A Checklist for Testing a Map
A map can pass one topological test and fail another. Continuity concerns preimages of open sets; openness and closedness concern images of open and closed sets. Bijectivity is a set-theoretic condition, and a homeomorphism requires a bijection whose map and inverse are both continuous. The examples below use these tests separately so that each conclusion has a specific justification.
Unless stated otherwise, \(\mathbb{R}\) has its standard topology. A real interval has the subspace topology inherited from \(\mathbb{R}\). A subset of \(\mathbb{R}^2\) has the subspace topology inherited from the Euclidean topology on \(\mathbb{R}^2\). For a map \(f:X\to Y\), the preimage of \(V\subseteq Y\) is \(f^{-1}(V)=\{x\in X:f(x)\in V\}\).
For a quick first pass, check the following separately: preimages of codomain open sets, images of domain open sets, images of domain closed sets, and whether distinct domain points can have the same image or some codomain point is missed. To establish a homeomorphism, check the map and its inverse, or use the criterion proved below.
An Inclusion Between a Real Interval and the Line
Proof. If \(V\) is open in \(X\), then \(i^{-1}(V)=A\cap V\), which is open in the subspace topology on \(A\). Thus \(i\) is continuous. Suppose \(A\) is open in \(X\). Every open \(U\subseteq A\) has the form \(U=A\cap W\) for some open \(W\subseteq X\), so \(i(U)=U=A\cap W\) is open in \(X\). Conversely, if \(i\) is open, then \(A=i(A)\) is open in \(X\), since \(A\) is open in itself. For the closed-map assertion, every closed \(C\subseteq A\) has the form \(C=A\cap F\) for some closed \(F\subseteq X\). If \(A\) is closed in \(X\), then \(i(C)=A\cap F\) is closed in \(X\). Conversely, if \(i\) is closed, then \(A=i(A)\) is closed in \(X\), since \(A\) is closed in itself. \(\square\)
Worked Example: Including an Open Interval in the Line
Let \(i:(0,1)\to\mathbb{R}\) be \(i(x)=x\), with the subspace topology on \((0,1)\). Determine its continuity, openness, closedness, injectivity, surjectivity, and homeomorphism status.
The map is continuous by the proposition. It is open because \((0,1)\) is open in \(\mathbb{R}\): any set open in the subspace \((0,1)\) is the intersection of \((0,1)\) with an open subset of \(\mathbb{R}\), hence is open in \(\mathbb{R}\). The map is injective because \(i(x)=i(y)\) means \(x=y\). It is not surjective, since \(0\in\mathbb{R}\) is not in its image.
It is not closed. The set \(C=(0,\tfrac12]\) is closed in \((0,1)\), because its complement in \((0,1)\) is \((\tfrac12,1)\), which is open in that subspace. But \(i(C)=(0,\tfrac12]\) is not closed in \(\mathbb{R}\): every neighborhood of \(0\) contains a point of \(C\), while \(0\notin C\). For example, given \(\varepsilon>0\), choose \(n\in\mathbb{N}\) with \(n>2\) and \(1/n<\varepsilon\); then \(1/n\in C\cap(-\varepsilon,\varepsilon)\). Thus the image of a closed set need not be closed.
Conclusion: \(i\) is continuous, open, and injective, but not closed or surjective. It is not a homeomorphism because it is not bijective.
Finite Spaces: A Continuous Bijection Can Still Fail
Worked Example: Testing Every Property on Two Finite Spaces
Let \(X=\{a,b\}\) have the discrete topology \(\mathcal{T}_X=\mathcal{P}(X)\), and let \(Y=\{u,v\}\) have topology \(\mathcal{T}_Y=\{\varnothing,\{u\},Y\}\). Define \(f:X\to Y\) by \(f(a)=u\) and \(f(b)=v\). Determine all six properties.
The map is bijective: \(a\) and \(b\) have distinct images, and both \(u\) and \(v\) occur. To test continuity, list the open subsets of \(Y\). Their preimages are \(f^{-1}(\varnothing)=\varnothing\), \(f^{-1}(\{u\})=\{a\}\), and \(f^{-1}(Y)=X\). All three are open in the discrete topology on \(X\), so \(f\) is continuous.
It is not open: \(\{b\}\) is open in \(X\), but \(f(\{b\})=\{v\}\) is not open in \(Y\). The closed subsets of \(X\) are all its subsets. In \(Y\), the closed subsets are the complements of its open sets, namely \(\varnothing,\{v\},Y\). Since \(\{a\}\) is closed in \(X\) but \(f(\{a\})=\{u\}\) is not closed in \(Y\), \(f\) is not closed.
The inverse is not continuous: the open set \(\{b\}\subseteq X\) has preimage \((f^{-1})^{-1}(\{b\})=\{v\}\), which is not open in \(Y\). Therefore this continuous bijection is not a homeomorphism.
Check: the map has two elements in its domain and two in its codomain, and the distinct images \(u\) and \(v\) exhaust \(Y\), confirming bijectivity. Its failure to be a homeomorphism is specifically the failure of inverse continuity.
A Homeomorphism onto a Subspace of the Plane
Worked Example: Parametrizing a Parabola
Let \(P=\{(x,y)\in\mathbb{R}^2:y=x^2\}\), with the subspace topology from the Euclidean topology on \(\mathbb{R}^2\). Define \(h:\mathbb{R}\to P\) by \(h(t)=(t,t^2)\). Test its properties.
