Use the basis of a topology consisting of open rectangles:It covers $X\times Y$, since $X\times Y\in \mathcal{B}$. Whenever a point lies in two basis rectangles, their intersection is another basis rectangle containing it:These are precisely the two basis of a topology axioms. Arbitrary unions of the rectangles therefore form a topology, the product topology. To check the finite-intersection axiom directly, distribute the intersection of two unions over all pairs of rectangles and use the displayed formula. The empty union gives the empty set.

If $f(A)=U\cup V$ were a separation, thenwould be disjoint nonempty relatively open sets covering $A$. They are relatively open by continuity of the restriction $f{|}_A:A\to f(A)$. This contradicts connectedness of $A$.This is the continuous image of a connected space theorem; no separation property of the target is required.

Suppose $A\cup B=U\cup V$ were a separation. By connectedness, all of $A$ lies on one side, say $U$: otherwise $A\cap U,A\cap V$ would separate $A$. A point of $A\cap B$ then lies in $B\cap U$, so connectedness of $B$ forces all of $B$ into $U$ as well. This makes $V$ empty, a contradiction.

Therefore the union of two intersecting connected subsets is connected. The argument also proves the connected union with a connected hub rule for arbitrarily many connected pieces meeting one connected subset.

A topological space is connected if it has no separation: two disjoint nonempty open sets whose union is the whole space. Equivalently, its only subsets that are both open and closed are the empty set and the whole space.

A subset is connected when its subspace topology has this property. Thus the sets in a proposed separation of a subset need only be relatively open, rather than open in the ambient space.

Apply the first extraction move to the two outer occurrences of $x$, then cyclically rotate the word to expose the two occurrences of $y^{-1}$:In the third line the repeated letter for the move is $y^{-1}$, and the intervening string is $z^{-1}{z}^{-1}$, whose inverse is $zz$. Rename $y^{-1}$ as a new edge label. The final word has the standard form $aabbcc$.

Each same-direction repeated pair represents a projective-plane summand, so the classification theorem for surfaces identifies this word with the connected sum of three real projective planes:Thus the surface has crosscap number three. As a check, the original identifications make all six polygon vertices equivalent, leaving one vertex, three edges and one face. Its Euler characteristic is $1-3+1=-1=2-3$, consistent with that crosscap count.

Label the endpoints around the original polygon by $v_0,v_1,v_2,v_3$ so that the successive boundary arcs areThe two occurrences of $x$ have the same boundary-traversal direction, as prescribed by the word. We use cutting along a new diagonal followed by gluing the old $x$ sides. Cutting introduces two copies of the diagonal with opposite orientations, so rejoining them later exactly recovers the original quotient.

First move. Draw a diagonal $d$ from $v_0$ to $v_2$. The two pieces have boundary words $xA{d}^{-1}$ and $dxB$. Glue them along their $x$ sides, turning one piece over in the planar drawing to match the arrows. Reading the outside boundary of the joined disk givesReverse the orientation of this entire disk and change the starting point of its boundary word. The same side-paired disk then has word $dd{A}^{-1}B$. Renaming the new paired edge $d$ as $x$ yields

Second move. Instead cut along $d$ from $v_1$ to $v_3$. The pieces now have words $xdB$ and $Ax{d}^{-1}$. Gluing along the old $x$ sides gives outside word $dB{A}^{-1}d$. A cyclic change of starting point gives $A^{-1}ddB$, and renaming $d$ yieldsAll side identifications inside the strings $A$ and $B$ are carried along unchanged; reversing a piece reverses the order of its string and inverts each letter. The construction therefore gives homeomorphic quotients, not merely an equality of Euler characteristics. This is the crosscap extraction move for a polygonal word.

The vertex stars form an open cover of $|K|$: every point has a nonempty positive-coordinate support, and therefore belongs to the star of at least one vertex. At a vertex $w$ itself, only its own barycentric coordinate is positive. Thus $w$ belongs to its own star and to no other vertex star.

Any finite selection of stars omits every vertex not among their finitely many labels. Infinitely many vertices therefore prevent a finite subcover, provingNo local-finiteness hypothesis is needed for this vertex-star cover obstruction to compactness.

The open star in a simplicial complex of a vertex $v$ is the union of the relative interiors of the simplices containing $v$. In barycentric coordinates it isLet $L$ be the subcomplex consisting of all simplices not containing $v$. The complement of the star is exactly $|L|$: the positive-coordinate support of a point misses $v$ precisely when it is a simplex of $L$. By the allowed closed-subcomplex theorem, $|L|$ is closed in $|K|$. Hence the star is open.

This is the open-star convention required by the question. The closed star, which also includes the opposite faces, is a different subset and need not be open.

An abstract simplicial complex $K=(V,\Sigma )$ consists of a vertex set $V$ and a collection $\Sigma$ of finite nonempty subsets containing every singleton and every nonempty subset of each member. A member with $n+1$ vertices is an $n$-simplex.

Its topological realization glues a standard geometric simplex for each member along common faces. A concrete model iswith finite support and the weak topology of a simplicial realization: a set is closed precisely when its intersection with every simplex is closed. This agrees with the quotient topology of the gluing construction. Specifying this topology matters for complexes with infinitely many vertices.

A boundary reparameterization gives an explicit homeomorphism. Let $Q=[0,1{]}^2$, with centre $c=(1/2,1/2)$, and parametrize its boundary once byHere $b(0)=b(4)$. The omitted top edge of $[0,1]\times [0,1)$ is $b([0,1])$. The omitted top and right edges of $[0,1{)}^2$ are $b([0,2])$.

