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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.

Use a topological graph consisting of the ray $[0,\infty )$ with vertices $v_n=n$ and one oriented graph loop ${\ell }_n$ attached at each $v_n$, for $n=0,1,2,\dots$. Set the base point to $b=v_0$. This is a connected, locally finite graph; it is deliberately infinite.

Define continuous maps $f,g:X\to X$ byHere each graph loop is parametrized by $s\in [0,1]$ with its endpoints at the attaching vertex. The formulas agree at every vertex, and their restrictions to every closed edge are continuous. The weak topology of a CW complex therefore makes $f,g$ continuous maps. Directly on the ray and on each graph loop,

We check the two fundamental group maps at exactly the specified base points. The ray is a spanning tree. At $b=v_0$ a free basis of a group consists of $a_n$, obtained by going along the ray to $v_n$, following ${\ell }_n$, and returning. At $f(b)=v_1$, use the corresponding free basis of a group $c_n$ with connecting paths starting at $v_1$; in particular $c_0$ travels backwards along the first edge before following ${\ell }_0$. The standard spanning tree description applies to this infinite topological graph as well: every loop and every homotopy between based loops has compact image, so lies in a finite subcomplex.

The induced group homomorphisms areThus $f_{\ast }$ misses $c_0$: words in $c_1,c_2,\dots$ cannot equal $c_0$ by the normal form for free groups and free product of groups. Consequently $f_{\ast }$ is not surjective. Meanwhile $c_0$ is a nontrivial free group element killed by $g_{\ast }$, so $g_{\ast }$ is not injective. Neither induced map is a group isomorphism, while functoriality of the fundamental group correctly gives $g_{\ast }{f}_{\ast }=\mathrm{id}$.

A homeomorphism is a bijection for which both the function and its inverse are continuous maps.

Let $f:X\to Y$ satisfy the given assumptions. For a closed subset $F$ of $X$, the closed subset of a compact space theorem makes $F$ compact. The continuous image of a compact space theorem makes $f(F)$ compact. Because $Y$ is Hausdorff, its compact subset $f(F)$ is closed. Thus $f$ is a closed map.

For $g=f^{-1}:Y\to X$, the inverse image of a closed set $F\subseteq X$ iswhich is closed in $Y$. The closed-set criterion for continuity therefore makes $g$ continuous. Hence $f$ is a homeomorphism, proving the compact-to-Hausdorff continuous bijection theorem.

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.

Write $Q=(Z\times I)/\mathcal{R}$. Since $Q$ is homeomorphic to the circle, it is Hausdorff. For fixed $t\in I$, the functionis continuous, being a coordinate slice of the quotient map. Part (a)(i) therefore makes it constant. Choose $x_0\in Z$ and setThen $g$ is continuous andSince the quotient map $p$ is a surjection, so is $g$. This is Hausdorff factorization through a cofinite product factor: a continuous map into a Hausdorff space cannot distinguish the points of the coordinate with the cofinite topology.

If $g$ were injective, it would be a continuous bijection from the compact closed interval $I$ to the Hausdorff space $Q$. Part (a)(ii) would make it a homeomorphism, contradicting the fact that the circle and interval are not homeomorphic.

To recall that obstruction explicitly, remove the interior point $1/2$ from $I$. The resulting topological subspace is separated by the nonempty relatively open sets $[0,1/2)$ and $(1/2,1]$, so it is not connected. The circle with any one point removed is connected: after a rotation, it is the image under a continuous map of the connected open interval $(0,2\pi )$ under $t\mapsto (\cos t,\sin t)$. A homeomorphism and its restriction preserve connectedness, giving a contradiction.

Thus $g$ is not injective. Choose distinct $a,b\in I$ with $g(a)=g(b)$. ThenIn particular the requested $x,a,b$ exist. The argument proves the stronger conclusion that the same two interval coordinates work for every $x$.

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.

