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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.

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.

A topological space is compact if every open cover has a finite subcover. It is connected if it cannot be written as the union of two disjoint nonempty open subsets. It is Hausdorff if every two distinct points have disjoint open neighborhoods. These definitions apply to the space itself and, with the subspace topology, to its subsets.

The difference map $(z,w)\mapsto z-w$ is continuous and $X$ is its inverse image of the open set $\mathbb{C}\setminus \{0\}$, so $X$ is open in ${\mathbb{C}}^2$. For fixed $z$, the equivalence class is $\{z\}\times (\mathbb{C}\setminus \{z\})$, homeomorphic to a punctured plane. It is path-connected: two points can be joined by a polygonal path avoiding the missing point. Thus all equivalence classes are connected.

The projection $\pi$ is continuous and onto, since $(z,z+1)\in X$. If $U$ is open in $X$, it is open in ${\mathbb{C}}^2$ because $X$ is open. Product-coordinate projection is open: around any $(z,w)\in U$ there is a product of open disks contained in $U$, whose first disk is a neighborhood inside $\pi (U)$. Hence $\pi (U)$ is open.

Its fibers are precisely the equivalence classes, so it induces a bijection $\bar{\pi }:X/R\to \mathbb{C}$. Continuity follows from the quotient topology and $\pi =\bar{\pi }\circ q$. If $V$ is open in $X/R$, then $q^{-1}(V)$ is open in $X$ and $\bar{\pi }(V)=\pi (q^{-1}(V))$ is open. Thus the inverse is continuous too, proving $\boxed{X/R\cong \mathbb{C}}$ as topological spaces. The plane is connected, so the connected-fiber equivalence in (b)(ii) gives $\boxed{X\text{ is connected}}$.

In any topological space a closed subset of a compact set is compact: add its open complement to an open cover and take a finite subcover of the compact set. However, compact subsets need not be closed in a general space. In the indiscrete two-point space a singleton is compact but not closed. Conversely a closed subset need not be compact: $\mathbb{R}$ itself is closed in $\mathbb{R}$ but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover.

In a Hausdorff space every compact subset $K$ is closed. Given $x\notin K$, separate $x$ and each $k\in K$ by disjoint neighborhoods; compactness selects finitely many neighborhoods covering $K$, and the intersection of their companion neighborhoods of $x$ misses $K$. Thus the complement is open. Closedness still need not imply compactness, as the same real-line example shows. It does imply compactness if the ambient space is compact.

In a metric space compactness implies boundedness: finitely many radius-one balls cover the set, so their centers have a finite maximal distance from any fixed center and the triangle inequality gives a global bound. The converse is false. An infinite set with discrete metric $d(x,y)=1$ for $x\ne y$ is bounded, but its singleton cover has no finite subcover. It is even closed and complete, so closedness plus boundedness alone does not suffice in general metric spaces.

Two continuous maps $f_0,f_1:X\to Y$ are homotopic if there is a continuous mapHere the Cartesian product has the product topology. Such an $H$ is a homotopy, written $f_0\simeq f_1$.

The topological spaces $X,Y$ are homotopy equivalent if there are continuous maps $f:X\to Y$ and $g:Y\to X$ such thatEach is then a homotopy inverse of the other.

For $A\subseteq X$ with its subspace topology with inclusion map $i:A\hookrightarrow X$, a homotopy retract, in the course's convention, is a continuous map $r:X\to A$ withThus $r$ is an actual retraction, and $i,r$ give a homotopy equivalence. Requiring the latter homotopy to fix $A$ pointwise at all times gives the stronger condition of a deformation retraction as used in this wiki. The course's definition of homotopy retract requires no such restriction on that homotopy.

Put $A=T^2\setminus B^2$ and let $E=\partial A=S^1$ denote the attaching circle. The version of the Seifert-van Kampen theorem used here states that if $X=U\cup V$ is an open cover with $U,V,U\cap V$ path-connected and a common base point in $U\cap V$, thenThe expression on the right means the pushout of groups of the two inclusion-induced group homomorphisms; they need not be injective.

