I want to prove that the suspension $\Sigma X$ of a CW-complex $X$ is a CW-complex, buy I'm starting with CW-complexes and I don't have a clue of how start, so I'd appreciate any help. Thanks.
$$\Sigma X=\frac{X \times I}{X \times \{0,1\}}$$
[Math] Suspension of a CW complex
algebraic-topologycw-complexes
Related Solutions
Here are some thoughts that could hopefully help.
First of all, trying to give the product of two simplices a simplicial (or $\Delta$-complex) structure is in general annoying. There's a reason that the relevant sections of Hatcher are rather technical (see for example the proof of Theorem 2.10 - page 112 here (PDF)).
Here's one thing that is neat about simplices (when I say simplices I mean $\Delta$-complexes but I'm being sloppy) though, and this construction actually underlies a lot of what you're doing.
Let's suppose that $\sigma = [v_0, \cdots, v_n]$ is an $n$-simplex. Then it has an $(-1)^n$-oriented codimension-1 face $\tau = [v_0, \cdots, v_{n-1}]$, and this is just an $(n-1)$-simplex.
Now let's add a point $v_{n+1}$ to everything - just put it at the end of every simplex. We boost $\sigma$ to an $(n+1)$-simplex $\tilde{\sigma} = [v_0, \cdots, v_n, v_{n+1}]$. And now it has an $(-1)^n$-oriented face $\tilde{\tau} = [v_0, \cdots, v_{n-1}, v_{n+1}]$ which is still codimension-1, and this is just an $n$-simplex.
This is actually super-neat. Let's suppose we have a description of a space $X$ realized as a collection of simplices $\{\sigma_i\}$. Then a simplicial description of $CX$ is given by $\{\sigma_i, [c], \tilde{\sigma}_i\}$, where $\tilde{\sigma}_i$ just denotes adding $c$ to the end of the list of vertices of $\sigma_i$.
If you want to see some poorly-drawn examples, here you go:
Hopefully these convince you that $C(\Delta^n) = \Delta^{n+1}$.
For the suspension, there's a really important point that your notation makes it unclear whether you get it. So I'll say it, and you may well already know it. You identify $Y \times \{0\}$ to one point, and $Y \times \{1\}$ to a different point. Otherwise the equality $\Sigma S^n = S^{n+1}$ wouldn't be true.
Perhaps it's better to think about the suspension as two cones smushed together. Given a space $X$ we can form cones $C_+X$ and $C_-X$, and these come with canonical inclusions $X \to C_{\pm}X$. Then the suspension is just the union of $C_+X$ and $C_-X$ with the two copies of $X$ identified. In the case of the sphere, these correspond to the north and south hemispheres.
In particular, if we start off with $S^0$ as two points (0-simplices), then $S^1 = \Sigma S^0$ is four 1-simplices, one for each quadrant in the standard embedding $S^1 \to \mathbb R^2$, $S^2 = \Sigma S^1$ is eight 2-simplices, one for each octant in $\mathbb R^3$, and so on and so forth.
If you want another poorly-drawn picture, here you go:
For any topological space $X$, the suspension $\Sigma X$ is path-connected and hence connected. To see this, suppose $[(x_1, a_1)], [(x_2, a_2)] \in \Sigma X$ where the square brackets denote the equivalence class of a pair from $X \times [0, 1]$ under the equivalence relation $\sim$.
The path $p_1 : [0, 1] \to \Sigma X$, $p_1(t) = [(x_1, (1-t)a_1 + t)]$ joins $p_1(0) = [(x_1, a_1)]$ and $p_1(1) = [(x_1, 1)]$.
The path $p_2 : [0, 1] \to \Sigma X$, $p_2(t) = [(x_2, 1 - t + ta_2)]$ joins $p_2(0) = [(x_2, 1)]$ and $p_2(1) = [(x_2, a_2)]$.
As $p_1(1) = [(x_1, 1)] = [(x_2, 1)] = p_2(0)$, the concatenation of the paths $p_1$ and $p_2$ gives a path $p := p_1\ast p_2$ joining $[(x_1, a_1)]$ and $[(x_2, a_2)]$.
On the other hand, $\Sigma X$ need not be simply connected. For example, if $X = S^0 = \{-1, 1\}$ (equipped with the discrete topology), then $\Sigma S^0 = S^1$ which is not simply connected. In general, $\Sigma S^n = S^{n+1}$. However, if $X$ is path-connected, then $\Sigma X$ is simply connected; this follows from the Seifert-van Kampen Theorem.
Best Answer
It is a general result (that I invite you to prove) that the product of CW-complexes (one of which is finite) is still a CW-complex, using the product of the cells involved. Hence $X \times I$ is a CW-complex. Moreover, $X \times \{0,1\}$ is a subcomplex thereof, so that $(X \times I,X \times \{0,1\})$ is a relative subcomplex (see this http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf , at page 73-74): this implies that $\Sigma X=(X\times I) /(X \times \{0,1\})$ is a CW-complex.