Smash product of X with $S^1$ (Example 0.10 in Hatcher’s)

Question

My first question/confusion is regarding the definition of reduced suspension $\sum X$ and suspension $SX$ of a space $X$.

Usually, we denote the quotient space of $X$ with subspace $A$ identified to a point as $X/A$. I understand that suspension of a space is like a 'double-cone': we take a space $X$ and consider its cone $CX=X\times I/X\times \{0\}$, then define $SX:=CX/X\times \{1\}$. Therefore, $\sum X:=SX/\{x_0\}\times I$. I understand that it is not correct to say that $\sum X=\frac{X\times I}{(X\times \{1\})\cup (X\times \{0\})\cup (\{x_0\}\times I)}$ because this would mean that the whole set $(X\times \{1\})\cup (X\times \{0\})\cup (x_0\times I)$ is identified to a point. Is my understanding correct? Thanks.

Now comes my second question/confusion:

This pertains to Example 0.10 of Hatcher's. I am trying to understand and prove the following statement:

'The reduced suspension $ΣX$ is actually the same as the smash product $X ∧ S^1$ since both spaces are the quotient of $X×I$ with $X×∂I ∪\{x_0\}×I$ collapsed to a point.'

How to prove this?

$X∧ S^1:=\frac{X\times S^1}{X\times \{0\}\cup \{x_0\}\times S^1}$

I think I should now use $I/\partial I=S^1$ somehow but I don't understand how to go from here to prove the statement.

Answer 1

1. It is correct to say that $\Sigma X=\frac{X\times I}{(X\times \{1\})\cup (X\times \{0\})\cup (\{x_0\}\times I)},$ i.e. that the whole set $(X\times \{1\})\cup (X\times \{0\})\cup (x_0\times I)$ is identified to a point. This is due to the fact that $\Sigma X=(X\times I)/\mathcal R$ where $\mathcal R$ is the equivalence relation generated by: $$(x,0)\mathcal R(x_0,0),\quad(x,1)\mathcal R(x_0,1),\quad(x_0,t)\mathcal R(x_0,0)\qquad(\forall x\in X,\forall t\in I).$$
2. $X\times S^1=(X\times I)/\mathcal S$ where $\mathcal S$ is the equivalence relation generated by: $$(x,0)\mathcal S(x,1)\qquad(\forall x\in X).$$ Therefore, $X∧S^1:=(X\times S^1)/(X\times \{0\}\cup \{x_0\}\times S^1)$ is $(X\times I)/\mathcal T$ where $\mathcal T$ is the equivalence relation generated by: $$(x,0)\mathcal T(x,1),\quad(x,0)\mathcal T(x_0,0)\mathcal T(x_0,t)\qquad(\forall x\in X,\forall t\in I).$$ You can easily check that $\mathcal T=\mathcal R.$