Continuous Functions and Homeomorphisms

In this lesson you’ll meet the preimage definition of continuity, check that it agrees with epsilon-delta, learn what a homeomorphism is and why the inverse must also be continuous, and see how to prove two spaces are not homeomorphic.

$f:X\to Y$ is continuous if $f^{-1}(V)$ is open in $X$ for every open $V\subseteq Y$.

That is the whole definition. No epsilons, no distances, no sequences - and it works in any topological space.

The figure pulls an open band back through two functions. For $x^2$, the band $(1,4)$ pulls back to $(-2,-1)\cup(1,2)$, an open set. For the step function, the band $(0.5,1.5)$ pulls back to a set containing its own left boundary point, which is not open. The figure decides openness by checking whether the function value at a run boundary lands strictly inside the band, so the verdict is computed rather than asserted.

Notice the direction. Continuity is about preimages, not images. A continuous function can easily map an open set to a non-open one: $f(x)=x^2$ sends the open set $(-1,1)$ to $[0,1)$, which is not open. Maps that do send open sets to open sets are called open maps, and that is a separate property.

For metric spaces the two definitions agree, and seeing why explains where the epsilons came from.

Suppose $f^{-1}(V)$ is always open, and fix $x_0$ and $\varepsilon>0$. The ball $B(f(x_0),\varepsilon)$ is open, so $f^{-1}(B(f(x_0),\varepsilon))$ is open and contains $x_0$. Being open, it contains a ball $B(x_0,\delta)$. That $\delta$ is exactly the one epsilon-delta asks for.

Running the argument backwards gives the converse. The epsilon-delta definition is the preimage definition with the balls written out.

All of these are the same condition, and different ones are convenient in different proofs:

• $f^{-1}(V)$ is open for every open $V$
• $f^{-1}(C)$ is closed for every closed $C$
• $f^{-1}(B)$ is open for every $B$ in a basis of $Y$ - usually the cheapest to check
• $f(\bar A)\subseteq\overline{f(A)}$ for every $A\subseteq X$
• For every $x$ and every neighbourhood $V$ of $f(x)$, there is a neighbourhood $U$ of $x$ with $f(U)\subseteq V$

The basis version is worth remembering: to show a map into $\mathbb{R}$ is continuous you need only check preimages of open intervals.

• Compositions of continuous maps are continuous, immediately from $(g\circ f)^{-1}(V) = f^{-1}(g^{-1}(V))$.
• Constant maps are continuous: the preimage is $\varnothing$ or $X$.
• The identity is continuous.
• Restrictions of continuous maps to subspaces are continuous.
• Any map from a discrete space is continuous, since every preimage is open.
• Any map into a trivial space is continuous, since the only open sets to pull back are $\varnothing$ and $Y$.

Those last two are worth pausing on. They say that continuity depends on the topologies as much as on the function - the same function can be continuous or not depending on which topologies you place on the two sets.

$f:X\to Y$ is a homeomorphism if $f$ is a bijection, $f$ is continuous, and $f^{-1}$ is continuous.

Equivalently, $f$ is a continuous bijection that is also an open map. Spaces related by a homeomorphism are homeomorphic, written $X\cong Y$, and they are indistinguishable by any topological property.

The inverse condition is not redundant. Consider

$$f:[0,2\pi)\to S^1, \qquad f(t) = (\cos t,\sin t)$$

This is a continuous bijection. Its inverse is not continuous: points on either side of $(1,0)$ on the circle are close to each other, but their preimages, near 0 and near $2\pi$, are far apart. The map glues the two ends of the interval together, and the inverse has to tear them.

So $[0,2\pi)$ and $S^1$ are not homeomorphic, which we can confirm independently: removing one point disconnects the circle never, and disconnects the interval sometimes.

There is one important case where the condition comes free:

A continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism.

That saves a great deal of work later, and it is why so many theorems specify compact and Hausdorff.

You cannot do it by failing to find a map. You exhibit a topological property one has and the other lacks:

Invariant	Distinguishes
Cardinality	trivially different sizes
Number of components	$(0,1)$ from $(0,1)\cup(2,3)$
Compactness	$[0,1]$ from $(0,1)$
Connectedness after removing a point	$[0,1]$ from $S^1$
Number of non-cut points	$[0,1]$ from $[0,1)$
Fundamental group	$S^2$ from the torus

The cut-point technique deserves highlighting because it is so cheap: if $X\cong Y$ then removing corresponding points must leave homeomorphic spaces, so any difference in what removal does is a proof.

