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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 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.

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.