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An open map sends every open subset of its source to an open subset of its target. This is an image property; continuity instead concerns preimages of open sets. A local diffeomorphism has both properties, but openness alone does not promise unique recovery of an input. In the periodic coordinate map under the inverse function theorem, open patches give open output regions while multiple angle values still give the same output.

Together, open and closed maps describe the two image-set preservation properties.

Because the image is integer-valued, the open interval $(n-\frac{1}{2},n+\frac{1}{2})$ contains exactly one possible image value, namely $n$. ThusThe right side is open in $A$ by continuity, soThis includes the possibility of an empty fiber. The source notation $f^{-1}(n)$ is understood as the preimage of the singleton $\{n\}$.

A Möbius transformation maps generalized circles to generalized circles on the Riemann sphere. A generalized circle is an ordinary circle or a straight line together with the point at infinity. Its image is a straight line precisely when it contains infinity, equivalently when the preimage of infinity lies on the extended real axis.

If $c=0$, the pole is infinity itself and the transformation is the affine map $(a/d)z+b/d$, taking the extended real axis to a line. If $c\ne 0$, the pole is $-d/c$, which is real exactly when $d/c$ is real. Thus the full condition isEquivalently, it is $\mathrm{Im}(d\bar{c})=0$, a condition also valid when $c=0$. If it fails, the image misses infinity and is therefore an ordinary circle. The nonzero determinant ensures that none of these generalized circles collapses to a point.

For a Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$, the preimage $f^{-1}(U)$ of every open set $U$ is a Lebesgue measurable set. For a Borel measurable function, each such preimage must instead be a Borel set. Equivalently, both definitions can test all Borel target sets: the collectionis a sigma-algebra, since preimages preserve complements and countable unions. Containing all open sets therefore makes it contain the Borel sigma-algebra.

Now let $U$ be open. Since $h$ is a Borel measurable function, $h^{-1}(U)$ is Borel. Since $g$ is a Lebesgue measurable function, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set. ButThus $h\circ g$ is a Lebesgue measurable function. The intermediate sigma-algebra is Borel; this is why Borel postcomposition preserves Lebesgue measurability.

An observed value $k$ must lie in the image of $f$. Choose either preimage $x_k$. The two-to-one function promise and $s\ne 0^n$ imply that its preimages are exactly $x_k$ and $x_k\oplus s$. A computational-basis measurement of the second quantum register projects the joint quantum state onto the unnormalized vectorBy the Born rule, $\mathrm{Pr}(k)=2^{1-n}$. Dividing by its square root gives the conditional first-register pure stateChoosing the other preimage merely interchanges the two summands. The result is a Simon coset state, retaining quantum coherence between the two inputs that share the measured function value. Outcomes outside the image have probability zero and hence no conditional state to normalize.

For compact connected oriented $n$-manifolds without boundary, the degree is characterized byfor every smooth top-degree form $\eta$. Equivalently, at any regular value $y$,The preimage is finite by compactness and the inverse function theorem. Taking a form supported in a small neighbourhood of a regular value and applying change of variables proves the signed-count formula. The top de Rham cohomology of a compact connected oriented manifold is one-dimensional, with integration giving its coordinate; pullback and Stokes theorem make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the regular value. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under homotopy, as follows by applying Stokes theorem to the pullback on $M\times [0,1]$.

For the polynomial assertion take $k\ge 1$. The map $P:\mathbb{C}\to \mathbb{C}$ extends smoothly to a map of the Riemann sphere fixing infinity: in reciprocal coordinates its target coordinate isThe homotopy $P_t(z)=z^k+t(a_1{z}^{k-1}+\cdots +a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the regular value one, each orientation-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\mathrm{deg}P=k\ne 0$. It is onto the sphere and must attain zero at a finite point. This proves every nonconstant complex polynomial has a root, without assuming the result to count roots of a general polynomial.

