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A bouquet of circles identifies one point of each of a family of circles to a common point, using the quotient topology. As a CW complex it has one vertex and one graph loop for each circle. The empty family means a single point. Its fundamental group is the free group on the family, as described by the fundamental group of a bouquet of circles.

For a nonempty subset $A$ of a topological space $X$, identify all points of $A$ and leave each point of $X\setminus A$ in its own equivalence class. The resulting quotient space is denoted $X/A$. For every subset $K\subseteq X$, its saturation of a subset isThe quotient topology therefore gives a useful test: $q_A(K)$ is a closed set if and only if the displayed saturation of a subset is a closed set of $X$. In particular, every closed set disjoint from $A$ has closed image, even if $A$ is not closed. When $A$ is closed, $q_A$ is a closed map, because both possibilities in the formula are then closed sets for closed $K$.

Take one copy of the standard geometric simplex ${\Delta }^{\sigma }$ for every face $\sigma$ of an abstract simplicial complex $K$, label its vertices by $\sigma$, and identify equally labelled common faces. Give the result the quotient topology from the disjoint union of all these simplexes. Equivalently, a point is a finitely supported family of barycentric coordinates ${\lambda }_v\ge 0$ with ${\sum }_v{\lambda }_v=1$ and $\{v:{\lambda }_v>0\}\in \Sigma$.

The weak topology of a simplicial realization makes $U$ open precisely when $U\cap |\sigma |$ is open in every closed simplex. This definition determines the topology even for infinite complexes, when a metric realization of a simplicial complex may give a different topology.^[ref]

For a topological space $X$, the displayed relation is an equivalence relation. Give its quotient the quotient topology. Every inverse image under a continuous map from $X$ to a Hausdorff space is a saturation of a subset. Distinct quotient classes can be distinguished by some such map; disjoint open neighborhoods of their images pull back to disjoint saturated open sets, whose quotient images are open. Thus the quotient is Hausdorff. Each continuous map from $X$ to a Hausdorff space factors uniquely through this quotient, and the factor is continuous by the quotient topology. This is its universal reflection property.

The indiscrete topology on a set $X$ has only the empty set and the whole set as open sets. When $X$ has more than one point it is not a Hausdorff space. For example the quotient topology on $\mathbb{R}/\mathbb{Q}$ is indiscrete: a nonempty open saturated subset contains an interval and all its rational translates, which cover $\mathbb{R}$.

The displayed metric on antipodal classes of the unit sphere induces its quotient topology. Triangle inequality follows by choosing the signs that realize each minimum and applying the Euclidean triangle inequality. Distinct classes have disjoint saturated unions of small balls, proving the quotient is Hausdorff; compactness follows from the continuous quotient map.

Choose $b_n\in (n,n+1/2)\setminus A$, possible because each interval is uncountable. Every tail $K_n=\{b_k:k\ge n\}$ is closed and disjoint from $A$. Therefore $U_n=\mathbb{R}\setminus K_n$ is an open saturated set for collapsing a subset to a point, so its image in the quotient topology is open. These increasing images cover the quotient, but no finite selection covers it, because the largest selected image still misses an infinite tail of distinct singleton classes. This is an open cover with no finite subcover, proving noncompactness. The argument does not require the countable subset to be closed or nondense. For the empty subset, read the quotient as the original line.

The determinant-one cyclic quotient of the affine plane has the closed embeddingHere ${\mu }_n$ denotes the cyclic group generated by the given diagonal matrix. We establish the presentation of the polynomial invariant ring and verify that its fibres really are the group orbits.

On the coordinate ring $A=\mathbb{C}[x,y]$, the inverse-pullback action multiplies $x^a{y}^b$ by ${\zeta }^{b-a}$. Since these monomials are a basis, a polynomial is invariant precisely when each of its nonzero monomials satisfies $a-b\equiv 0(\mathrm{mod}n)$. If $a\ge b$, such a monomial is $(x^n{)}^{(a-b)/n}(xy{)}^b$; if $b\ge a$, it is $(y^n{)}^{(b-a)/n}(xy{)}^a$. ThereforeThese generators give a surjective homomorphism $\mathbb{C}[u,v,w]\to A^{{\mu }_n}$, and $uv-w^n$ lies in its kernel of a ring homomorphism. To prove there are no further relations, reduce any monomial using $uv=w^n$. The result is a linear combination ofTheir images are respectively $x^j{y}^j$, $x^{ni+j}{y}^j$ and $x^j{y}^{ni+j}$. All have distinct exponent pairs, so they are linearly independent in $A$. Any polynomial in the kernel of a ring homomorphism has zero reduced expression and hence belongs to $(uv-w^n)$. ConsequentlyThis proves the claimed closed subvariety embedding of the affine quotient.

