Difference between revisions of "Cone"

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Let $(X,\tau)$ be a [[topological space]]. We define the cone over $X$ taking $X \times [0,1]$ and shrinking $X \times \{1\}$ to a point. More precisely we define $\mathrm{Cone}(X)$ to be the [[quotient space]] $X \times [0,1] / \sim$, where $\sim$ is an [[equivalence class]] defined by $(x_1,t_1) \sim (x_2,t_2)$ if and only if $t_1=t_2=1$. The point $X \times \{1\} \in \mathrm{Cone}(X)$ is called the vertex of the cone while $X \times \{0\}$ is called the base of $\mathrm{Cone}(X)$.
 
Let $(X,\tau)$ be a [[topological space]]. We define the cone over $X$ taking $X \times [0,1]$ and shrinking $X \times \{1\}$ to a point. More precisely we define $\mathrm{Cone}(X)$ to be the [[quotient space]] $X \times [0,1] / \sim$, where $\sim$ is an [[equivalence class]] defined by $(x_1,t_1) \sim (x_2,t_2)$ if and only if $t_1=t_2=1$. The point $X \times \{1\} \in \mathrm{Cone}(X)$ is called the vertex of the cone while $X \times \{0\}$ is called the base of $\mathrm{Cone}(X)$.
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=Properties=
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Proposition:</strong> Let $X$ be the [[circle with a spiral]], then $C(X)$ is [[homeomorphic]] to $\mathrm{Cone}(X)$, where $C(X)$ denotes the [[hyperspace|hyperspace of continua]] of $X$.
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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Revision as of 07:58, 5 April 2015

Let $(X,\tau)$ be a topological space. We define the cone over $X$ taking $X \times [0,1]$ and shrinking $X \times \{1\}$ to a point. More precisely we define $\mathrm{Cone}(X)$ to be the quotient space $X \times [0,1] / \sim$, where $\sim$ is an equivalence class defined by $(x_1,t_1) \sim (x_2,t_2)$ if and only if $t_1=t_2=1$. The point $X \times \{1\} \in \mathrm{Cone}(X)$ is called the vertex of the cone while $X \times \{0\}$ is called the base of $\mathrm{Cone}(X)$.

Properties

Proposition: Let $X$ be the circle with a spiral, then $C(X)$ is homeomorphic to $\mathrm{Cone}(X)$, where $C(X)$ denotes the hyperspace of continua of $X$.

Proof: