\mathcal{T}_1\) and \(\mathcal{T}_2\), respectively. The identity function still sends each point to itself, but its continuity now depends on which subsets are open in each topology.
Proof. Suppose first that the identity function is continuous. For each \(V\in\mathcal{T}_2\), continuity says that \(\operatorname{id}_X^{-1}(V)\in\mathcal{T}_1\). Since the function fixes every point, \(\operatorname{id}_X^{-1}(V)=V\). Therefore every member of \(\mathcal{T}_2\) belongs to \(\mathcal{T}_1\), giving \(\mathcal{T}_2\subseteq\mathcal{T}_1\). Conversely, suppose \(\mathcal{T}_2\subseteq\mathcal{T}_1\). For every open set \(V\in\mathcal{T}_2\), its preimage under the identity is \(V\), and the assumed inclusion makes \(V\) open in the domain topology \(\mathcal{T}_1\). Thus the identity is continuous. Both implications follow directly from the definition, including when \(X=\varnothing\). \(\square\)
When \(\mathcal{T}_2\subseteq\mathcal{T}_1\), one says that \(\mathcal{T}_1\) is finer than \(\mathcal{T}_2\): it has at least as many open sets. The identity map in the theorem goes from the finer topology to the coarser one. The theorem is a direct test for this particular map; it does not say that an arbitrary function is continuous just because its domain has many open sets.
Worked Example: From the Discrete Real Line to the Standard Real Line
Let the domain be \(\mathbb{R}\) with the discrete topology, and let the codomain be \(\mathbb{R}\) with the standard topology. Consider the identity function \(f(x)=x\). For any open set \(V\) in the standard topology, \(f^{-1}(V)=V\). The set \(V\), whatever its form, is a subset of the domain \(\mathbb{R}\); every subset is open in the discrete topology. Thus every codomain open set has an open preimage, and \(f\) is continuous.
Check: for example, the standard-open interval \((0,1)\) has preimage \((0,1)\). This preimage is open in the discrete domain because every subset of that domain is open. The argument works for every standard-open \(V\), not only this interval.
Worked Example: From the Standard Real Line to the Discrete Real Line
Reverse the topologies: let the domain be \(\mathbb{R}\) with its standard topology and the codomain be \(\mathbb{R}\) with the discrete topology. Again use the identity function \(f(x)=x\). The singleton \(\{0\}\) is open in the discrete codomain. Its preimage is \(f^{-1}(\{0\})=\{0\}\). This is not open in the standard topology: any standard-open set containing \(0\) must contain an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and that interval contains \(\varepsilon/2\ne 0\). Therefore \(\{0\}\) is not open in the domain, and \(f\) is not continuous.
Check: the failed condition is exactly the required one: an open codomain set, \(\{0\}\), has a preimage that is not open in the domain. In terms of the theorem, the discrete codomain has the open set \(\{0\}\), but the standard domain does not; the inclusion of topologies fails.
A Common Mistake: Testing Images Instead
Continuity is a preimage condition. It does not require the image \(f(U)\) of every open domain set \(U\) to be open. Confusing these two tests can lead to rejecting a continuous function.
Common mistake. In the identity map from the discrete real line to the standard real line, take the domain-open set \(U=\{0\}\). Its image is \(f(U)=\{0\}\), which is not open in the standard real line. A mistaken argument concludes that \(f\) is not continuous. The correction is to test preimages of codomain-open sets instead: for every standard-open \(V\subseteq\mathbb{R}\), \(f^{-1}(V)=V\), and \(V\) is open in the discrete domain. Thus the map is continuous, even though this particular open set has a non-open image.
The distinction is useful whenever topologies differ. To test continuity, start with an open set in the codomain, take its preimage, and decide whether that preimage is open in the domain topology. Keep the two roles separate: changing the codomain topology changes which sets must be tested, while changing the domain topology changes which preimages count as open.
Check Your Understanding
For each question, identify the topology used on both the domain and codomain before applying the preimage definition.
- Prove directly from the definition that the identity map from \((X,\mathcal{T})\) to itself is continuous.
- Let \(f:X\to Y\) be constant with value \(y_0\). What are the two possible forms of \(f^{-1}(V)\) as \(V\) ranges over open subsets of \(Y\)?
- On the same set \(X\), suppose \(\mathcal{T}_1\subseteq\mathcal{T}_2\). Is the identity from \((X,\mathcal{T}_1)\) to \((X,\mathcal{T}_2)\) necessarily continuous? Justify your answer using an open set.
- Let \(X\) have the indiscrete topology \(\{\varnothing,X\}\), and let \(Y\) be the same set with the discrete topology. Is the identity from \(X\) to \(Y\) continuous when \(X\) has at least two points?
- Explain why finding an open set \(U\) in the domain whose image is not open does not, by itself, show that a function is discontinuous.