For a proper smooth $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$, choose a compactly supported $n$-form $\eta$ with $\int \eta =1$ and defineProperness makes its pullback compactly supported. If ${\eta }^{\prime }$ is another choice, the given result gives $\eta -{\eta }^{\prime }=d\beta$ with compact support. The difference of the defining integrals is $\int d(F^{\ast }\beta )=0$, proving independence. Sard's theorem supplies a regular value. A proper map has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-function neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an integer.

Finally $x\mapsto x^k$ is proper for $k\ge 1$. For odd $k$, a positive regular value has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. ConsequentlyFor $k=0$ it is the constant map one, whose inverse image of the compact singleton $\{1\}$ is all of $\mathbb{R}$. It is not proper, so its degree is not defined by the proper-map construction requested here.

Use the characterization of measurability by preimages of open half-lines, allowing intermediate extended-real-valued functions. Put $u_m={\inf }_{n\ge m}{f}_n$. For each real $a$,Thus every $u_m$ is measurable. Countable suprema are measurable too, since $\{{\sup }_m{u}_m>a\}={\bigcup }_m\{u_m>a\}$ and the upper half-lines also generate the Borel sigma-algebra. Thereforeis measurable. Its values are real by the stated hypothesis, so it is the required ordinary real-valued measurable function, rather than only an extended-real-valued one.

The two child intervals partition their parent:Their preimages under $g$ satisfy the same identity. Since all $A$ sets lie inside $f(M)$ and $f$ is onto its image, equality of their preimages under $f$ impliesOn the first child, $h_{n+1}=2k{2}^{-n-1}=k{2}^{-n}=h_n$; on the second it is $h_n+2^{-n-1}$. Outside the image both functions are zero. ThusFor $m>n$, summing these bounds givesThis tends to zero uniformly in $m>n$, proving that each real sequence $h_n(t)$ is Cauchy and has a finite limit.

measurability with respect to $\sigma (f)$ gives a Borel set $C_{n,k}$ with $g^{-1}(B_{n,k})=f^{-1}(C_{n,k})$. Restrict it to the actual image:This is Borel by the assumed Borel property of $f(M)$, and its preimage under $f$ is unchanged.

The intervals $B_{n,k}$ partition the real line, so their preimages under $g$ partition $M$. If two of the restricted sets overlapped, a point in their intersection would equal $f(x)$ for some $x$, making $g(x)$ lie in two disjoint intervals, a contradiction. Similarly any $t\in f(M)$ has a preimage $x$, whose real value $g(x)$ belongs to one of those intervals; that puts $t$ in the corresponding $A_{n,k}$. Thus the restricted sets are disjoint and their union is exactly $f(M)$. This restriction is what removes arbitrary overlaps of the initially chosen Borel sets outside the image.

The closure $\mathrm{cl}(A)$ is the intersection of all closed subsets containing $A$; equivalently it consists of the points whose every open neighbourhood meets $A$.

Suppose $f$ is continuous and $x\in \mathrm{cl}(A)$. For any open neighbourhood $V$ of $f(x)$, continuity makes $f^{-1}(V)$ an open neighbourhood of $x$, so it meets $A$. Hence $V$ meets $f(A)$, showing $f(x)\in \mathrm{cl}(f(A))$. This proves $f(\mathrm{cl}(A))\subseteq \mathrm{cl}(f(A))$.

Conversely suppose this inclusion holds for every subset $A$. For a closed subset $C\subseteq Y$, let $A=f^{-1}(C)$. Then $f(A)\subseteq C$ and therefore $\mathrm{cl}(f(A))\subseteq C$. By the assumed inclusion, $f(\mathrm{cl}(A))\subseteq C$, so $\mathrm{cl}(A)\subseteq f^{-1}(C)=A$. Since $A\subseteq \mathrm{cl}(A)$ always, $A$ is closed. Thus every closed set has closed preimage, equivalently every open set has open preimage. This is continuity, proving the claimed equivalence.

True. For each $a\in A$, the definition of preimage givesThus $\boxed{f^{-1}(Z\cap W)=f^{-1}(Z)\cap f^{-1}(W)}$. Unlike the corresponding statement for images of a function, this intersection identity requires no injectivity.

