Let's define $SX = X \times I /(X \times \partial I \cup \lbrace x_0 \rbrace \times I)$ and the smash product $X \wedge Y = X \times Y /(\lbrace x_0 \rbrace \times Y \cup X \times \lbrace y_0 \rbrace)$.
I'd like to prove that $SX \simeq \mathbb{S}^1 \wedge X$ , is this true in general or depends on $X$? According to wikipedia on reduced suspension this should be true.
So I took the following diagram :
$$X \times I \overset{f}{\longrightarrow} \mathbb{S}^1 \times X \overset{q}{\longrightarrow} S^1 \wedge X$$
Where $q$ is the projection and $f$ is given by $f(x,t) = (\cos(2\pi t),\sin(2\pi t),x)$. It's clear that $q \circ f$ is continuos, surjective, and called $p$ the projection $p : X \times I \longrightarrow SX$, that $q \circ f$ is costant on the fiber of $p$, so exists a continuos and bijective map $\tilde{f} : SX \longrightarrow \mathbb{S}^{1} \wedge X$.
What about finding an inverse? I should find a continuos map from $\mathbb{S}^1 \times X \longrightarrow X \times I$ first that works, but unable to find a simple one. Any easier methods or solution to prove that $\tilde{f}$ is an homeomorphism would be appreciated.
Edit : I'm not sure if this is true, here's why : It's a fact that "the k-fold iterated reduced suspension of X is homeomorphic to the smash product of X and a k-sphere" according to this. So it should be true that $\Sigma X \simeq \mathbb{S}^{1} \wedge X$ and $SX \simeq \mathbb{S}^{1} \wedge X$. So it should be true in general that $SX \simeq \Sigma X$. If this is true I think excludes the homeomorphism, but feels necessary some hypothesis on $X$ to prove the latter as well, as $\lbrace x_0 \rbrace \times I$ being contractible
It is impossible to obtain the continuous map $S^1\times X \to X\times I$ you desired. I think the easier way is to find the homeomorphism $$ X\times I / X\times \partial I \cong X\times S^1/ X\times \{1\}. $$ You can write the function explicitly, since in the quotient space, you can still use the coordinate system. We define a function $f$ as $$ f(x,t) = (x,e^{2\pi i t}). $$ You need to check that (1) this is well-defined, (2) this is continuous, (3) this is bijective, (4) inverse function is also continuous. Once you check them, you can pass this continuous map to the quotient, $$ \require{AMScd} \begin{CD} X\times I / X\times \partial I @>{\cong}>> X\times S^1/ X\times \{1\} \\ @VVV @VVV\\ SX @>{\cong}>> S^1 \wedge X \end{CD} $$
P.S. It looks safer to assume that $X$ is locally compact Hausdorff. See comments.