An embedding is a map that is a homeomorphism onto its image.

This is the right notion when you care how a space sits inside another. A knotted circle and an unknotted circle in $\mathbb{R}^3$ are homeomorphic as spaces but are not equivalent embeddings, and knot theory studies exactly that difference.

Example 1: Show $f(x)=x^2$ is continuous using preimages.

Solution. By the basis criterion it suffices to check preimages of open intervals $(c,d)$.

If $d\le0$ the preimage is empty. If $c<0<d$ it is $(-\sqrt d,\sqrt d)$. If $0\le c<d$ it is

$$(-\sqrt d,-\sqrt c)\cup(\sqrt c,\sqrt d)$$

Every case is open, so $f$ is continuous ✓

For $(1,4)$ specifically this gives $(-2,-1)\cup(1,2)$, matching the figure. ∎

Example 2: Show a continuous map need not send open sets to open sets.

Solution. With $f(x)=x^2$, the image of the open set $(-1,1)$ is $[0,1)$, which is not open, since 0 has no interval around it inside the image.

Continuity constrains preimages only. ∎

Example 3: Show the step function is not continuous by a preimage.

Solution. Let $g(x)=0$ for $x<0$ and $g(x)=1$ for $x\ge0$. Then

$$g^{-1}\big((0.5,1.5)\big) = [0,\infty)$$

which is not open, since no interval around 0 stays inside it.

Answer: not continuous. ∎

Example 4: Show any map from a discrete space is continuous.

Solution. Let $X$ carry the discrete topology and let $f:X\to Y$ be arbitrary. For any open $V\subseteq Y$, the set $f^{-1}(V)$ is a subset of $X$, and every subset of a discrete space is open.

Answer: continuous. ∎

Continuity is a statement about the topologies, not only the formula. The same function becomes discontinuous if $X$ is given a coarser topology.

Example 5: Exhibit a continuous bijection that is not a homeomorphism.

Solution. $f:[0,2\pi)\to S^1$ with $f(t)=(\cos t,\sin t)$ is a continuous bijection.

The inverse fails continuity at $(1,0)$. Take the open set $[0,1)\subseteq[0,2\pi)$, which is open in the subspace. Its image is an arc containing the point $(1,0)$ but no full neighbourhood of it in $S^1$, so the image is not open - meaning $f$ is not an open map and $f^{-1}$ is not continuous.

Answer: not a homeomorphism. ∎

Example 6: Show $(0,1)\cong\mathbb{R}$.

Solution. The map

$$f(x) = \tan\left(\pi x - \frac{\pi}{2}\right)$$

is a continuous bijection from $(0,1)$ to $\mathbb{R}$, and its inverse $f^{-1}(y) = \frac{1}{\pi}\arctan y+\frac12$ is continuous too.

$$(0,1)\cong\mathbb{R}$$

∎

Boundedness is not topological. A bounded space and an unbounded one can be homeomorphic, which is exactly why compactness rather than boundedness is the topological notion.

Example 7: Show $[0,1]\not\cong[0,1)$ by cut points.

Solution. Call $p$ a non-cut point if $X\setminus\{p\}$ is still connected. In $[0,1]$ the non-cut points are 0 and 1, so there are two. In $[0,1)$ only 0 qualifies, so there is one.

A homeomorphism would carry non-cut points to non-cut points, so the counts would have to agree.

$$[0,1]\not\cong[0,1)$$

∎

Coordinate charts. A manifold is defined by homeomorphisms from patches of it onto open subsets of $\mathbb{R}^n$. Map projections are exactly such charts for the Earth, and the fact that no single chart covers the sphere is a topological statement.

Data embedding. Dimensionality reduction methods aim to embed data homeomorphically into a low-dimensional space. When the method distorts the topology - merging separate clusters, or tearing a loop - the failure is a failure of homeomorphism.

Mesh parameterization. Flattening a 3D surface patch for texturing requires a homeomorphism onto a planar region, and it is impossible without cutting when the patch has the wrong topology. Graphics tools detect this and introduce seams.

Robot workspace equivalence. Two mechanisms with homeomorphic configuration spaces admit the same motion-planning strategies, even if the machines look nothing alike. Recognizing the equivalence saves reimplementation.

Circuit layout. Whether a circuit can be drawn on a plane without crossings is a topological question about the graph, and it determines whether a single-layer board suffices.

Protein structure. Comparing folds uses topological equivalence rather than exact geometry, because the same functional shape appears with quite different coordinates.