A proper filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes $\varnothing$, is closed under finite intersections, and is upward closed under inclusion. In particular $X\in \mathcal{F}$. In a topological space, convergence of a filter $\mathcal{F}\to x$ means that every neighbourhood of $x$ belongs to $\mathcal{F}$. An ultrafilter is a proper filter maximal under inclusion; equivalently, it contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To justify the equivalence, if $A\notin \mathcal{U}$ and every member of $\mathcal{U}$ met $A$, adjoining $A$ would generate a larger proper filter. Maximality therefore supplies $F\in \mathcal{U}$ with $F\cap A=\varnothing$, and upward closure gives $X\setminus A\in \mathcal{U}$. Both complementary sets cannot belong to a proper filter. Conversely, a filter with this dichotomy cannot acquire any new set without also containing its complement, which would give an empty intersection.

The ultrafilter characterization of compact Hausdorff spaces, stated without proof, is: a topological space is compact and Hausdorff exactly when every ultrafilter on it converges to a unique point.

For a finite set partition $X=A_1\cup \cdots \cup A_n$, if no cell belonged to $\mathcal{U}$, the ultrafilter dichotomy would put all $X\setminus A_j$ in $\mathcal{U}$. Their finite intersection is empty, a contradiction. Thus some $A_j\in \mathcal{U}$, and at most one cell can belong because distinct cells are disjoint. The converse is: a proper filter on a set meeting every finite partition by containing one cell is an ultrafilter. Apply the partition hypothesis to $X=A\cup (X\setminus A)$ to obtain the dichotomy, then the maximality argument above. Empty cells can be omitted, and the cases $A=\varnothing$ or $A=X$ are automatic.

For a function $f:X\to Y$, the image of a filter isSince preimages preserve inclusions and finite intersections, this is a proper filter on a set $Y$. Equivalently, it is the filter generated by $f(F)$ for $F\in \mathcal{F}$: if $f^{-1}(B)\in \mathcal{F}$, then its image is contained in $B$, while $f(F)\subseteq B$ implies $F\subseteq f^{-1}(B)$. For an ultrafilter $\mathcal{U}$, the identity $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$ shows that $f_{\ast }\mathcal{U}$ contains exactly one of $B,Y\setminus B$. Therefore the pushforward filter of an ultrafilter is an ultrafilter.

Now let $x\in {\ell }^{\infty }$, the l-infinity sequence space, and put $a_x(n)=n^{-1}{\sum }_{j=1}^n{x}_j$. All these Cesàro means lie in the closed disk $K_x=\{z:|z|\le \Vert x{\Vert }_{\infty }\}$, a compact Hausdorff space. The pushforward filter $a_{x\ast }\mathcal{U}$ contains $K_x$; restricting it to that disk and applying the ultrafilter theorem gives a unique limitFor real sequences the disk can be replaced by a closed interval. This is the ultrafilter construction of a Banach limit.

The identities $a_{\alpha x+\beta y}(n)=\alpha a_x(n)+\beta a_y(n)$ give linearity of $l$. Explicitly, the two sets on which $|a_x(n)-l(x)|<\delta$ and $|a_y(n)-l(y)|<\delta$ both belong to $\mathcal{U}$, hence so does their intersection. On that intersection the difference from $\alpha l(x)+\beta l(y)$ is at most $(|\alpha |+|\beta |)\delta$, proving $l(\alpha x+\beta y)=\alpha l(x)+\beta l(y)$. If $x_n\ge 0$, its means lie in $[0,\Vert x{\Vert }_{\infty }]$, so $l(x)\ge 0$. Thus $l$ is a positive linear functional. The bound $|l(x)|\le \Vert x{\Vert }_{\infty }$ gives $\Vert l\Vert \le 1$, while the constant-one sequence satisfies $l(1)=1$. Hence $\boxed{\Vert l\Vert =1}$.

