Example 1

Let X be a topological space. One denotes by C(X), and calls the cone of the space X, the cone of the mapping Id$_X$; it is the space obtained from $X\times \mathbf{I}$ by contracting $X\times \{0\}$. Let $\pi$ be the canonical mapping of $X\times \mathbf{I}$ into C(X); the image $C'(X)$ of $X\times [0,1[$ under $\pi$ is called the open cone of the space X.

Assume that X is not empty. Then the cone C(X) is not empty and its base point is again called the vertex of the cone.

The mapping $((x, t), u)\mapsto (x, t(1-u))$ of $(X\times \mathbf{I})\times \mathbf{I}$ into $X\times \mathbf{I}$ is a homotopy joining the identity mapping of $X\times \mathbf{I}$ to the mapping $(x, t)\mapsto (x,0)$. One deduces from this, by passing to quotient sets, a mapping $\sigma : C(X)\times \mathbf{I}\rightarrow C(X)$ such that $\sigma (\pi (x, t), u) =\pi (x, t(1-u))$ for $x\in X,t, u\in \mathbf{I}$. The mapping $\sigma$ is continuous by I, p. 19, prop. 10. The mapping $\sigma$ is then a pointed homotopy at $s$ joining the identity mapping of C(X) to the constant mapping with image $\{s\}$. Consequently, the cone C(X) is contractible to its vertex $s$. Since $\sigma (C'(X)\times \mathbf{I})$ is contained in $C'(X)$, the mapping $\sigma$ defines, by passing to subspaces, a pointed homotopy at $s$ joining the identity mapping of $C'(X)$ to the constant mapping with image $\{s\}$; the open cone $C'(X)$ is therefore contractible to $s$.