Choose a thin collar neighbourhood $C_A$ of $E$ inside $A$, and a thin equatorial band $C_S$ inside $S^2$. For example, before the square's side identifications take $C_A=\{v:1\le |v|<1+ϵ\}$ with $0<ϵ<1/2$, and take $C_S=\{(x,y,z)\in S^2:|z|<ϵ\}$. In the glued topological space setThese are open subsets: their preimages in the two pieces are open and respect the gluing. They cover $X$, and the three deformation retractionsfix $E$. Thus all three spaces are path-connected, and a base point $b\in E$ is preserved.

For completeness, the fundamental group of a once-punctured torus is $F(a,b)$, and its boundary loop of a once-punctured torus represents $[a,b]=ab{a}^{-1}{b}^{-1}$, up to orientation and a change of base point. Here is the geometry behind both facts. In the square model, move each point of $A$ radially outwards to the square boundary:The denominator never vanishes on $A$. The homotopy stays outside the missing open disk, stays inside the square, and fixes the square boundary, so it descends through the side identifications. The resulting retract is a wedge of two circles. Its fundamental group is the free group on the two oriented side based loops $a,b$. The inner boundary loop travels around the square once under this deformation retraction; its successive sides read $a,b,a^{-1},b^{-1}$.

The sphere $S^2$ is simply connected. Hence the group homomorphism from ${\pi }_1(E)\cong \mathbb{Z}$ to ${\pi }_1(V)$ is trivial, while its map to ${\pi }_1(U)$ sends a group generator to $[a,b]$. Applying the Seifert-van Kampen theorem givesThe final group isomorphism follows because the single relator makes $a,b$ commute, leaving the free abelian group on two group generators. The two hemispheres kill the same boundary loop; they do not create another group generator.

For a covering map over a nonempty connected space, the degree of a covering map isthe cardinality of any fibre of a function. To show independence of $x$, choose an evenly covered neighbourhood $U$. Each covering sheet contains exactly one point over every $y\in U$, so $|p^{-1}(y)|$ equals the number of covering sheets over $U$. The fibre of a function cardinality is therefore locally constant. For a fixed cardinal number $\kappa$, the points with that cardinality form an open subset, and their complement is also open. Connectedness forces the value to be constant everywhere.

For a path-connected base there is also an explicit fibre bijection by path lifting: lift a path from $x$ to $y$ starting at each point over $x$ and take its endpoint. The path lifting theorem gives existence and uniqueness; path reversal gives the inverse bijection. The bijection may depend on the path, but the resulting equality of cardinalities does not.

A connected base is needed for one global degree. Without it, use a degree on each connected component on which the cardinality is constant. For example, a covering map of a discrete two-point topological space can have one point over the first point and two over the second. Thus the unqualified printed definition needs the usual connected-base convention, or the explicit assumption that every fibre of a function has the same cardinality.

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.

A topological space is compact if every open cover has a finite subcover:There is no Hausdorff requirement in this definition.

A topological space is Hausdorff when any two distinct points have disjoint open neighbourhoods.

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is a finite union of finite sets and cannot be all of $Z$. Thus $Z$ is a hyperconnected space.

If the continuous map $f$ took distinct values at $s,t\in Z$, choose disjoint open neighbourhoods $G,H$ of $f(s),f(t)$ in the Hausdorff space $Y$. By continuity, $f^{-1}(G)$ and $f^{-1}(H)$ would be nonempty disjoint open sets of $Z$, a contradiction. ThereforeThe proof establishes the general fact that continuous maps from an infinite cofinite space to a Hausdorff space are constant.

Fix $x_0,y_0\in Z$. For a continuous function $f:Z\times Z\to \mathbb{R}$, every coordinate slice is continuous, because the maps $x\mapsto (x,y)$ and $y\mapsto (x_0,y)$ into the product topology are continuous. Their target $\mathbb{R}$ is Hausdorff. Part (a)(i) givesTherefore every such $f$ is a constant function. This is the two-factor case of continuous maps from products of cofinite spaces to Hausdorff spaces are constant.

