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With local path connectedness, the intended conclusion holds. Choose a base point $x_0$ and the standard covering mapThe fundamental group of the circle is $\mathbb{Z}$. Since the image of a finite group under a group homomorphism is finite and $\mathbb{Z}$ has no nontrivial finite subgroups,is trivial. If $X$ is also locally path-connected, the lifting criterion for a covering space supplies a continuous map $\tilde{f}:X\to \mathbb{R}$ with $p\tilde{f}=f$, because $p_{\ast }{\pi }_1(\mathbb{R})=0$. The homotopyruns from $f$ to the constant function $f(x_0)$. This proves the claimed null-homotopy under that additional hypothesis, in particular for CW complexes and the topological graphs considered later in the paper.

For arbitrary path-connected spaces, the printed assertion is false. Here is an explicit Warsaw circle counterexample, including the obstruction that the usual lifting argument misses. In the Euclidean plane takeJoin $P=(1,0)$ to $Q=(0,-1)$ by the polygonal arc $A$ passing successively through $(2,0),(2,-2),(0,-2)$, and put $W=G\cup L\cup A$ with its subspace topology. The arc meets $G\cup L$ only at $P,Q$. The topologist's sine curve $G\cup L$ has two path components, but the added arc joins them, making $W$ path-connected.

Every loop in $W$ is contained infor some $\delta >0$. To justify the assertion, a path cannot reach $L$ along the oscillating tail: the intermediate value theorem for its $x$-coordinate would force its $y$-coordinate to attain both $1$ and $-1$ arbitrarily close to the supposed arrival time, contradicting continuity. The only alternative route from the tail to $L$ uses $P$ and the joining arc. If a loop visited tail points with $x\to 0$, a convergent subsequence of their parameter values would approach a time at which the loop lies in $L$. On a sufficiently small parameter interval around that time its image avoids $P$. That would give a forbidden path from $L$ into the tail without passing through $P$. Thus those tail coordinates have a positive lower bound. The space $T_{\delta }$ is a finite tree, in fact a single arc, so every such loop contracts. ConsequentlyNow define a continuous map $h:W\to S^1$ to be $1$ on $G\cup L$, and to make one full turn on $A$: for a parametrization $a:[0,1]\to A$ from $P$ to $Q$, set $h(a(t))=e^{2\pi it}$. The pieces $G\cup L$ and $A$ are closed in $W$, and the formulas agree at $P,Q$, so the pasting lemma for closed subspaces proves continuity.

There is no continuous lift $\tilde{h}:W\to \mathbb{R}$. Any such lift is integer-valued on $G\cup L$, so it is constant on each of the connected spaces $G,L$. Since $G$ accumulates on $L$, continuity forces those two constants to be the same. But along $A$ a lift of $e^{2\pi it}$ has the form $t+k$, giving values at $Q$ and $P$ differing by $1$, a contradiction. Finally, if $h$ had a null-homotopy, the homotopy lifting theorem for a covering map, starting with a lift of the final constant function and reversing time, would produce a continuous lift of $h$. This theorem needs no local path connectedness of the parameter space. Hence $h$ is not null-homotopic, although $W$ has finite, indeed trivial, fundamental group.

The native PDF explicitly assumes only path connectedness here. The missing local path connectedness cannot be inferred from the finiteness of the fundamental group.

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.

Let $F$ be a closed subset of the compact space $X$. Express an arbitrary open cover of $F$ in its subspace topology as $\{F\cap U_i:i\in I\}$, where the $U_i$ are open sets of $X$. Since $X\setminus F$ is open, the familyis an open cover of $X$. Take a finite subcover and discard $X\setminus F$ if it occurs. The remaining finitely many $U_i$ still cover $F$, since the discarded set contains no point of $F$. Their traces on $F$ give the required finite subcover. Hence every closed subset of a compact space is compact, with no Hausdorff assumption.

