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Invariance of domain says that a continuous map that is an injective function between topological manifolds without boundary of the same dimension is an open map. It is useful in the Penrose singularity theorem: projecting a compact achronal boundary along timelike curves onto a Cauchy hypersurface gives a nonempty subset that is both open and closed, forcing that connected Cauchy hypersurface to be compact.

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 diffeomorphism is a differentiable map that restricts near every point to a diffeomorphism onto an open subset of its target. The local inverse functions need not combine into a single global inverse function: different parts of the domain of a function can map onto the same output region.

For continuously differentiable maps between open subsets of ${\mathbb{R}}^n$, being a local diffeomorphism is equivalent to having an invertible derivative at every point. The inverse function theorem proves one direction. For the other, the chain rule for each local inverse function gives two inverse linear maps. The periodic angle example under the inverse function theorem has this property everywhere, even though it has infinitely many inputs for every point of the punctured plane. The noninjective map with constant Jacobian determinant gives another useful distinction: even a constant nonzero Jacobian determinant need not give global injectivity.

Every local diffeomorphism is an open map. To see this, take an open subset of the source and intersect it with the neighbourhoods where the map is a diffeomorphism. Each local image is open, and their union is the full image. Thus nearby outputs remain available locally, despite the possible global ambiguity.

An open map sends each open set to an open set; a closed map sends each closed set to a closed set. These properties concern images, whereas continuity concerns inverse images of open sets. Neither openness nor closedness alone implies continuity.

A continuous surjective linear map between Fréchet spaces is an open map. The Baire category theorem first shows that the closure of the image of every convex balanced zero-neighbourhood is a zero-neighbourhood: the target is the countable union of scalar multiples of this image, and an interior point of its closure can be translated to zero using convexity and symmetry.

To remove the closure, fix a domain zero-neighbourhood $U$ and choose convex balanced zero-neighbourhoods $U_n$ so that every series ${\sum }_n{u}_n$, $u_n\in U_n$, converges to a point of $U$. With an increasing defining sequence of seminorms $p_j$, one can ensure $p_j(u_n)<2^{-n}$ for $j\le n$ and make the finitely many seminorms defining $U$ summable within its bounds. Completeness gives convergence. Choose target zero-neighbourhoods $V_n\subset \overline{A(U_n)}$ shrinking to zero. For $y\in V_1$, choose $u_n\in U_n$ successively so that $y-A{\sum }_{j\le n}{u}_j\in V_{n+1}$. Density permits each correction. Continuity then gives $y=A{\sum }_n{u}_n\in A(U)$, proving openness.

A quotient operator is a surjective bounded linear operator that is an open map. For a linear map, being an open map is equivalent to the existence of $c>0$ such thatThe inclusion gives a neighbourhood of $Tx$ inside the image of every sufficiently small ball about $x$, by translation and scaling. Conversely, an open map sends $B_X(0,1)$ to a neighbourhood of zero. A linear open map onto $Y$ is automatically surjective, since its image contains a ball and is a linear subspace.

The Banach-space adjoint is $T^{\ast }g=g\circ T$. For $g\in Y^{\ast }$, the ball inclusion givesAlso $\Vert T^{\ast }g\Vert \le \Vert T\Vert \Vert g\Vert$. Thus the Banach-space adjoint is an isomorphic embedding, with lower bound $c$.

The open mapping theorem states that a surjective bounded linear operator $T:X\to Y$ between Banach spaces is an open map: $T(U)$ is an open set in $Y$ whenever $U$ is an open set in $X$. Equivalently, there is $\delta >0$ such thatwhere both displayed balls are open. In particular, a bijective bounded linear operator between Banach spaces has a bounded inverse, the bounded inverse theorem.

Coordinate projections are open maps, so the local form of the submersion proves that $f$ is open. If $X$ is compact and nonempty, its continuous image is compact and hence closed in the Hausdorff manifold $Y$. Therefore $f(X)$ is a nonempty set both open and closed in connected $Y$, giving $f(X)=Y$.

True. Let $g=f^{-1}$. For every open $U\subseteq X$,which is open in $Y$ because $f$ is an open map. Thus $g$ is continuous. Together with the given continuity and bijectivity of $f$, this establishes a homeomorphism. This is inverse continuity from an open bijection.

False. The constant continuous function $f:\mathbb{R}\to \mathbb{R}$, $f(x)=0$, sends the open set $\mathbb{R}$ to $\{0\}$, which is not open. Therefore continuity alone does not imply an open map.

Let $X=\mathbb{R}$ with the usual metric and let $Y=\mathbb{R}$ with the discrete metric, and take $f(x)=x$. Every subset of $Y$ is an open set, so the image of every open subset of $X$ is open: $f$ is an open map.

It is not continuous at any point. For $ϵ=1/2$, every $\delta >0$ allows a distinct $y$ with $|y-x|<\delta$, but the distance between $f(y)$ and $f(x)$ in the discrete metric is 1. Equivalently, the preimage of the open singleton $\{x\}\subset Y$ is not open in $X$. Thus $\boxed{\text{an open map need not be continuous}}$ even between metric spaces.

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

For the quotient map $\pi$, a subset $U$ of the quotient is open exactly when ${\pi }^{-1}(U)$ is open in $X$. This definition immediately makes $\pi$ continuous. If $g$ is continuous then $g\circ \pi$ is continuous. Conversely, continuity of the composition gives, for every open $V\subseteq Z$, that ${\pi }^{-1}(g^{-1}(V))$ is open. By the defining criterion $g^{-1}(V)$ is open, so $g$ is continuous.

A map $h$ constant on each equivalence class defines a unique map $\bar{h}([x])=h(x)$. It is well-defined by the assumed class constancy, and its composition with $\pi$ is $h$. The preceding criterion proves that $\bar{h}$ is continuous.

For the integer lattice quotient, defineThis continuous map is onto the product of the two unit circles, and its fibres are exactly the lattice equivalence classes. It therefore induces a continuous bijection $\bar{F}:{\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{T}}^2$. Every class has a representative in $[0,1{]}^2$, so the quotient is the continuous image of that compact square and is compact. The target is Hausdorff as a subspace of ${\mathbb{C}}^2$. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, provingFinally, if $U\subseteq {\mathbb{R}}^2$ is open, then ${\pi }^{-1}(\pi (U))={\bigcup }_{m\in {\mathbb{Z}}^2}(U+m)$ is open, being a union of translates. The quotient criterion makes $\pi (U)$ open. Thus the quotient map is an open map.

For the canonical surjection $p:X\to X/\sim$, define a subset $W$ of $X/\sim$ to be open exactly when $p^{-1}(W)$ is open in $X$. This is the quotient topology. The empty set and the whole quotient have open inverse images. Moreover inverse images commute with arbitrary unions and finite intersections, so the corresponding axioms of a topology follow directly from those on $X$.

The map $p$ need not be an open map. For example, on $X=\mathbb{R}$ identify every point of $[0,1]$ to one point and leave all other classes singletons. The set $U=(-1/2,1/2)$ is open, butis not open. The definition of the quotient topology therefore shows that $p(U)$ is not open. The obstruction is that saturating an open set by equivalence classes need not preserve openness.

A quotient operator is a surjective bounded linear operator that is an open map. It is equivalent to requiring a ball about zero in the image of the open unit ball. Translation and scaling then prove being an open map at every point. For a Banach space domain, a bounded linear operator is a quotient operator exactly when its Banach-space adjoint is an isomorphic embedding: a lower bound on the adjoint gives approximate ball-surjectivity by Hahn-Banach separation, and successive approximation for quotient operators gives exact preimages.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.