For surjectivity of $\pi$, if $u\ne 0$ choose $x^n=u$ and set $y=w/x$. Then $y^n=w^n/u=v$. If $u=0$, the relation forces $w=0$; choose $x=0$ and any $n$th root of $v$ for $y$. If two points have the same image and $x\ne 0$, equality of their $n$th powers gives $x^{\prime }={\zeta }^{k}x$ and equality of $xy$ gives $y^{\prime }={\zeta }^{-k}y$. If $x=0$, both first coordinates vanish, and the same argument using $y^n$ gives the orbit relation; the case $x=y=0$ has only that one point in its fibre. Thus every fibre is exactly one group orbit.

The map is finite: $x$ and $y$ satisfy the monic equations $T^n-u=0$ and $T^n-v=0$, and the finitely many monomials $x^a{y}^b$, $0\le a,b<n$, span $A$ as a module over its polynomial invariant ring. A finite morphism is closed when it is surjective as here and therefore gives the quotient topology. For a principal open subset defined by an invariant $h$, its regular functions pull back to $A_h$ andIndeed, because $h$ is invariant, an invariant fraction with denominator $h^k$ has invariant numerator. These principal open subsets cover the affine hypersurface, proving the required local equality of invariant regular functions. Together with the orbit fibres, this verifies the geometric quotient, not merely a bijection of underlying sets.

A monad on a category $\mathcal{C}$ consists of an endofunctor $T$, a unit natural transformation $\eta :1_{\mathcal{C}}\Rightarrow T$, and a multiplication $\mu :T^2\Rightarrow T$, satisfyingAn algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aTa=a{\mu }_A$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ satisfies $ha=bTh$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

For an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:U$ with $F⊣U$, the monad induced by an adjunction is $T=UF$, with $\mu =U\epsilon F$. The Eilenberg-Moore comparison functor is $K(D)=(UD,U{\epsilon }_D)$ and $K(h)=Uh$. The functor $U$ is monadic when this comparison is an equivalence of categories; this is a monadic adjunction.

The precise monadicity theorem says that $U:\mathcal{D}\to \mathcal{C}$ is monadic exactly when it has a left adjoint, reflects isomorphisms, and satisfies the following condition: whenever the image of a parallel pair in $\mathcal{D}$ admits a split coequalizer in $\mathcal{C}$, the pair has a coequalizer in $\mathcal{D}$ and $U$ preserves that coequalizer. Equivalently, these coequalizers are created up to the canonical isomorphism identifying their underlying coequalizer objects. Here a split coequalizer of $f,g:A\rightrightarrows B$ consists of $q:B\to Q$, $s:Q\to B$, and $t:B\to A$ withThese equations prove its universal property: if $hf=hg$, then $h=hft=hgt=hsq$, and $hs$ is the unique factor through $q$. Every functor preserves these equations, so every functor preserves split coequalizers.

First consider the forgetful functor $U^T:{\mathcal{C}}^T\to \mathcal{C}$. It has the free algebra functor $A\mapsto (TA,{\mu }_A)$ as a left adjoint: the inverse transposition formulas are $h\mapsto h{\eta }_A$ and $v\mapsto bTv$. An monad algebra morphism whose underlying arrow is invertible has an algebra-morphism inverse, by rearranging its structure equation. Thus $U^T$ reflects isomorphisms.