The second coordinate projection ${\pi }_2:Z\times \mathbb{R}\to \mathbb{R}$ is continuous and nonconstant. If a homeomorphism $H:Z\times Z\to Z\times \mathbb{R}$ existed, ${\pi }_2\circ H$ would be a nonconstant continuous function, contradicting the preceding conclusion. Hence

For the half-plane, put $H_{+}=[0,\infty )$ as a set, so $K=\mathbb{R}\times H_{+}$. Its two ambient topologies give different answers:In $Z\times Z$, if $K$ were closed, its inverse image under the continuous map $y\mapsto (x_0,y)$ would be closed in $Z$. That inverse image is $H_{+}$, which is neither a finite set nor all of $Z$, so it is not closed in the cofinite topology. Thus $K$ is not closed.

To establish compactness, note that the subspace topology on any subset $A$ of a topological space with the cofinite topology is again cofinite. An open cover of a nonempty such $A$ has a nonempty member omitting at most finitely many points; finitely many more members cover those points. Hence every subset of a cofinite space is compact. In particular both $Z$ and $H_{+}$ with its cofinite subspace topology are compact. The subspace topology on $K\subseteq Z\times Z$ is their product topology, as one checks by tracing basis rectangles. The product of two compact spaces is compact, so $K$ is compact here.

For clarity, the finite-product compactness result needs no Hausdorff condition. Given an open cover of $A\times B$, cover each point by a basis rectangle lying in one covering member. For fixed $a\in A$, use compactness of $B$ to select finitely many rectangles whose second factors cover $B$, and intersect their first factors to get an open neighbourhood $U_a$ of $a$. Those finitely many original members cover $U_a\times B$. Finally finitely many $U_a$ cover $A$, producing a finite subcover of $A\times B$.

In $Z\times \mathbb{R}$, the second coordinate projection hasSince the ray is closed in the usual real line, $K$ is closed. It is not compact: its image under a continuous map under ${\pi }_2$ is $[0,\infty )$, which is not compact. For an explicit obstruction, the open intervals $(-1,n)$, $n\ge 1$, cover this ray but have no finite subcover.

Finally consider the diagonal $W$ in $Z\times \mathbb{R}$. DefineThis is a bijection. The identity function from the usual real line to $Z$ is continuous: each nonempty cofinite open set is the complement of a finite set, and finite sets are closed in the usual real line. Both coordinates of $\phi$ are therefore continuous, so $\phi$ is continuous for the product topology and the subspace topology of $W$. Its inverse is the restricted second coordinate projection, which is continuous. ThereforeFor the second conclusion, $Z$ is compact whereas $W\cong \mathbb{R}$ is not compact; homeomorphisms preserve compactness. The diagonal in a product with cofinite and Euclidean topologies thus inherits the usual topology, despite one cofinite factor.

The usual real line is connected. In contrast,is a separation of a topological space. Both pieces are nonempty and disjoint; the first is open by part (c), and the second is the open union ${\bigcup }_{n\ge 1}[0,n)$. Thus the Sorgenfrey line is not connected. Since a homeomorphism preserves connectedness, the two topological spaces are not homeomorphic. Merely showing that their topologies differ would not by itself exclude some other homeomorphism between them.

False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers $\mathbb{Q}$ with the usual subspace topology from the real line.

Every connected subset of $\mathbb{Q}$ has at most one point. Indeed, if it contained $r<s$, choose an irrational number $t\in (r,s)$. The two relatively open setspartition $\mathbb{Q}$ and separate any subset containing both $r$ and $s$. Hence the connected components of $\mathbb{Q}$ are singletons.

If a topological disjoint union of nonempty connected spaces were homeomorphic to $\mathbb{Q}$, the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But $\mathbb{Q}$ is not a discrete space: for each $r\in \mathbb{Q}$, the distinct rational numbers $r+1/n$ approach $r$, so $r$ is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.

Write $Z={\coprod }_{n\in \mathbb{N}}{X}_n$, where each $X_n$ is a labelled copy of the topological space $X$, with its topological disjoint union topology. Define the bijectionFor a subset $U\subseteq \mathbb{N}\times X$, let $U_n=\{x\in X:(n,x)\in U\}$. By definition of the topological disjoint union, $F^{-1}(U)$ is open in $Z$ exactly when each $U_n$ is open in $X$.

The same condition characterizes open sets in the product topology on $\mathbb{N}\times X$. If $U$ is open, the map $x\mapsto (n,x)$ is continuous, so each $U_n$ is open. Conversely, if each $U_n$ is open, thenis open, since the discrete topology makes every $\{n\}$ open in $\mathbb{N}$. Hence $F$ and its inverse function are continuous, proving the required homeomorphism. This also covers the case $X=\varnothing$.

The quotient set $X/\sim$ consists of the equivalence classes $[x]=\{y\in X:y\sim x\}$. Its collapsing map is the canonical projection of a quotient setEquip this set with the quotient topology:This is a topology because preimages preserve arbitrary unions, finite intersections, the empty set, and the whole set. The resulting topological space is the quotient space. The displayed rule immediately implies that the preimage of every open set is open, so $p$ is continuous; it is also surjective.

If a topological space is the union of finitely many closed subspaces and continuous maps on those subspaces agree on every overlap, they combine into a continuous map on the whole space. For a closed set in the target, its preimage is a finite union of closed sets, proving continuity.

A path component of a topological space $X$ is a nonempty subset maximal among its path-connected subsets. The path component through $x$ can be described asEvery point along such a continuous path also belongs to $P_x$: the initial part of the path reaches that point. Thus the paths witnessing membership stay inside $P_x$. This makes path components useful for deciding which allowed positions can be reached from one another by continuous motion.

Every path component lies inside one connected component, because every path-connected space is connected. Thus each connected component is a union of path components. In a locally path-connected space the two classifications agree, by connected locally path-connected spaces are path-connected. For a subset of the real line with its usual subspace topology, the classifications also agree: every connected component is a real interval, and interpolation supplies a continuous path within it.

The topologist's sine curve shows why the distinction matters in general. Use specifically the closed versionwith the subspace topology from the usual Euclidean plane. It has one connected component, $T$, but two path components, $G$ and $L$. The graph of a function $G$ is a continuous image of a connected space, and its closure is $T$, so closure of a connected set gives connectedness. Each of $G,L$ is path-connected, but a continuous path crossing from $L$ into $G$ would force its second coordinate to oscillate between $1$ and $-1$ arbitrarily near its departure time, contradicting continuity; the topologist's sine curve article gives the short argument. Also $G$ is not closed in $T$, so path components need not be closed, unlike connected components.^[ref]

Being joined by a continuous path is an equivalence relation. Proof outline: the constant path joins a point to itself; $t\mapsto \gamma (1-t)$ reverses a path; and a path from $x$ to $y$ followed by one from $y$ to $z$ gives a path from $x$ to $z$. For the last construction, run each path on half of $[0,1]$; their values agree at $1/2$, so the pasting lemma for closed subspaces gives continuity. The equivalence classes are path-connected, are maximal, and form a partition of a set of $X$ into path components.^[ref]

A topological space $X$ is path-connected if, for every $a,b\in X$, there is a continuous mapThis continuous path must stay within $X$ for its entire duration. Its parameter need not represent physical time, and continuity does not require a derivative or finite arc length. ^[ref]

In the closed interval $[0,1]$, any $a,b$ are joined by $\gamma (t)=(1-t)a+tb$. For $a=1/4$ and $b=3/4$, this becomes $\gamma (t)=1/4+t/2$, so $\gamma (0)=1/4$, $\gamma (1/2)=1/2$, and $\gamma (1)=3/4$. All its values stay in $[0,1]$. In the union $D=[-2,-1]\cup [1,2]$, however, a continuous path from $-1$ to $1$ would attain $0$ by the intermediate value theorem, although $0\notin D$. Thus $D$ is not path-connected.

Every path-connected space is a connected space. Proof outline: if $X=U\cup V$ admitted a separation of a topological space, choose $a\in U$ and $b\in V$. A continuous path between them would have nonempty open inverse images ${\gamma }^{-1}(U),{\gamma }^{-1}(V)$ with empty intersection and union $[0,1]$. This contradicts the connectedness of that closed interval.

The converse fails: the topologist's sine curve is connected but has two path components. Path connectedness is useful when we need an actual continuous path of allowed positions, rather than only the absence of a separation of a topological space. For instance, when modelling allowed positions as a topological space, membership in the same path component answers whether one can move continuously between two positions while remaining in that topological space.