# The cone of a topological space is contractible and simply connected

I would like help in verifying my proof and making it rigorous.

Question:

Let $$X$$ be a topological space. The cone on $$X$$, denoted $$CX$$, is the quotient space $$(X × [0, 1])/(X × \{0\})$$. Prove that $$CX$$ is contractible and simply connected?

(note: $$X/A$$ is the quotient space $$X/\mathord{∼}$$ for an equivalence relation $$∼$$ on $$X$$ such that the equivalence classes are $$A$$ itself, and the singletons $$\{x\}$$ such that $$x \notin A$$.)

A previous answer by kobe shows that $$CX$$ is contractible:

Consider the map $$H: X \times [0,1] \times [0,1] → X \times [0,1]$$ defined by $$H((x,t),s) = (x(1-s)t)$$. Then $$H$$ is continuous and $$H((x,0,s) = (x,0)$$ for all $$x∈X$$ and $$s∈[0,1]$$. $$H$$ induces a continuous map $$\hat{H}$$: $$CX \times [0,1] → CX$$ such that $$\hat{H}([(x,t)],s) = [(x,(1-s)t)]$$. Now $$\hat{H}[(x,t)],0) = [(x,t)]$$ and $$\hat{H}[(x,t)],1) = [(x,0)] = X \times \{0\}$$. So $$\hat{H}$$ is a homotopy in $$CX$$ from the identity on $$CX$$ to the point $$X \times \{0\}$$. Consequently, $$CX$$ is contractible.

$$CX$$ is simply connected as a result of being contractible. To show this, we show that all contractible spaces are simply connected. Let $$X$$ by contractible. We will show that any loop in $$X$$ is homotopic, relative to $$\{ 0,1 \}$$, to the constant loop. Observe that the homotopy $$F$$ from the identity on $$X$$ to the constant map at $$x∈X$$ gives us a homotopy $$G(s,t) = F(α(s),t)$$ from any loop $$α$$ based at $$x$$ to the constant loop at $$x$$. We must show that the homotopy is relative to $$\{0,1\}$$. We use the fact that the square $$[0,1] \times[0,1]$$ is convex, so there is a straight line homotopy $$H$$ between any two paths from $$a$$ to $$b$$ in $$[0,1] \times [0,1]$$, and so any two such paths are homotopic relative to $$\{0,1\}$$. Using this, we see that the path along the left edge of the square is homotopic, relative to $$\{0,1\}$$, to the path along the bottom edge, up the right edge and back along the top edge of the square. Composing the homotopies $$H$$ and $$G$$ gives us the homotopy relative to $$\{0,1\}$$. We have shown that any loop in $$X$$ is homotopic, relative to $$\{0,1\}$$, to the constant loop and so it is path connected and the fundamental group $$π_1(X,b)$$ for a loop based at $$b$$ has only one element and $$π_1(X,b) = \{e\}$$ where $$e$$ is a constant. $$X$$ is path-connected and has a trivial fundamental group. Then it is simply connected. Contractible spaces are simply connected. Then $$CX$$ is simply connected.

• I think you mean $\hat{H}([(x,t)],s)=[(x,(1-s)t)]$. You have $x$ instead of the first $s$, which makes no $s$ on the left hand side. Dec 18, 2014 at 3:37
• You probably mean "... and singletons $\{x\}$ such that $x\not\in A$" in the second My question-paragraph. Dec 18, 2014 at 14:40
• It appears that the proof in the question body was plagiarized from the answer to this question... Jan 5, 2015 at 3:08
• – user191141
Jan 5, 2015 at 5:14

I'm going to deal with the first statement, that $CX$ is contractible.
Consider the map $H: X \times{[0,1]} \times{[0,1]}$ $→$ $X \times{[0,1]}$ defined by $H((x,t),s) = (x(1-s)t)$. Then $H$ is continuous and $H((x,0,s) = (x,0)$ for all $x∈X$ and $s∈[0,1]$.
Here you show that, whenever two points get identified, namely when they are of the form $(x,0,s)$ and $(x',0,s)$ for $x,x'\in X,s\in I$, then they have the same image $[(x,0)]$.
$H$ induces a continuous map $\hat{H}$: $CX \times [0,1] → CX$ such that $\hat{H}([(x,t)],s) = [(x,(1-s)t)]$. Now $\hat{H}[(x,t)],0) = [(x,t)]$ and $\hat{H}[(x,t)],1) = [(x,0)] = X \times${$0$}. So $\hat{H}$ is a homotopy in $CX$ from the identity on $CX$ to the point $X \times${$0$}. Consequently, $CX$ is contractible.
All your computations are correct. The function $\hat H$ starts with the identity and ends with a retraction $r:CX\to \{X\times\{0\}\}$ (Note that this is also relative $X\times\{0\}$, it is independent of the time $s$ on that subspace, so you actually have a deformation retraction $r$). The only problem here is that $\hat H$ is continuous only when $q\times\text{id}_I:X\times I\times I\to CX\times I$ is a quotient map. But in general, a product map $q\times\text{id}_Y:X\times Y\to Z\times Y$, where $q$ is a quotient map, is not necessarily a quotient map. Luckily, in your case it is, and this is because $I$ is locally compact and $q\times\text{id}_Y$ is a quotient map for each locally compact $Y$. For a proof see theorem 4.3.2 in the book Topology and Groupoids.