Assume $K$ is compact, and let the closed subsets $F_i$ of $K$ have the finite intersection property. If their total intersection were empty, their complements $K\setminus F_i$ would form an open cover of $K$. A finite subcover would givecontradicting the finite intersection property. Therefore the total intersection is nonempty.

Conversely, assume the stated nonempty-intersection property for every family of closed subsets of $K$. Given an open cover $\{V_i:i\in I\}$ of $K$ in its subspace topology, set $F_i=K\setminus V_i$. Their total intersection is empty. By the contrapositive of the assumed property, this family fails the finite intersection property; hence some finite $J\subseteq I$ satisfiesThis is a finite subcover, proving compactness. Both directions are instances of De Morgan's laws. Thus compactness is equivalent to the finite intersection characterization of compactness.

For completeness, the intersection of the empty family of subsets of $K$ is $K$. If $K$ is empty, no family has the finite intersection property when the empty finite subfamily is included, so the implication remains valid.

A subset $K$ is compact if every open cover has a finite subcover. Explicitly, if $U_i$ are open sets of $X$ and $K\subseteq {\bigcup }_{i\in I}{U}_i$, there are finitely many indices $i_1,\dots ,i_m$ such thatEquivalently, the subspace topology makes $K$ a compact space. An open set in this subspace topology is $K\cap U$ for an ambient open set $U$, which proves equivalence of the two open cover formulations. The empty set is compact: the empty selection is a finite subcover.

A subset $A$ is connected when it is a connected space in its subspace topology: there is no separation of a topological spaceinto two nonempty disjoint open subsets of $A$. Equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Since the given $A$ is not connected, choose such $P,Q$. Each is the complement of the other in $A$, so both are also closed subsets of $A$. They are compact by the closed subset of a compact space result. We now separate these two compact subsets by ambient open sets.

For each $p\in P$, part (b), applied to $Q$ and $p$, gives disjoint ambient open sets $G_p,H_p$ with $p\in G_p$ and $Q\subseteq H_p$. Choose $p_1,\dots ,p_m$ such that the $G_{p_j}$ cover $P$, using its compactness. DefineThen $P\subseteq U$, $Q\subseteq V$, and $U,V$ are disjoint open sets. Indeed a point of $U$ lies in some $G_{p_j}$, whereas every point of $V$ lies in the corresponding $H_{p_j}$. ConsequentlyThis proves the requested ambient separation, using the more general separation of disjoint compact subsets. The initially given relatively open sets $P,Q$ need not themselves be open in $X$; the finite construction supplies ambient open neighbourhoods.

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.

With the quotient realization defined in part (i), the answers are no and no. Write $W=|K|$ as countably many closed arms $[0,1{]}_n$ with all their zero endpoints identified to $v$.

First, the outer vertices $D=\{u_n:n\ge 1\}$ form a closed discrete subset. Its intersection with every closed arm is the closed singleton at coordinate one, so $D$ is closed by the weak topology. The terminal half-arm $(1/2,1{]}_n$ is an open neighbourhood of $u_n$ meeting $D$ only there. Thus $D$ is infinite, closed and discrete. Since closed discrete subsets of compact spaces are finite, $W$ is not compact. A homeomorphism to a compact subset of the plane would make $W$ compact, which is impossible.

For the stronger obstruction, suppose $(U_n)$ were a countable neighbourhood basis at $v$. Each $U_n$ contains an initial interval in arm $n$, so choose $x_n\in U_n$ on that arm with coordinate $t_n\in (0,1)$. Define $V$ to meet arm $n$ in $[0,t_n/2)$. It is an open neighbourhood of $v$, because each of these intersections is open in the closed interval $[0,1]$. But $x_n\in U_n\setminus V$ for every $n$, so no $U_n$ is contained in $V$. This contradicts the basis condition.