Let $q_S:S^2\to S^2/{\sim }_S$ and $q_D:D^2\to D^2/{\sim }_D$ be the quotient maps. The continuous mapis a homeomorphism onto the upper hemisphere; its inverse function is the restriction of the projection map $(x,y,z)\mapsto (x,y)$. Every antipodal pair on $S^2$ meets that hemisphere. Moreover, two points of this hemisphere are antipodal only if both lie on the equator. ThusBy the universal property of the quotient topology, $q_S\circ h$ induces a continuous bijection

To justify continuity of the inverse function, the source is compact, being a continuous image of a compact space $D^2$; the Heine-Borel theorem gives compactness of $D^2$. The target is Hausdorff, as follows. For distinct equivalence classes $[u]\ne [v]$, putIn $S^2$, take the union $U$ of the two relative open balls of radius $\delta /3$ about $u,-u$, and similarly take $V$ about $v,-v$. Every distance between their respective centres is at least $\delta$, so the triangle inequality gives $U\cap V=\varnothing$. These open sets contain entire antipodal pairs, hence $q_S^{-1}(q_S(U))=U$ and similarly for $V$. The quotient topology therefore makes $q_S(U)$ and $q_S(V)$ disjoint open neighbourhoods of the two classes.

The compact-to-Hausdorff continuous bijection theorem now gives the required homeomorphism $H$.

Use positive indices $n\ge 1$ and write $p=(0,-1)\in S^1$. SetFor fixed $n$, this function is a homeomorphism from the unit circle to $C_n$: it scales its radius to $1/n$ and translates its centre to $(0,1/n)$. Thus it takes values in the Hawaiian earring $X$ and is surjective onto $X$. It is continuous on each open slice $\{n\}\times S^1$; since these slices form an open cover of $\mathbb{N}\times S^1$, $f$ is continuous on the whole product topology.

Distinct circles $C_n,C_m$ meet only at $o=(0,0)$. To see this directly, their equations areFor $m\ne n$, subtraction gives $y=0$, then $x=0$. Each slice contains exactly one preimage of $o$, namely $(n,p)$. Therefore the only nonsingleton fibre of $f$ is $A$.

Let $q:\mathbb{N}\times S^1\to (\mathbb{N}\times S^1)/A$ be the quotient map collapsing $A$. The universal property of the quotient topology supplies a unique continuous map $\bar{f}$ with $f=\bar{f}\circ q$. The fibre calculation proves injectivity, and surjectivity follows from that of $f$. Hence $\bar{f}$ is the required continuous bijection.

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 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.

For a continuous map $f:X\to Y$ between Tychonoff spaces, extend it to $F:\beta X\to \beta Y$ by the Stone-Čech extension property. If $f$ is perfect, compactness of a fiber and closedness of $f$ show that no remainder point can map to a point of $Y$: separate such a point from the fiber by a neighborhood in $\beta X$, and use closedness to obtain a target neighborhood whose preimage stays inside it, contradicting density. Conversely, if $F^{-1}(Y)=X$, each fiber is compact, and for closed $A\subseteq X$ the equality $f(A)=F({\overline{A}}^{\beta X})\cap Y$ proves closedness. Equivalently $F(\beta X\setminus X)\subseteq \beta Y\setminus Y$.

For pointed topological spaces $(X,x_0)$ and $(Y,y_0)$, the pointed mapping space ${\mathrm{Map}}_{\ast }(X,Y)$ consists of continuous maps $f:X\to Y$ with $f(x_0)=y_0$. It carries the subspace topology from the compact-open topology on the ordinary mapping space, commonly followed by kification when working with compactly generated spaces. Its basepoint is the constant map with value $y_0$.

For the pointed circle, ${\mathrm{Map}}_{\ast }(S^1,Y)$ is the loop space $\Omega Y$: a circle map is a path returning to its starting point. More generally, for well-behaved pointed spaces, path components of ${\mathrm{Map}}_{\ast }(S^n,Y)$ correspond to ${\pi }_n(Y,y_0)$. Thus a mapping space turns families of maps and homotopies into points and paths of another topological space. The choice of mapping-space topology ensures that this correspondence works continuously.

A continuous map on a nonempty topological space is point-transitive if it has a dense forward orbit. This is one convention for topological transitivity, but on spaces with isolated points it must be distinguished from the open-set intersection definition. On a nonempty compact metric space, the positive-time open-set intersection property implies point transitivity: for each member $U_j$ of a countable basis of a topology, ${\bigcup }_{n\ge 1}{f}^{-n}(U_j)$ is open and dense, and the Baire category theorem supplies a point in their intersection. Conversely, a surjective point-transitive continuous map has every tail of its dense orbit dense, since $f^k(X)=X$ and continuity sends its orbit closure into the closure of that tail. The map $f(1/j)=1/(j+1)$, $f(0)=0$ on $\{0\}\cup \{1/j:j\ge 1\}$ is point-transitive but fails the open-set intersection property.