Because $\mathcal{U}$ contains no finite sets, its dichotomy puts every cofinite set in $\mathcal{U}$. If $x_n\to L$ ordinarily, the Cesàro theorem for convergent sequences follows directly by splitting the sum: for a fixed $N$ with $|x_j-L|<\epsilon$ for $j>N$,The finite head tends to zero, so $a_x(n)\to L$ ordinarily. Every sufficiently long tail belongs to $\mathcal{U}$, making its ultralimit also $L$. Thus $\boxed{x_n\to L⟹l(x)=L}$.

Finally, for the left shift on bounded sequences $S$,Taking ultralimits along the same cofinite-containing ultrafilter gives $\boxed{l(Sx)=l(x)}$. This establishes every defining property of a Banach limit.

The preimage isIt contains $0$, and if $u,v$ belong to it and $a,b\in \mathbb{R}$, then $T(au+bv)=aT(u)+bT(v)\in X$, since $X$ is a vector subspace. Thus $T^{-1}(X)$ is a vector subspace. It contains the kernel of $T$ because $0\in X$.

To establish the dimension of the preimage of a vector subspace formula without assuming the theorem to be deduced, choose a basis $k_1,\dots ,k_{n_T}$ of $\ker T$ and a basis $y_1,\dots ,y_d$ of $X\cap \mathrm{im}T$, where $d=\dim (X\cap \mathrm{im}T)$. For each $j$, choose $u_j\in U$ with $T(u_j)=y_j$. These $u_j$ lie in $T^{-1}(X)$.

The family $k_1,\dots ,k_{n_T},u_1,\dots ,u_d$ is linearly independent: applying $T$ to a vanishing linear combination forces all $u_j$ coefficients to vanish by linear independence of the $y_j$, and then all remaining coefficients vanish by linear independence of the $k_i$. It also spans $T^{-1}(X)$: for $u$ in that preimage, write $T(u)={\sum }_j{b}_j{y}_j$. Then $u-{\sum }_j{b}_j{u}_j\in \ker T$, so it is a linear combination of the $k_i$. ConsequentlySet $X=V$. Then $T^{-1}(V)=U$ and $V\cap \mathrm{im}T=\mathrm{im}T$, giving the rank-nullity theorem:

A map of topological spaces is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ of $Y$. Equivalently, the preimage of every closed subset of $Y$ is closed in $X$.

A basis for a topology is a family of open sets such that every open set is a union of members of the family. Equivalently, for every point in an open set there is a basis member containing that point and contained in the open set.

For a basis $\mathcal{B}$ on $Y$, continuity is equivalent to $f^{-1}(B)$ being open for every $B\in \mathcal{B}$. Necessity follows because basis sets are open. Conversely, taking preimages of a union of basis sets gives a union of open preimages, so every open set has an open preimage.

The map is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ in $Y$. Equivalently the preimage of every closed set in $Y$ is closed in $X$, since preimages commute with complementation.

Let $S=\{x:f(x)\ne f(a)\}$, which is countable. For any closed set $C\subseteq \mathbb{R}$, consider its preimage. If $f(a)\in C$, then $a\in f^{-1}(C)$ and that preimage is closed by definition. If $f(a)\notin C$, then $f^{-1}(C)\subseteq S$, so it is countable and again closed. Therefore every closed set has a closed preimage, proving continuity.

Let $(U_i{)}_{i\in I}$ be an open cover of $f(X)$, with the sets open in $Y$. The preimages $f^{-1}(U_i)$ form an open cover of $X$, because $f$ is continuous. By compactness, finitely many of these preimages cover $X$. Their corresponding sets $U_i$ cover $f(X)$, providing a finite subcover. Hence the continuous image is compact.

True in both cases. For every open $U$, continuity of $f$ makes $f^{-1}(U)$ open. Thenis Lebesgue measurable by measurability of $g$. The value $f(0)$ has no bearing on this preimage argument.

For a fixed centre $a$, the triangle inequality gives $|d(a,x)-d(a,y)|\le d(x,y)$, so the distance-to-centre function is continuous. Its preimage of $(-\infty ,r]$ is the closed ball, hence is closed. Since that closed set contains the corresponding open ball, it contains its closure:Here the bar on the right denotes a closed ball, while the bar on the left denotes closure.