The map is bijective. If \(h(s)=h(t)\), equality of the first coordinates gives \(s=t\), so \(h\) is injective. Every \((x,y)\in P\) satisfies \(y=x^2\), and hence equals \(h(x)\), so \(h\) is surjective.
To verify continuity, fix \(t\in\mathbb{R}\) and \(\varepsilon>0\). If \(|s-t|<1\), then \(|s+t|\leq 2|t|+1\), and \[ \|h(s)-h(t)\|^2=(s-t)^2+(s^2-t^2)^2 =(s-t)^2\bigl(1+(s+t)^2\bigr). \] Set \(M=\sqrt{1+(2|t|+1)^2}\), which is positive, and choose \(\delta=\min(1,\varepsilon/M)\). If \(|s-t|<\delta\), the displayed identity gives \(\|h(s)-h(t)\|\leq M|s-t|<\varepsilon\). Thus \(h\) is continuous at every \(t\).
The inverse is \(g:P\to\mathbb{R}\), \(g(x,y)=x\). For \(p=(x,y)\) and \(q=(x',y')\) in \(P\), \[ |g(p)-g(q)|=|x-x'|\leq\sqrt{(x-x')^2+(y-y')^2}=\|p-q\|. \] Therefore, if \(p\) is within distance \(\varepsilon\) of \(q\), then \(g(p)\) is within \(\varepsilon\) of \(g(q)\). This proves continuity of \(g\).
Conclusion: \(h\) is a homeomorphism. In particular, it is also open and closed; the criterion in the next section explains why both properties follow.
When Openness or Closedness Completes the Test
Proof. Suppose first that \(f\) is a bijective continuous open map. For any open \(U\subseteq X\), the preimage of \(U\) under \(f^{-1}:Y\to X\) is \(f(U)\), which is open in \(Y\). Thus \(f^{-1}\) is continuous, and \(f\) is a homeomorphism. If instead \(f\) is a bijective continuous closed map, take a closed \(C\subseteq X\). Its preimage under \(f^{-1}\) is \(f(C)\), which is closed in \(Y\). A map is continuous exactly when preimages of closed sets are closed: this follows by taking complements in the open-set definition. Therefore \(f^{-1}\) is continuous in this case as well.
Conversely, suppose \(f\) is a homeomorphism. If \(U\subseteq X\) is open, then \(f(U)=(f^{-1})^{-1}(U)\) is open in \(Y\), since \(f^{-1}\) is continuous. If \(C\subseteq X\) is closed, the same complement argument shows that \((f^{-1})^{-1}(C)=f(C)\) is closed in \(Y\). Hence \(f\) is both open and closed. \(\square\)
The theorem gives a useful shortcut, but its hypotheses matter: a continuous bijection alone is not enough, as the finite example showed. Conversely, checking that a map is open does not establish continuity; the two tests use images and preimages, respectively.
Common pitfall. For the finite map above, one might see that \(f(\{b\})=\{v\}\) is not open and conclude that \(f\) is not continuous. This uses the open-map test, not the continuity test. Continuity asks for preimages of open subsets of \(Y\): in particular, \(f^{-1}(\{u\})=\{a\}\) is open in \(X\), and the other open subsets of \(Y\) have preimages \(\varnothing\) and \(X\), also open. The correct conclusion is that \(f\) is continuous but not open.
Worked Example: A Constant Map
Let \(c:\mathbb{R}\to\mathbb{R}\) be \(c(x)=0\), with the standard topology on both copies of \(\mathbb{R}\). It is continuous: for any open \(V\subseteq\mathbb{R}\), \(c^{-1}(V)=\mathbb{R}\) if \(0\in V\), and \(c^{-1}(V)=\varnothing\) if \(0\notin V\). Both possibilities are open.
It is not open, since \(\mathbb{R}\) is open and \(c(\mathbb{R})=\{0\}\) is not open in the standard topology. It is closed: for every closed \(C\subseteq\mathbb{R}\), \(c(C)\) is either \(\varnothing\) or \(\{0\}\), both closed. It is neither injective nor surjective: \(c(0)=c(1)\), while \(1\) is not in its image. Consequently it is not a homeomorphism.
Check: the image of the closed set \(\mathbb{R}\) is \(\{0\}\), which is closed; the image of the open set \(\mathbb{R}\) is that same singleton, which is not open. This verifies directly that closedness and openness are distinct tests.
Check Your Understanding
For each question, keep preimage tests distinct from image tests and state the topology on every space.
- For the inclusion \([0,1]\to\mathbb{R}\), determine whether the map is continuous, open, closed, injective, surjective, and a homeomorphism.
- For the finite map in the example, compute the preimage of each open subset of \(Y\), then identify an open subset of \(X\) whose image is not open.
- Explain why the parabola parametrization is injective and why every point of \(P\) is in its image.
- Give a closed subset of the domain of \(i:(0,1)\to\mathbb{R}\) whose image is not closed in \(\mathbb{R}\), and verify both claims.
- State the extra hypothesis that, together with a continuous bijection, guarantees a homeomorphism in the theorem. Give both available alternatives.