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For topological spaces, continuity means that $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$. Equivalently, inverse images of closed sets are closed. A homeomorphism is a bijection for which both the map and its inverse are continuous.

Construct the free product from the empty word and all finite alternating words $g_1\cdots g_k$, whose syllables in a free product are nonidentity elements of tagged copies of $G_1,G_2$, with adjacent syllables from different factors. Multiply by concatenating, multiplying adjacent elements in the same factor, and deleting identities until the word is reduced. The normal form theorem for a free product gives a unique result; reducing three concatenated words gives the same result under either parenthesization, so multiplication is associative. The empty word is the identity, and the inverse reverses the word and inverts each syllable. Each factor embeds as words of length one.

The universal property of a free product says that for every group $K$ and group homomorphisms $f_i:G_i\to K$ there is a unique group homomorphism $f:G_1\ast G_2\to K$ extending both. Explicitly, send a reduced word to the product of its syllable images; reduction preserves this product. This gives existence, while generation by the two factors gives uniqueness.

A standard form of Klein's combination theorem, or the ping-pong lemma, is the following. Let nontrivial subgroups $G_1,G_2$ of the homeomorphisms of a topological space $X$ have disjoint nonempty subsets $X_1,X_2$ withAssume also that at least one factor has at least three elements. Then the generated subgroup is $\boxed{⟨G_1,G_2⟩\cong G_1\ast G_2}$. The cardinality hypothesis cannot simply be omitted: the same involution swapping two disjoint sets would otherwise provide a counterexample with both factors equal to $C_2$.

Here is the ping-pong lemma proof. A reduced word of odd length begins and ends in the same factor, so repeated application of the displayed inclusions sends the other factor's domain into that factor's domain. Disjointness shows that the word is not the identity. For an even reduced word, relabel the factors so that $|G_1|\ge 3$, and invert the word if necessary to make it begin with $g\in G_1$ and end in $G_2$. Choose $s\in G_1\setminus \{1,g^{-1}\}$. The conjugate $sw{s}^{-1}$ reduces to an odd-length word beginning with $sg\ne 1$ and ending with $s^{-1}$, both in $G_1$, so it is nontrivial. Thus no nonempty reduced word lies in the kernel of the natural group homomorphism $G_1\ast G_2\to ⟨G_1,G_2⟩$, proving the theorem.

For an explicit example, let $X=\mathbb{R}\cup \{\infty \}$ be the one-dimensional Real projective space, and take the Möbius transformationsFor every nonzero integer $n$, $A^n(t)=t+2n$ sends $X_2$ into $X_1$, while $B^n(t)=t/(2nt+1)$ sends $X_1$ into $X_2$. Indeed $|2nt+1|>|t|$ when $|t|>1$, and $B^n(\infty )=1/(2n)$. Both transformations have infinite order. Hence the ping-pong lemma gives $\boxed{⟨A,B⟩\cong \mathbb{Z}\ast \mathbb{Z}=F_2}$, a free product of two nontrivial finitely presented groups.

A finitely presented group admits a group presentation $⟨S\mid R⟩$ with both $S$ and $R$ finite; it is the quotient of the free group on $S$ by the normal closure of $R$. Ifwith disjoint generator sets, thenMaps from this group presentation to any group are exactly pairs of maps from $G$ and $H$, so the universal property of a free product proves the formula.

For a group homomorphism $\varphi :H\to \mathrm{Aut}(G)$, choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each $t\in T,s\in S$. The presentation of a semidirect product isThe presentation maps onto the specified semidirect product. Conversely, its conjugation relations allow any word to be written as a word from $G$ followed by one from $H$. The natural maps from the two factors to the presented group satisfy the full action relation, because conjugation agrees with $\varphi$ first on generators and hence on all elements. They define the reverse group homomorphism $(g,h)\mapsto gh$. The two maps are inverse on every generator, proving the group isomorphism. There are finitely many cross-relations, so the result is again a finitely presented group.

Apply this to the specified permutation action. The presentation isEliminate $y=cx{c}^{-1}$ and $z=c^{2}x{c}^{-2}$. The last cross-relation becomes $c^{3}x{c}^{-3}=x$, already implied by $c^3=1$. ThusBoth factors are nontrivial finitely presented groups, as required.

A quotient topological space of a topological space $X$ is obtained by an equivalence relation $\sim$: its points are the classes $X/\sim$, and the canonical surjection is $q(x)=[x]$. The quotient topology declares a subset $U$ of the quotient open exactly when $q^{-1}(U)$ is open in $X$. More generally a surjection onto a set specifies the same topology by this condition.