Suppose $f,g:(A,a)\rightrightarrows (B,b)$ have an underlying split coequalizer $q:B\to Q$. Applying $T$ gives a coequalizer of $Tf,Tg$. The equation $qbTf=qfa=qga=qbTg$ therefore gives a unique $c:TQ\to Q$ with $cTq=qb$. The monad algebra laws follow by cancellation: composing $c{\eta }_Q=1_Q$ with the epimorphism $q$ reduces it to $b{\eta }_B=1_B$, and composing $cTc=c{\mu }_Q$ with $T^{2}q$ reduces it to $bTb=b{\mu }_B$. Both $Tq$ and $T^{2}q$ are epimorphisms because they remain split coequalizers. Thus $(Q,c)$ is an algebra for a monad and $q$ is an monad algebra morphism. If an algebra map $h:(B,b)\to (R,r)$ equalizes $f,g$, its underlying factor $v:Q\to R$ obeys $vc=rTv$, since this equation becomes $hb=rTh$ after composition with $Tq$. This proves that $U^T$ creates the required coequalizers. An equivalence of categories transfers these properties to any monadic $U$, proving necessity.

Conversely assume the three conditions and write $T=UF$. Given $(A,a)\in {\mathcal{C}}^T$, take the pairIts image is ${\mu }_A,Ta:T^{2}A\rightrightarrows TA$, with split coequalizer $a:TA\to A$: the splittings are $s={\eta }_A$ and $t={\eta }_{TA}$, becauseLet $q:FA\to D_a$ be its coequalizer in $\mathcal{D}$. Preservation gives a canonical isomorphism $U{D}_a\cong A$ under which $Uq=a$. Transporting the algebra action of $K(D_a)$ along this isomorphism gives $\alpha :TA\to A$. Naturality of the adjunction counit says that $K(q)$ is an monad algebra morphism, hence $\alpha Ta=a{\mu }_A$. But $aTa=a{\mu }_A$ too, and $Ta$ is a split epimorphism, with section $T{\eta }_A$. Therefore $\alpha =a$. Every algebra is consequently isomorphic to some $K(D_a)$.

It remains to establish full faithfulness, rather than merely construct objects. For $B\in \mathcal{D}$, put $b=U{\epsilon }_B$. The adjunction counit ${\epsilon }_B:FUB\to B$ coequalizes ${\epsilon }_{FUB},Fb$. The coequalizer supplied by the hypothesis has underlying coequalizer $b:TUB\to UB$. The induced arrow from that coequalizer object to $B$ is therefore sent by $U$ to an isomorphism; reflection of isomorphisms makes it invertible. Thus ${\epsilon }_B$ itself is this coequalizer.

Now let $h:K(B)\to K(C)$ be an monad algebra morphism, and put $c=U{\epsilon }_C$. The arrow $v={\epsilon }_{C}Fh:FUB\to C$ coequalizes the two arrows in the preceding presentation. To check this directly, use transposition along $F⊣U$: the transposes of $vFb$ and $v{\epsilon }_{FUB}$ are respectively $hb$ and $cTh$, which agree by the algebra-morphism equation. Hence a unique $\bar{h}:B\to C$ satisfies $\bar{h}{\epsilon }_B=v$. Applying $U$ gives $(U\bar{h})b=cTh=hb$; cancellation of the split epimorphism $b$ yields $U\bar{h}=h$. Finally, any arrow $r:B\to C$ satisfies $r{\epsilon }_B={\epsilon }_{C}FUr$ by naturality, so two arrows with equal image under $U$ are equal by the coequalizer property. The comparison is full and faithful and essentially surjective, hence an equivalence of categories. This proves the theorem.

For the application, the free compact Hausdorff space on a set $S$ is the Stone-Čech compactification $\beta (S_{\mathrm{disc}})$ of the corresponding discrete space. Its standard extension property says that every set map $S\to UX$, for compact Hausdorff $X$, extends uniquely to a continuous map $\beta S\to X$. This constructs the required left adjoint. Also a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so $U$ reflects isomorphisms.

Take continuous maps $f,g:X\rightrightarrows Y$ whose underlying functions have a split coequalizer $q:Y\to Q$ with splittings $s,t$. The splittings need not be continuous. LetThe set $E$ is compact and hence closed. Since $qf=qg$, it is contained in $R$. Conversely $(y,sq(y))=(ft(y),gt(y))\in E$. ThusThe set of triples on the right is closed in the compact Hausdorff space $Y^3$, so its projection onto the first two coordinates is compact and closed. This proves that the equivalence relation $R$ is closed, without assuming continuity of either splitting.

