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I was recently looking through J. J. Rotman's book: An Introduction to Algebraic Topology, where on page 23 he has the following result:

Theorem 1.11. For every space $X$, the cone $CX$ is contractible.

Proof. Define $F:CX\times I\to CX$ by $F([x,t],s)=[x,(1-s)t+s]$. $\square$

Now it is clear to me that if $F$ is continuous, then it is a homotopy from $1_{CX}$ to the constant map which sends each point of $CX$ to to the "top" point of the cone; hence, $1_{CX}$ is null-homotopic and so $CX$ is contractible as claimed.

This is all well and good, but it is not immediately apparent to me which result Rotman is relying on to ensure the continuity of $F$. Does anyone have any ideas about how to guarantee the continuity of $F$?

Thanks in advance.

dwar
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1 Answers1

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Let $q: X \times I \to CX$ denote the quotient map. Since $I$ is compact the map $q \times 1: X \times I \times I \to CX \times I$ is a quotient map. Therefore by the universal property of the quotient map, $F$ is continuous iff $F \circ (q \times 1)$ is continuous.

Define $G: X \times I \times I \to X \times I$ by $G(x,t,s) = (x, (1-s)t +s)$ then $F \circ (q \times 1) = q \circ G$. Since $q \circ G$ is clearly continuous so is $F \circ (q \times 1)$ hence so is $F$.

Mark Sandey
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R. Frankhuizen
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  • could you, or anyone else, explain why $F$ is continuous iff $F\circ(q\times 1)$ is continuous please? – TopologyGetsMeTired May 08 '15 at 20:24
  • It is the universal property of quotient maps that q : X -> Y quotient map <=> (f : Y -> Z continuous iff fq : X -> Z continuous). This follows from the universal property of the topology coinduced by a family of maps, in this case just q. The coinduced topology is the unique topology with the universal property. – Not Legato Sep 22 '21 at 18:47