Hence $W$ is not first countable. Every metric space is first-countable, with local basis of balls of radius $1/n$, and this property passes to subsets with the subspace topology. Thus $W$ is not homeomorphic to any subset of ${\mathbb{R}}^2$. This is the weak realization of a countable star.

The realization convention matters. For the metric realization of a simplicial complex using Euclidean distance on barycentric coordinates, the second answer would be yes: map the point on arm $n$ at parameter $t$ to $t{e}^{i/n}$. Each ray is isolated away from zero, and at zero both spaces have neighbourhood bases given by a single radius on all arms. This gives the metric realization of a countable star as a planar subset, still noncompact. It does not give the quotient topology used in part (i).

The only isolated point of the subset $[0,1]$ is $1$. Indeed,so $\{1\}$ is open in the subspace topology. If $0\le x<1$, every basic neighbourhood $[a,b)$ containing $x$ also contains a point strictly between $x$ and $\min (b,1)$, hence another point of $[0,1]$. Such an $x$ is not an isolated point of the subset. Here isolation refers to $[0,1]$ with its subspace topology; $1$ is not an isolated point of the ambient Sorgenfrey line.

$X$ does not have sequential compactness. Take the sequenceEvery subsequence has usual limit of a sequence $1$. If a subsequence converged in the subspace topology of the Sorgenfrey line, it would also converge in the usual subspace topology, since part (c) makes the identity function from the former to the latter continuous. Uniqueness of a limit of a sequence in the usual Hausdorff space therefore forces that limit of a sequence to be $1$.

But $\{1\}$ is an open neighbourhood of $1$ in $X$, and no term $x_n$ belongs to it. Thus no subsequence converges to $1$ in $X$, proving the assertion.

$X$ is not compact. In the subspace topology on $X=[0,1]$, consider the open coverThe intervals are basic open sets in the ambient Sorgenfrey line, and $\{1\}=X\cap [1,2)$ is open in $X$. Every $x<1$ belongs to one of those intervals, while the singleton covers $1$.

A finite subcover would have to include $\{1\}$. Among its finitely many other members, let $N$ be the largest index. Their union is $[0,1-1/N)$, leaving $1-1/N$ uncovered. Selecting no such member also fails to cover $0$. Hence no finite subcover exists.

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.

Let $W=(\mathbb{N}\times S^1)/A$, and choose in each circle the point opposite $o$:In the subspace topology from the Euclidean plane, the sequence $z_n$ converges to $o$, which does not belong to $B$. Thus $B$ is not closed in $X$.

On the other hand, let $D={\bar{f}}^{-1}(B)\subseteq W$. With the quotient map $q$ from part (i),This is closed in $\mathbb{N}\times S^1$: its complement is the open set $\mathbb{N}\times (S^1\setminus \{(0,1)\})$. The definition of the quotient topology, applied to complements, says that $D$ is closed in $W$. A homeomorphism would carry this closed set to the closed set $B$, a contradiction. Therefore $\bar{f}$ is not a homeomorphism.

The obstruction occurs at the common point. The open set $W\setminus D$ contains the collapsed point, but its image $X\setminus B$ contains no neighbourhood of $o$, since $z_n\to o$. The wedge sum topology allows a separate small neighbourhood on each circle; a neighbourhood of $o$ in the Hawaiian earring must contain all sufficiently small whole circles.

Give the nonempty subset $A\subseteq X$ its subspace topology. It is compact if every open cover admits a finite subcover. Equivalently, whenever $A\subseteq {\bigcup }_{j\in J}{U}_j$ with each $U_j$ open in $X$, there are $j_1,\dots ,j_m$ such that $A\subseteq U_{j_1}\cup \cdots \cup U_{j_m}$.

It is connected if it has no separation of a topological space: there are no disjoint nonempty open subsets $B,C$ of $A$ with $A=B\cup C$. Here openness is relative to the subspace topology; equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.