Give $Q\cong Y/R$ the quotient topology. The allowed closed-relation criterion makes it a compact Hausdorff space. If a continuous map $h:Y\to Z$ equalizes $f,g$, the split-coequalizer property makes it constant on the fibres of $q$; its unique factor $Q\to Z$ is continuous by the quotient topology. Hence this is a coequalizer in compact Hausdorff spaces, preserved by $U$. Its topology is forced uniquely, since any continuous surjection from a compact space to a Hausdorff space is a quotient map. All three conditions hold, and therefore the forgetful functor from compact Hausdorff spaces to sets is monadic.

Adjoin ${\mathbb{P}}^1(\mathbb{Q})$ to the complex upper half-plane. Ordinary points retain their usual neighborhoods. At a rational boundary point $a=\sigma \infty$, a neighborhood basis consists of $a$ together with $\sigma \{z\in \mathbb{H}:\mathrm{Im}z>Y\}$, with $Y$ arbitrarily large. Different choices of $\sigma \in S{L}_2(\mathbb{Z})$ change it by a translation and sign, so give the same neighborhood system. Give $X(1)$ the quotient topology. The modular group acts properly discontinuously on $\mathbb{H}$; its modular cusp neighborhoods descend to punctured neighborhoods with their modular cusp point filled in. This quotient is Hausdorff: interior orbits can be separated by proper discontinuity, and a sufficiently high modular cusp neighborhood avoids any prescribed compact interior neighborhood.

Away from fixed points of the effective group $PS{L}_2(\mathbb{Z})$, a small disk is disjoint from its nonidentity translates, so its coordinate gives a holomorphic quotient coordinate. At an elliptic fixed point ${\tau }_0$, useIts effective stabilizer rotates $u$ by an $e$th root of unity, where $e=2$ at $i$ and $e=3$ at $\rho$. The quotient coordinate is $w=u^e$, not $u$. All rational boundary points lie in one modular cusp class. At infinity the stabilizer acts by $\tau \mapsto \tau +1$, andis a punctured-disk coordinate; adjoining $q=0$ fills in the modular cusp. These charts and their holomorphic transition maps define the complex structure of the compactified modular curve.

For compactness use the standard regionIts part with $\mathrm{Im}\tau \le Y$, $Y>1$, is compact: the imaginary part is at least $\sqrt{3}/2$. The quotient of its upper tail, with its two vertical sides identified and infinity added, is the closed disk $|q|\le e^{-2\pi Y}$. These two compact images cover $X(1)$, including the boundary-edge identifications. Thus $X(1)$ is a compact Riemann surface.

The $j$ function also extends over the modular cusp with one simple pole, since $j=q^{-1}+O(1)$. Its local expressions at elliptic points descend to holomorphic functions of $u^e$. By Question 1 it maps the remaining quotient bijectively onto $\mathbb{C}$, so its extension is a degree-one holomorphic map $X(1)\to {\mathbb{P}}^1$. A degree-one map of compact Riemann surfaces is biholomorphic: any local ramification would already give degree at least two. Therefore $X(1)$ is the Riemann sphere; this will also identify the base genus in Question 5.

Use three compactness facts: a closed subset of a compact space is compact; a continuous image of a compact space is compact; and a compact subset of a Hausdorff space is closed. For a closed $C\subseteq X$, these facts show that $f(C)$ is closed in $Y$. Therefore the bijection $f$ is a closed map, and $(f^{-1}{)}^{-1}(C)=f(C)$ proves continuity of its inverse. Thus $f$ is a homeomorphism. This proves the compact-to-Hausdorff continuous bijection theorem without requiring $X$ itself to be Hausdorff.

