The adjunction space $Y{\cup }_{f}X$ glued along a continuous map, and, for a nonempty space, the cone and the suspension as quotients of $X\times [0,1]$

Throughout, $[0,1]$ denotes the closed unit interval (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with the subspace topology inherited from the usual topology of $\mathbb{R}$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and $X\times [0,1]$ the binary product with the product topology (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). All three constructions below are quotients (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection) of a space already built on this page, and none of them needs anything further. The one exception is the degenerate case $X=\mathrm{\varnothing }$: then $X\times [0,1]$ is empty, so the one-point cone and the two-point suspension stipulated below are not quotients of it, and they are fixed by that stipulation rather than by the construction.

Adjunction space. Let $X$ and $Y$ be topological spaces, let $A\subseteq X$ carry the subspace topology, and let $f:A\to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Form the disjoint union $Y\bigsqcup X$ with its canonical injections ${\kappa }_Y$ and ${\kappa }_X$ (The disjoint union (coproduct) ${⨆}_i{X}_i$ with the final topology of the canonical injections: a set is open exactly when each of its traces is) and let ${\sim }_f$ be the relation on $Y\bigsqcup X$ whose classes are

$C_y:=\{{\kappa }_Y(y)\}\cup \{{\kappa }_X(a):a\in A,f(a)=y\}(y\in Y),\{{\kappa }_X(x)\}(x\in X\setminus A)\mathrm{.}$

These sets are pairwise disjoint and their union is $Y\bigsqcup X$, since every element of $Y\bigsqcup X$ is ${\kappa }_Y(y)$ for exactly one $y$, or ${\kappa }_X(x)$ for exactly one $x$, and in the latter case lies in $C_{f(x)}$ when $x\in A$ and in its own singleton otherwise. A family of pairwise disjoint nonempty sets covering a set is the family of classes of exactly one equivalence relation, so ${\sim }_f$ is well defined. The adjunction space is the identification space

$Y{\cup }_{f}X:=(Y\bigsqcup X)\mathrm{/}\text{\hspace{0.17em}⁣}{\sim }_f$

with the quotient topology of its canonical projection. It is said to be obtained by gluing $X$ to $Y$ along $f$: each point $a\in A$ is identified with its image $f(a)$, and nothing else is identified.

Collapsing a subset. For a space $Z$ and a nonempty $B\subseteq Z$, write $Z\mathrm{/}B$ for the quotient of $Z$ by the equivalence relation whose classes are $B$ and the singletons $\{z\}$, $z\mathrm{\notin }B$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection): all of $B$ becomes one point and nothing else is identified.

Cone. The cone on a space $X$ is

$CX:=(X\times [0,1])/(X\times \{1\}),$

the product of $X$ with the unit interval, with the whole top face collapsed to a single point, called the apex. The definition presupposes only that $X\times \{1\}$ is nonempty, which holds exactly when $X$ is nonempty; for $X=\mathrm{\varnothing }$ the product is empty, the quotient of the empty space is empty, and no apex is produced; this library nonetheless takes $C\mathrm{\varnothing }$ to be the one-point space by convention, so that the apex always exists and every cone is nonempty. The convention is a stipulation, not a consequence of the description above, which gives the empty space.

Suspension. The suspension of $X$ is

$\mathrm{\Sigma }X:=(X\times [0,1])/\text{\hspace{0.17em}⁣}\sim ,$

where $\sim$ has as classes $X\times \{0\}$, $X\times \{1\}$, and the singletons $\{(x,t)\}$ for $0<t<1$. So both faces are collapsed, each to its own point, and the two resulting points are distinct as soon as $X$ is nonempty. For $X=\mathrm{\varnothing }$ this library takes $\mathrm{\Sigma }\mathrm{\varnothing }$ to be the two-point discrete space, one apex from each end, matching the convention that a suspension is two cones glued along $X$; the alternative stipulation of a single point is also in use in the literature, and nothing here depends on the choice.

Mapping cone. For a continuous $f:X\to Y$ the mapping cone is the adjunction space $Y{\cup }_{f^{\mathrm{\prime }}}CX$, where $CX$ is the cone, $X$ is identified with the subspace $X\times \{0\}$ of $X\times [0,1]$ and thence with its image in $CX$, and $f^{\mathrm{\prime }}$ is the corresponding map into $Y$. It is recorded here as the standard instance of the two constructions used together, and nothing below depends on it.

What is deliberately not asserted. These constructions produce spaces, and this page proves nothing about which of them are homeomorphic to which. The invariants that separate them, connectedness, compactness and the homotopy notions, are not available for general topological spaces at this point in the reading order: connectedness and compactness are developed here only for subsets of $\mathbb{R}$ (Separated sets, disconnection, and connected subset of $\mathbb{R}$) and for metric spaces (Open cover, subcover, compact metric space, and compact subset of a metric space), and neither development applies to a quotient that has not been shown metrizable, while the homotopy notions are absent altogether. So no statement here says that two of these spaces are different, and none says that the cone or the suspension of a familiar space is any particular familiar space.

• For a nonempty space every construction above is a quotient of a coproduct or of a product, and that is the whole point (the stipulated empty-space cone and suspension are the sole exception, as the opening records). The adjunction space needs the disjoint union (The disjoint union (coproduct) ${⨆}_i{X}_i$ with the final topology of the canonical injections: a set is open exactly when each of its traces is) so that $X$ and $Y$ start out unattached, and then one quotient to attach them; the cone and the suspension need the product with $[0,1]$ and then one quotient. So the constructions of this page suffice, and the universal properties already proved apply verbatim: a continuous map out of $Y{\cup }_{f}X$ is exactly a pair of continuous maps out of $Y$ and out of $X$ agreeing along $f$, by A map out of a disjoint union is continuous iff each of its restrictions is; the canonical injections are open and closed embeddings; and each summand is clopen in the union together with For a quotient map $q:X\to Y$, a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$; and a composite of quotient maps is a quotient map.

• Why a partition rather than a generated relation. The classes of ${\sim }_f$ are written down explicitly instead of taking "the equivalence relation generated by ${\kappa }_X(a)\sim {\kappa }_Y(f(a))$". The two agree, but the explicit form makes the verification that ${\sim }_f$ is an equivalence relation a one-line check on a partition, and it makes the saturated sets easy to recognise, which is what every computation with the quotient topology needs.

• The interval is the only piece of $\mathbb{R}$ used. Nothing here needs the order or the arithmetic of $[0,1]$ beyond the fact that it is a topological space with two distinguished points $0$ and $1$; the choice of $[0,1]$ rather than another such space is conventional, and the companion page's cylinder and Mobius band are quotients of $[0,1]\times [0,1]$ for the same reason.