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Define a continuous map $F:S\to T$ by the standard embedded torus of revolution parametrizationIts image lies in $T$ becauseThe periodicity of the trigonometric functions makes $F(s,0)=F(s,1)$ and $F(0,t)=F(1,t)$, so $F$ is constant on every $\sim$-equivalence class. The universal property of the quotient topology therefore gives a unique continuous map

Every point of $T$ has cylindrical radius between $1$ and $3$, so its polar angle determines $s$ modulo one, while the pair $(\sqrt{x^2+y^2}-2,z)$ on the unit circle determines $t$ modulo one. Hence $\overline{F}$ is surjective, and two parameters have the same image exactly when they differ by the endpoint identifications defining $\sim$; thus it is injective.

The quotient $S/\sim$ is compact as the continuous image of the given compact space $S$ under the quotient map. The surface $T$ is Hausdorff because it has the subspace topology inherited from ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now shows that $\overline{F}$ is a homeomorphism. Therefore

If $f:\tilde{X}\to Y$ is continuous, then $f\circ \pi$ is continuous because the quotient map $\pi$ is continuous by definition.

Conversely, suppose $f\circ \pi$ is continuous. For every open set $V\subseteq Y$,is open in $X$. The definition of the quotient topology then says that $f^{-1}(V)$ is open in $\tilde{X}$. HenceThis is the universal property of the quotient topology.

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.

DefineThis is continuous and takes values on the unit sphere. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the universal property of the quotient topology gives a continuous map

For $0<y<1$, the third coordinate $\cos (\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline{f}$ is injective. The spherical-coordinate formula also shows that it is surjective.

The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now makes $\overline{f}$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the suspension of a topological space $S^1$, which is $S^2$. This proves the square quotient model of the two-sphere.

A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.