For a Hausdorff source with a non-Hausdorff quotient, use $X=\mathbb{R}$ and $x\sim y$ when $x-y\in \mathbb{Q}$. Any nonempty open inverse image is a nonempty open subset of $\mathbb{R}$ invariant under rational translations. It contains an interval $I$, and for every real $t$ there is a rational $r$ with $t-r\in I$; hence it contains every $t$. Thus $\mathbb{R}/\mathbb{Q}$ has the indiscrete topology.

It has distinct points, for instance the classes of $0$ and $\sqrt{2}$, but neither has a proper nonempty open neighbourhood. The quotient is not Hausdorff, even though $\mathbb{R}$ is a Hausdorff space.

For a point $x\in X$ and an abelian group $A$, the skyscraper sheaf $i_{\ast }A$, with $i:\{x\}\hookrightarrow X$, isRestrictions are identities or the map to zero, so it is a flasque sheaf. The global sections equal $A$ and higher cohomology vanishes:No closed-point assumption is needed for this calculation of skyscraper sheaf cohomology. On an arbitrary topological space, its nonzero stalks are at points of $\overline{\{x\}}$ when $A\ne 0$; only for a closed point is the nonzero stalk confined to $x$.

Treat $0$ as the trivial group if $G$ is not abelian. The restriction maps satisfy the presheaf identities. To verify the sheaf gluing axiom, take an open cover $(V_i)$ of $U$. If $x\notin U$, every section group is trivial. If $x\in U$, at least one $V_i$ contains $x$. Every two such opens have an overlap containing $x$, so compatibility forces all their sections to have the same value $g\in G$. Opens not containing $x$ have their unique trivial section. The value $g\in \mathcal{G}(U)$ is exactly the unique glued section. Thus the stated presheaf is a skyscraper sheaf, including for nonabelian groups.

For the stalk of a sheaf, take the direct limit over neighbourhoods of $y$. If some neighbourhood $V$ of $y$ omits $x$, neighbourhoods inside $V$ are cofinal and all their groups are trivial. If every neighbourhood of $y$ contains $x$, every group in the system is $G$ and every transition map is the identity. Consequently, on the arbitrary topological space in the question,Membership in this closure means exactly that every neighbourhood of $y$ contains $x$. In particular ${\mathcal{G}}_x=G$. If $x$ is a closed point, the closure is just $\{x\}$ and the familiar point-supported answer results. The separation assumption is not present in the printed question, so it cannot be imposed silently.

Let $\{O_i\}$ be an open cover of $Z$ in its subspace topology. Write $O_i=Z\cap U_i$ with each $U_i$ open in $X$. Since $Z$ is a closed set, $X\setminus Z$ is open, and $\{U_i\}\cup \{X\setminus Z\}$ is an open cover of the compact space $X$. A finite subcover restricts to a finite subcover of $Z$. Hence the closed subset $Z$ is compact.

If $\{V_j\}$ is an open cover of $f(Z)$ in its subspace topology, the inverse images under the restricted continuous map $f{|}_Z$ form an open cover of $Z$. Take a finite subcover; the corresponding $V_j$ cover $f(Z)$. Thus the continuous image $f(Z)$ is compact. Neither argument assumes a Hausdorff space or asserts that a compact set must be closed set in an arbitrary topological space.

A topological space is connected if it is not the union of two disjoint nonempty open sets, or equivalently has no nontrivial subset that is both an open set and a closed set.

Let $K_x$ be the union of all connected subspaces containing $x$. A union of connected subsets with a common point is connected: if a separation of a topological space existed, every member containing $x$ would lie wholly in the same side. Hence $K_x$ is connected and contains every connected subspace through $x$, so it is the connected component of $x$.

If $K_x\cap K_y\ne \varnothing$, their union is connected and maximality gives $K_x=K_y$. Otherwise they are disjoint. Since every $x$ belongs to $K_x$, the connected components partition $X$.

A topological space is connected when it is not the union of two disjoint nonempty open subsets. It is path-connected when every two points $x,y$ can be joined by a path, meaning a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.

To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. LetWhichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the supremum or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.

If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$, contradicting the connectedness just proved. Hence

Now let $X\subseteq {\mathbb{R}}^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an open ball is path-connected by straight line segments. It follows that every path component of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.

The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is locally path-connected, its path components are open, and connectedness forces there to be only one.

For the final example, putThe set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the closure of $Y$, because $(1/n,y)\to (0,y)$ for every $y\in [1/2,1]$. A connected set together with a connected subset of its closure is connected, so

It is not path-connected. The image of the first coordinate of any path in $X$ lies inbut, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup \{1/n:n\ge 1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\le y<1/2$. Hence no path joins $B$ to $Y$, and

A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

Let $L$ be any distributive lattice. Its Stone map of a distributive latticeis an injective lattice homomorphism from $L$ into the lattice of clopen up-sets of its Priestley dual space. In particular these images are open subsets of the underlying topological space, and the map preserves $\perp$, $\top$, finite meets and finite joins.