Use the increasing boundary parameter mapIt takes $[0,4]$ bijectively onto itself, respects the identified endpoints, and maps the first omitted arc exactly onto the second. Extend it radially:Every noncentral point has a unique boundary ray and a unique radial parameter $r=2\Vert q-c{\Vert }_{\infty }$; the only duplicate boundary parameters are zero and four, whose images agree. The inverse uses ${\varphi }^{-1}$ by the same construction. Both maps are continuous off the centre, and at the centre their radial distances tend to zero. Thus $H$ is a homeomorphism of the whole closed square.

It carries the first omitted edge precisely onto the second omitted pair of edges, so restriction givesThis is the radial extension of a square boundary homeomorphism; it avoids any need to treat corners as topologically distinguished points.

The interval $[0,1]$ is compact by the Heine-Borel theorem, so its square is compact by part (a). The half-open interval $[0,1)$ is not compact because it is not closed in the real line.

Each of the other two products has a continuous surjective coordinate projection onto $[0,1)$. If either product were compact, that image would be compact, a contradiction. Compactness is preserved by homeomorphisms, so neither product is homeomorphic to the closed square.

First assume both factors are nonempty. If the product is compact, its continuous coordinate projections are onto, so each factor is compact by the continuous image of a compact space theorem.

For the converse, let $X,Y$ be compact and let $(W_{\lambda })$ be an open cover of their product. Fix $x\in X$. For every $y\in Y$, choose a basis rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a cover member $W_{\lambda (x,y)}$. Compactness of $Y$ gives finitely many $V_{x,y_j}$ covering $Y$. Thenis an open neighbourhood of $x$, and $U_x\times Y$ is covered by the corresponding finitely many $W_{\lambda (x,y_j)}$.

The sets $U_x$ cover $X$. Compactness gives finitely many of them covering $X$, and the finitely many cover members associated to these choices now cover $X\times Y$. Hence the product is compact. This is the finite-product Tychonoff theorem proof, with the finite intersection creating a uniform neighbourhood along each fibre.

As printed, the assertion omits the nonempty hypothesis. If $X=\varnothing$ and $Y=\mathbb{R}$, the product is compact while $Y$ is not. The fully correct version without any nonemptiness assumption is

The product topology has basisOpen subsets of the product are arbitrary unions of these rectangles. The rectangles cover the product and their finite intersections are rectangles, so they satisfy the basis axioms. Equivalently, this is the smallest topology making both coordinate projections continuous.

For a nonempty open subset $U$ of the quotient, $p^{-1}(U)$ is nonempty and open in the cofinite space. Thereforeis finite. Surjectivity implies that $(X/\mathcal{R})\setminus U$ is finite as well: it is the image of that finite preimage. Hence every quotient-open set is cofinite or empty, proving the desired inclusion.

Equality is not necessary. Give $X=\mathbb{Z}$ its cofinite topology and use two equivalence classes, the even and odd integers. Neither singleton of the quotient is open, since its preimage and its complement are both infinite. The quotient is the two-point indiscrete topology, whereas the cofinite topology on a finite two-point set is discrete.This is the quotient topology of a cofinite space phenomenon, not a failure of continuity of the projection.

The quotient set $X/\mathcal{R}$ consists of the equivalence classes. Let $p(x)=[x]$. Equip that set with the quotient topologyPreimages preserve the empty set, whole set, arbitrary unions and finite intersections, so this is a topology. The canonical projection is continuous and surjective, and the displayed equivalence specifies the quotient space completely.

Any two nonempty open sets of $Y$ are cofinite by part (iii). Their complements have finite union, so the two sets must intersect because $Y$ is infinite. Thus $Y$ is a hyperconnected space.

A Hausdorff space with at least two points has disjoint nonempty open neighbourhoods of two distinct points. This is impossible here. Therefore $Y$ is not Hausdorff.

Any proper closed subset $F$ of $Y$ must be finite: if it were infinite, part (ii) would give $F=\overline{F}=Y$. Therefore every nonempty open subset of $Y$, being the complement of a proper closed set, is cofinite.The empty set belongs to both topologies. Inclusion need not be equality; the target may have fewer open sets.

Because $f$ is surjective, $f^{-1}(B)$ is infinite whenever $B$ is infinite; a finite preimage could have only finitely many image points. By part (i), its closure is $X$. Apply part (a)(ii):The other inclusion is automatic, soBoth continuity and surjectivity enter the argument.

A proper closed subset in the cofinite topology is finite, so cannot contain the infinite subset $A$. Its only closed superset is therefore the whole space:In particular, every infinite subset is dense in this topology.

Suppose $f$ is continuous and $x\in \overline{A}$. Every open neighbourhood $V$ of $f(x)$ has an open preimage containing $x$, which meets $A$. Hence $V$ meets $f(A)$, proving $f(x)\in \overline{f(A)}$.

Conversely, suppose the indicated inclusion holds for every subset. To prove continuity, take any closed set $C\subseteq Y$ and put $A=f^{-1}(C)$. Since $f(A)\subseteq C$ and $C$ is closed,Thus $\overline{A}\subseteq f^{-1}(C)=A$, so $A$ is closed. Every closed set has closed preimage, which is the closed-set characterization of continuity.This is the closure characterization of continuity; it requires no Hausdorff or metric hypothesis.