# 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.

• Upvotes are approval. You get points with upvotes, which, when you accumulate enough of them, allows you to do certain things you can't do initially. For this question, you probably got upvotes because you've shown a lot of work. Too often, people post here with obvious homework problems and zero effort on their part, so voters appreciate a question with so much thought expressed. – Thomas Andrews Dec 18 '14 at 3:53
• That's not what the upvotes mean. It just means appreciation of effort, not that you got it right. – Thomas Andrews Dec 18 '14 at 4:01
• Many upvotes and no correcting answer do not necessarily mean that you are clear and correct. Even if you were completely wrong, I would still have heartily upvoted for investing effort into the problem and showing your work. – Neal Dec 18 '14 at 4:02
• It appears that the proof in the question body was plagiarized from the answer to this question... – Milo Brandt Jan 5 '15 at 3:08
• asdfdas: Please, no dot vandalize posts. – Pedro Tamaroff Jan 5 '15 at 5:08

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.