Homeomorphism ($f:X\stackrel{\sim }{\to }Y$)

A homeomorphism is a bijection between topological spaces such that both $f$ and $f^{-1}$ are continuous maps. Equivalently, for every subset $U\subseteq X$, $U$ is an open set if and only if $f(U)$ is an open set in $Y$. A homeomorphism therefore transports the entire topology, including connectedness and compactness. It need not preserve numerical metric values, and a bounded set need not map to a bounded set.

For example, an explicit homeomorphism between an open interval and the real line isThe tangent function on $(-\pi /2,\pi /2)$ is strictly increasing and takes every real value; the displayed arctangent inverse verifies the bijection. Both formulas define continuous maps on their stated domains. Computed values are $f(1/4)=-1$, $f(1/2)=0$, and $f(3/4)=1$. Points approach infinity as $x$ approaches an excluded endpoint. Thus a bounded set can have the same topology as an unbounded one; neither of these topological spaces is compact.

A useful shortcut is the compact-to-Hausdorff continuous bijection theorem: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Its images of closed sets are compact sets, hence closed sets in the target, which proves continuity of the inverse. The indiscrete topology example shows why the Hausdorff assumption matters. ^[ref]

A homeomorphism $g:\partial Q\to \partial Q$ of the boundary of a closed square $Q$ extends to a homeomorphism of $Q$ by fixing its centre $c$ and setting $H(c+r(p-c))=c+r(g(p)-c)$ for $p\in \partial Q$ and $0<r\le 1$. Each noncentral point lies on a unique ray from $c$ meeting the boundary once, so the formula is well defined and its inverse is obtained from $g^{-1}$. Both maps are continuous off the centre by the radial parametrization, and at the centre because their distances from $c$ are bounded by $r$ times the radius of the square. It follows that complements of any two closed boundary arcs are homeomorphic: first map the arcs to one another by a boundary homeomorphism, then extend radially.

The circle $S^1$ and the closed interval $[0,1]$ are both compact and connected, so those two properties alone do not distinguish them. Instead examine connectedness after removing a point. Every punctured circle is connected: after rotating the missing point to $(1,0)$, the continuous map $t\mapsto (\cos t,\sin t)$ from the connected open interval $(0,2\pi )$ maps onto it. Continuous maps preserve connectedness. ^[ref]

If a homeomorphism $h:S^1\to [0,1]$ existed, choose $p=h^{-1}(1/2)$. Its restriction would give a homeomorphism between $S^1\setminus \{p\}$ and $[0,1]\setminus \{1/2\}$. But the latter splits into two nonempty disjoint open sets in its subspace topology, $[0,1/2)$ and $(1/2,1]$, contradicting connectedness of the punctured circle. Removing an endpoint of $[0,1]$ leaves a connected space; the proof deliberately chooses the preimage of an interior point. This illustrates how a simple connectedness test can rule out every possible homeomorphism.

A continuous bijection that is an open map is a homeomorphism. If $g=f^{-1}$ and $U$ is open in the original domain, then $g^{-1}(U)=f(U)$ is open by the open-map property. This proves continuity of the inverse directly.

A local homeomorphism is a continuous map $f:X\to Y$ such that every point of $X$ has a neighbourhood on which $f$ restricts to a homeomorphism onto an open subset of $Y$.

A closed map sends closed subsets of its domain to closed subsets of its codomain. A bijection is a homeomorphism exactly when it is continuous and closed.

A perfect map is a continuous closed map whose fibers are compact. Surjectivity is sometimes an additional convention; here it is not imposed, and empty fibers are compact. For maps between Tychonoff spaces, the perfect-map characterization by Stone-Čech remainders identifies this property using the unique extension to their compactifications.

For a continuous map $f:X\to Y$ between Tychonoff spaces, extend it to $F:\beta X\to \beta Y$ by the Stone-Čech extension property. If $f$ is perfect, compactness of a fiber and closedness of $f$ show that no remainder point can map to a point of $Y$: separate such a point from the fiber by a neighborhood in $\beta X$, and use closedness to obtain a target neighborhood whose preimage stays inside it, contradicting density. Conversely, if $F^{-1}(Y)=X$, each fiber is compact, and for closed $A\subseteq X$ the equality $f(A)=F({\overline{A}}^{\beta X})\cap Y$ proves closedness. Equivalently $F(\beta X\setminus X)\subseteq \beta Y\setminus Y$.

For metric spaces $X,Y$ with $X$ compact, the projection with a compact factor is closed: if $F\subset X\times Y$ is a closed set and $y_n\in {\pi }_Y(F)$ converges to $y$, choose $x_n$ with $(x_n,y_n)\in F$. Sequential compactness supplies $x_{n_k}\to x$, so $(x_{n_k},y_{n_k})\to (x,y)\in F$, proving $y\in {\pi }_Y(F)$. The projection map is also continuous for the sum metric. Without compactness, the closed set $\{(x,y)\in {\mathbb{R}}^2:xy=1\}$ projects to the nonclosed set $\mathbb{R}\setminus \{0\}$.

Let $f:X\to Y$ be a continuous map that is a closed map and whose fibres are compact. Then $f^{-1}(K)$ is compact for every compact set $K\subseteq Y$, without a Hausdorff hypothesis. Given an open cover, cover each fibre by a finite union $V_y$ of its members. The set $W_y=Y\setminus f(X\setminus V_y)$ is open, contains $y$, and has $f^{-1}(W_y)\subseteq V_y$. A finite subcover of $K$ by the $W_y$ uses only finitely many original cover members and covers $f^{-1}(K)$. Empty fibres cause no exception.