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 $x \notin A$.)
My Answer:
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.
Best Answer
I'm going to deal with the first statement, that $CX$ is contractible.
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 $[(x,0)]$.
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.