Let $q:[0,1]\to [0,1]/\sim$ be the quotient map and $h(t)=(\cos 2\pi t,\sin 2\pi t)$. The map $h$ is continuous and identifies precisely the equal parameters and the endpoint pair. Hence it induces a bijection $\bar{h}$ from the quotient space to the unit circle, with $h=\bar{h}\circ q$; the definition of the quotient topology makes $\bar{h}$ continuous. The quotient is compact as a continuous image of $[0,1]$, while the circle is Hausdorff as a subspace of ${\mathbb{R}}^2$. Applying the preceding theorem gives

Throughout, rings are commutative with identity and ring homomorphisms preserve identity. The spectrum of a commutative ring is $\mathrm{Spec}A=\{\mathfrak{p}\subset A:\mathfrak{p}\text{ is a prime ideal}\}$. For an ideal $I$, put $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$; these are the closed sets of the Zariski topology. Its axioms follow fromFor the last equality, a prime ideal containing $IJ$ must contain $I$ or $J$: if it contained neither, choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. The other equalities follow directly from containment. Thus arbitrary intersections and finite unions of closed sets are closed, proving that this is a topology.

The Zorn lemma says that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. For $A\ne 0$, apply it to proper ideals ordered by inclusion. The zero ideal is proper. The union of a nonempty chain is an ideal and is proper, because containing one would put one in a member of the chain; the empty chain has the zero ideal as an upper bound. There is therefore a maximal ideal $\mathfrak{m}$. If $a\notin \mathfrak{m}$, then $\mathfrak{m}+(a)=A$, so the class of $a$ in $A/\mathfrak{m}$ is invertible. This quotient ring is a field, making $\mathfrak{m}$ a prime ideal. Hence $A\ne 0⟹\mathrm{Spec}A\ne \varnothing$.

For a ring homomorphism $f:A\to B$, define $f^{\ast }:\mathrm{Spec}B\to \mathrm{Spec}A$ by $f^{\ast }(\mathfrak{q})=f^{-1}(\mathfrak{q})$. The inverse image is proper because $f(1)=1$, and primality follows from $f(ab)=f(a)f(b)$. Moreoverwhere $IB$ is the ideal generated by $f(I)$. This proves continuity. Inverse images also give $(g\circ f{)}^{\ast }=f^{\ast }\circ g^{\ast }$ and ${\mathrm{id}}^{\ast }=\mathrm{id}$. Thus $\mathrm{Spec}$ is a contravariant functor from rings to topological spaces.

For the specified inclusion, the fundamental theorem of algebra gives the points $0$ and $(x-z)$ of $\mathrm{Spec}\mathbb{C}[x]$. Over $\mathbb{R}$, the nonzero prime ideals are generated either by $x-r$ with $r\in \mathbb{R}$, or by $(x-a{)}^2+b^2$ with $a\in \mathbb{R}$, $b>0$; all irreducible real polynomials have degree one or two. Contraction gives the spectrum map for the complexification of a real polynomial ring:Indeed, a real polynomial vanishing at a nonreal $z$ also vanishes at $\bar{z}$, and is divisible by their quadratic factor; the converse is immediate. This map is surjective. Its fibers contain one point over a real linear factor, two conjugate points over an irreducible quadratic factor, and only the generic point over zero. It identifies complex-conjugate pairs. Since each proper Zariski-closed set in either one-variable polynomial ring is a finite set of closed points, images of closed sets are closed; hence the Zariski topology on the target is the quotient topology for this identification.

The quotient map $q:S^2\to X$ is continuous and surjective by the quotient topology, so $X$ is compact as a continuous image of a compact space. To prove the Hausdorff property explicitly, let $[u]\ne [v]$ and set $d=\min (\Vert u-v\Vert ,\Vert u+v\Vert )>0$. Choose $r<d/3$. The saturated open setsare disjoint by the triangle inequality. Each contains both antipodes of every one of its points, so $q^{-1}(qU)=U$ and similarly for $V$. Consequently $qU,qV$ are disjoint open quotient neighborhoods separating the classes. Thus the quotient is compact and Hausdorff. It is the real projective plane; equivalently the metric on the antipodal sphere quotient is $d_X([u],[v])=\min (\Vert u-v\Vert ,\Vert u+v\Vert )$ and induces this topology.

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 the torus quotient, a point has the same image as a point of $B(u,r)$ precisely when they differ by an integer lattice vector. ThereforeThis union is open in ${\mathbb{R}}^2$, so $U$ is open by the quotient topology. Distinct integer lattice vectors are at least distance one apart, whereas $2r<1$. Thus these open balls are disjoint, with a positive gap between any two.

