Cone over $X$, equivalence relation

Question

I have a question about the definition of the cone over $X$.

If $X$ is a space, define an equivalence relation $X\times [0,1]$ by $(x,t)\sim (x',t')$ if $t=t'=1$. Denote the equivalence class of $(x,t)$ by $[x,t]$. The cone over $X$, denoted by $CX$, is the quotient space $X\times [0,1]/\sim$.

I acutally do not get this definition of the equivalence relation.... Sure, two points $(x,t), (x',t')$ are equivalent if $t=t'=1$. But which points are equivalent to (for example) $(x,\tfrac12)$? The relation does not tell anything about the cases when $t\neq 1$ or $t\neq t'$, which feels incomplete. But I think I am doing a horrible mistake here.

How does this relation include every pair $(x,t)\in X\times [0,1]$, when the relation is only defined for $t=1$?

I am currently studying "Introduction to algebraic topology" by Joseph J. Rotman. An exercise goes as follows:

For fixed $t$ with $0\leq t<1$, prove that $x\mapsto [x,t]$ defines a homeomorphism from a space $X$ to a subspace of $CX$.

Which revealed my misunderstanding.

So I am not understanding which equivalence classes there are. For every point $(x,t)$ with $t\neq 0$, the equivalcence class should just contain this one point.

Can you elaborate more? Thanks in advance.

Answer 1

When discussing equivalence relations, one often uses the following fact of set theory:

For every relation $R \subset X \times X$ on a set $X$ there is a unique smallest equivalence relation $E \subset X \times X$ on the set $X$ such that $xRy \implies x E y$. This equivalence relation $E$ is the intersection (in $X \times X$) of all equivalence relations $S$ which have the property $x R y \implies x S y$. We say that $E$ is the equivalence relation generated by $R$.

You can construct $E$ rather concretely. First take the reflexive closure, by adding all pairs $(x,x) \in X \times X$ to the relation $R$. Then take the symmetric closure, by adding all pairs $(y,x)$ for which $(x,y)$ is already in the relation. Finally take the transitive closure: for all sequences $x_0,x_1,x_2,...,x_n$ such that each of the pairs $(x_0,x_1), (x_1,x_2), ..., (x_{n-1},x_n)$ is already in the relation, add $(x_0,x_n)$ to the relation.

So, when you get an equivalence relation on a set (such as $X \times [0,1]$) which appears to only be partially defined, what you are supposed to do is to use that partial definition to define a relation $R$ on the set, and then you should take the equivalence relation generated by $R$.

In particular, since the point $(x,1/2) \in X \times [0,1]$ has not even been mentioned in the definition of $R$, it will follow that point $(x,1/2)$ is the only point in its equivalence class, i.e. its equivalence class is $\{(x,1/2)\}$.

Answer 2

The short answer to your question "which points are equivalent to $(x,1/2)$?": Only $(x,1/2)$.

The cone should really look like the geometric cone you know. For a generic example, start with a cylinder and identify the points at the top face. This is the motivation for the name cone for $CX$. Up to pathological examples, the other examples look similar. You start with a topological space $X$, make a cylinder with $X$ as a face, so we get $X \times [0,1]$. Then, we identify the points of one face.