Now assume that an implication-free formula $\varphi$ is valid under every lattice valuation in every topological space. Given any valuation of its variables in any distributive lattice $L$, compose it with the Stone map. Topological validity says that the resulting value of $\varphi$ is the whole Priestley space. Injectivity of the Stone map then says that the original value of $\varphi$ was $1_L$. Hence $\varphi$ is valid in every distributive lattice.

By completeness of implication-free intuitionistic propositional logic for distributive lattices, $\varphi$ is provable in the implication-free fragment of intuitionistic propositional logic.

A regular category has finite limits, regular-epimorphism--monomorphism image factorizations, and regular epimorphisms stable under pullback. A relation $A⇝B$ in $\mathbf{Rel}(\mathcal{C})$ is a subobject of $A\times B$; composition forms the pullback over the middle object and then takes its image.

Suppose $R:A⇝B$ has a right adjoint relation $S$, so $1_A\le SR$ and $RS\le 1_B$. In the internal regular logic, the first inequality says that for every $a$ there is a $b$ with $R(a,b)$ and $S(b,a)$. If also $R(a,b^{\prime })$, the second inequality forces $b=b^{\prime }$. Thus $R$ is total and single-valued. Categorically, if $R\hookrightarrow A\times B$ has projections $p:R\to A$ and $q:R\to B$, totality makes $p$ a regular epimorphism and single-valuedness makes it a monomorphism. Hence $p$ is an isomorphism and $R$ is the graph of a morphism as a relation $q{p}^{-1}:A\to B$. Conversely, the graph of any morphism is left adjoint to its converse relation, as the two required inequalities follow directly from equality. This proves the characterization.

Let $L$ be a frame. Composition in the category of matrices valued in a frame isIf $f⊣g$, the diagonal part of $1_A\le gf$ impliesso every row of $f$ joins to $1$. For $b\ne b^{\prime }$, distribute $f(a,b)\wedge f(a,b^{\prime })$ over the displayed join. Every term vanishes by $fg\le 1_B$, first using the factor with column $b$ and then the one with column $b^{\prime }$. HenceConversely, if the rows of $f$ join to $1$ and have pairwise disjoint entries, define $g(b,a)=f(a,b)$. Thenwhile $(fg)(b,b^{\prime })=0$ for $b\ne b^{\prime }$. Thus $1_A\le gf$ and $fg\le 1_B$, proving the stated criterion.

Finally take $L=\Omega (X)$ for a connected topological space $X$. For fixed $a$, the opens $f(a,b)$ are pairwise disjoint and cover $X$. Connectedness forces exactly one of them to be $X$ and all the others to be empty. Hence a left adjoint matrix determines a unique function $\phi :A\to B$ by $f(a,\phi (a))=X$. Conversely every function gives this matrix, and matrix composition agrees with function composition. The left adjoints in $\mathbf{Mat}(\Omega (X))$ therefore form a category isomorphic to $\mathbf{Set}$.

A topological space is connected if it cannot be expressed as a union of two disjoint nonempty sets open in that space. Suppose $f(X)$ had such a separation $U\cup V$. Here $U,V$ are open in the subspace topology of $f(X)$. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$. They are disjoint, nonempty because $U,V$ contain image points, and cover $X$. This contradicts connectedness. Hence a continuous image of a connected space is connected.

A topological space is connected when it cannot be written as the union of two disjoint nonempty open sets. Equivalently, the only subsets that are simultaneously open and closed are the empty set and the whole space. For a subset, connectedness refers to its subspace topology.

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 $X$.

An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

In an arbitrary topological space, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb{R}$ is closed in itself but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a compact space $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.

In a Hausdorff space, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint open sets $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.

In every metric space, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the triangle inequality bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus compact implies closed and bounded in a metric space, while closed and bounded need not imply compact.

A subset $K$ of a topological space is compact if every cover of $K$ by open subsets of the ambient space has a finite subcover. This is equivalent to compactness of $K$ with its subspace topology.

In a metric space,The sequence may equivalently be required to have pairwise distinct terms. The metric hypothesis makes a countable sequence sufficient; arbitrary topological spaces need not have this sequential characterization.

A topological space is compact if every open cover of it has a finite subcover. That is, if $X={\bigcup }_{i\in I}{U}_i$ with all $U_i$ open, finitely many of those same open sets already cover $X$.

A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, its only subsets that are both open and closed are the empty set and the whole space. Under this definition the empty space is connected.