Each ball is connected, since it is convex. It is both open and closed relative to the displayed union: its complement there is the union of all the other open balls. No connected component can meet two such relatively separated pieces. Consequently the components are exactly the infinitely many balls $B(u+m,r)$. Translation by $-m$ is a homeomorphism from each component onto $B(u,r)$. The statement uses the intended positive radius $0<r<1/2$.

Let $q$ denote the quotient map and write $[X]$ for a point of Real projective space. For $i=1,\dots ,n+1$, take the open setIt is open in the quotient topology because its inverse image under $q$ is the open set $\{X:X_i\ne 0\}$. These sets cover, since a nonzero vector has at least one nonzero coordinate. Define the manifold chartThe hat means that the $i$th coordinate is omitted. Ratios are unchanged by multiplying $X$ by any nonzero real number, so the chart is well defined. Its inverse inserts $1$ into position $i$ and takes the resulting equivalence class. The ratios and this inverse are continuous, giving a homeomorphism $U_i\to {\mathbb{R}}^n$.

For a compact description of the smooth transition maps, write $a_i=1$ and let $a_k=X_k/X_i$ for $k\ne i$. On $U_i\cap U_j$ one has $a_j\ne 0$, andEach coordinate is a rational smooth function on the open domain $a_j\ne 0$; reversing $i,j$ gives a smooth inverse. Thus these $n+1$ charts form the standard affine atlas of real projective space.

For completeness, the underlying space is Hausdorff: normalizing $X$ identifies it with the antipodal quotient of the unit sphere, and the continuous injective map $[X]\mapsto X{X}^T/(X^{T}X)$ into the space of symmetric matrices separates any two distinct classes by disjoint open neighborhoods. It is also a second-countable space: pull back the countable bases of rational balls in these finitely many charts, and take their union. Therefore

If $A\subseteq X$, the subspace topology on $A$ isIf $\sim$ is an equivalence relation on $X$ and $q:X\to X/\sim$ is the natural projection, the quotient topology declares $V\subseteq X/\sim$ open exactly when $q^{-1}(V)$ is open in $X$. For topological spaces $X$ and $Y$, the product topology on $X\times Y$ has as a basis all sets $U\times V$ with $U$ open in $X$ and $V$ open in $Y$.

Let $i:Y\hookrightarrow M_f$ be the canonical inclusion and defineThis is well-defined because $[0,x]=f(x)$ in the mapping cylinder, and continuity follows from the universal property of the quotient topology. Clearly $r\circ i={\mathrm{id}}_Y$.

DefineThe formula respects the gluing relation at $t=0$, so it descends from the disjoint union to a continuous homotopy. It satisfies $H_0={\mathrm{id}}_{M_f}$ and $H_1=i\circ r$. Thusindeed $Y$ is a deformation retract of $M_f$.

The quotient topology on $\tilde{X}$ isSince inverse images preserve arbitrary unions and finite intersections,Also ${\pi }^{-1}(\varnothing )=\varnothing$ and ${\pi }^{-1}(\tilde{X})=X$. The displayed collection therefore satisfies all the topology axioms.

Choose a contractionApply the homotopy extension property to the initial map ${\mathrm{id}}_X:X\to X$ and to $h$, regarded as a homotopy into $X$. It suppliessuch that

At $t=1$, the map $H_1$ sends every point of $A$ to $a_0$. It is therefore constant on the equivalence classes of the quotient topology $X/A$. By the universal property of the quotient topology, there is a continuous mapsuch thatwhere $q:X\to X/A$ is the quotient map. The homotopy $H$ immediately gives

For every $t$, the map $q\circ H_t$ is constant on $A$, because $H(a,t)=h(a,t)\in A$. It therefore descends toThis descended map is continuous: the product $q\times {\mathrm{id}}_I$ is a quotient map because $I$ is compact Hausdorff, and $q\circ H$ is constant on its fibres.
At the endpoints,ThusThe maps $q$ and $g$ are homotopy inverses, soThis proves the theorem on collapsing a contractible cofibration.