Here $D^2$ must denote the closed disc $\{(x,y):x^2+y^2\le 1\}$, as its boundary points participate in the identification. For $r=\sqrt{x^2+y^2}$, defineThe Pythagorean trigonometric identity gives $\Vert F(x,y){\Vert }^2=1$, so the image lies in the sphere $S^2$. Away from the origin the formula is continuous. At the origin, the first two coordinates have absolute value at most $|\sin (\pi r)|\to 0$ and the third tends to $1$, proving continuity there as well.

Every boundary point maps to $(0,0,-1)$, while the origin maps to $(0,0,1)$. For a point $(u,v,w)\in S^2$ with $-1<w<1$, there is a unique $r\in (0,1)$ with $\cos (\pi r)=w$, and its unique preimage isThis point has radius $r$, because $u^2+v^2=1-w^2={\sin }^2(\pi r)$. The two poles have the preimages already described. Thus $F$ is surjective and its fibers are exactly the prescribed equivalence classes.

The closed disc is compact by the Heine-Borel theorem, and $S^2$ is Hausdorff with the subspace topology from Euclidean space. The preceding result gives the homeomorphism

True. Choose the irrational number $\alpha =\sqrt{2}$, and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover $\mathbb{Q}$. Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on $\mathbb{Q}$.

Let two rational numbers be equivalent exactly when they belong to the same $C_n$. The bijectionis well-defined. For every subset $B\subseteq \mathbb{Z}$, the preimage of $h^{-1}(B)$ under the canonical projection of a quotient set is ${\bigcup }_{n\in B}{C}_n$, which is open in $\mathbb{Q}$. Thus every subset of $\mathbb{Q}/\sim$ is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because $\{n\}=\mathbb{Z}\cap (n-1/2,n+1/2)$. Consequently $h$ is a homeomorphism.

False. Take the metric space $X=\mathbb{R}\times \{0,1\}$ with the Euclidean distance and subspace topology inherited from ${\mathbb{R}}^2$. For every $t\ne 0$, identify $(t,0)$ with $(t,1)$; leave the two points $(0,0)$ and $(0,1)$ in separate equivalence classes. All equivalence classes have one or two points, so all are closed in this metric space.

Let $o_0,o_1$ be the two distinct images of the origins. If $U_0,U_1$ are any open neighbourhoods of $o_0,o_1$ in the quotient space, their preimages contain respectivelyfor some $\epsilon ,\delta >0$. Choose $0<t<\min (\epsilon ,\delta )$. The common image $p(t,0)=p(t,1)$ belongs to both $U_0$ and $U_1$. Therefore the origins have no disjoint open neighbourhoods and the quotient space is not Hausdorff. This is the line with two origins.

An abstract simplicial complex consists of a set $V$ of vertices of a simplicial complex and a family $K$ of finite nonempty subsets of $V$, called abstract simplexes, such that every singleton $\{v\}$ belongs to $K$ and every nonempty subset of a member of $K$ also belongs to $K$. One may include the empty set as an additional face by convention. A member with $k+1$ vertices is a $k$-simplex; its nonempty subsets are its faces of a simplex.

The geometric realization has the following explicit description using a barycentric coordinate system:Here the support is $\mathrm{supp}(t)=\{v:t(v)>0\}$. Each $\rho \in K$ gives a standard geometric simplexwith its usual Euclidean topology. Common faces of a simplex are identified using the same barycentric coordinate system.

Precisely, the topology on $|K|$ is defined by declaring $A\subseteq |K|$ closed if and only if $A\cap |\rho |$ is closed in $|\rho |$ for every $\rho \in K$. Equivalently, use the quotient topology on the disjoint union of all the standard geometric simplexes, identifying identical points of common faces of a simplex. For a finite simplicial complex, this agrees with the subspace topology in ${\mathbb{R}}^V$. For an infinite simplicial complex, the stated rule specifies the realization topology without presuming a finite-dimensional ambient space.

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]