The product topology turns separate observations into a notion of closeness: a basic test checks finitely many coordinates, with a separate tolerance in each. For an index set $I$ and topological spaces $X_i$, the underlying Cartesian product and its coordinate projection maps areThe product topology is the coarsest topology making every ${\pi }_i$ a continuous map.^[ref]

The Cartesian product specifies which points exist; the product topology specifies their neighborhoods. Finite observations motivate its finite-coordinate basis for the product topology, while convergence in a product space and continuity into a product space explain why it is useful. The compactness result for this construction is Tychonoff's theorem.

A projection map from a Cartesian product selects specified coordinates. For the coordinate indexed by $j$,For example, $p_1(a,b)=a$ forgets the second component of an ordered pair.

The product topology makes each coordinate projection map a continuous map. In the two-factor case $X\times Y\to X$, surjectivity depends on whether the omitted factor $Y$ has an element, subject to the empty-set cases for Cartesian projections. For an arbitrary infinite family, simultaneous existence of the other coordinates can also involve the axiom of choice.

In a compact metric space with a continuous map $T$, points $x,y$ are proximal ifThis property is independent of the compatible metric. If $T$ is injective and $x\ne y$, arbitrarily small proximal distances must occur at arbitrarily large times, since each finite collection of distances is strictly positive. Proximal points need not be equal or have dense orbits.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.

The quotient topology turns an instruction to glue points together into a precise topological space. Start with a topological space $(X,{\tau }_X)$ and an equivalence relation $\sim$. The quotient set $X/\sim$ consists of equivalence classes, and its canonical projection of a quotient set isThe quotient topology is defined by the rule using preimages of open sets:Thus a subset of the quotient space is open if and only if its entire preimage is open in $X$. Each fiber $q^{-1}(\{[x]\})$ is the whole equivalence class $[x]$, so the preimage includes every point glued to any of its members.^[ref]

This is the finest topology making $q$ a continuous map: every topology with that property can contain only subsets whose preimages are open. It really is a topology, because preimages preserve the empty set, the whole set, arbitrary unions, and finite intersections. The gluing example below shows why checking a complete preimage matters, especially at the seam.

The quotient topology also makes continuous maps out of a glued space easy to construct. A continuous map $f:X\to Z$ descends to a unique continuous map $\bar{f}:X/\sim \to Z$ exactly when it is constant on every fiber of $q$, meaning $x\sim y\Rightarrow f(x)=f(y)$. Define $\bar{f}([x])=f(x)$; the equivalence class representative cannot affect the answer. For every open set $V\subseteq Z$,The right side is open, so the quotient topology makes ${\bar{f}}^{-1}(V)$ open. Conversely, if $f=\bar{f}\circ q$ for a continuous map $\bar{f}$, then composition of continuous functions makes $f$ continuous, and equal equivalence classes force equal values. This is the universal property of the quotient topology. ^[ref]

A surjective continuous map is not automatically a quotient map. On the same two-point set $D=\{a,b\}$, take the identity map from the discrete topology to the indiscrete topology. It is continuous and surjective, but $\{a\}$ is not open in the target even though its preimage is open in the domain. The quotient topology for this identity map would be the discrete topology. Thus continuity supplies only one direction of the rule using preimages of open sets; a quotient map requires both.

Removing any interior point $a$ from the Euclidean closed unit ball allows a retraction onto its unit sphere. For $v=x-a\ne 0$, the positive root of $|a+tv{|}^2=1$ isThen $r(x)=a+\tau (x)v$ is a continuous map on the punctured closed unit ball, has norm one, and fixes its boundary because $\tau (x)=1$ there. The quadratic equation has one positive and one negative root since $|a|<1$, so the chosen root is unambiguous.

For $R>0$ in a normed vector space, define $P_R(0)=0$ and use the displayed formula elsewhere. This continuous map fixes the closed ball of radius $R$ and maps every point outside it to the boundary along its radial line. If a continuous map $T$ is defined only on that closed ball, $T\circ P_R$ is a continuous extension to the whole space. This extension is useful when applying a theorem whose hypotheses concern a globally defined vector field.