The problem, although arising from some deeper facts, is quite simple. I would like to visualise the quotient space $A$ given by the cylinder $I\times S^{1}$ ($S^{1}$ is the circle in $\mathbb{R}^{2}$ and $I$ is the unit interval $[0,1]\subseteq \mathbb{R}$), where the subspace $(\{0\}\times S^{1})\cup (\{1\}\times S^{1})$ is killed to a point. Here, by visualising A, I mean to find a subspace of $\mathbb{R}^{n}$ for some $n\in\mathbb{N}$ (if possible, with $n\leq 3$) which is homeomorphic to $A$.
In my opinion such an $A$ should be $S^{2}\vee S^{1}$ (here $S^{2}$ is the $2-$sphere) but I am tempted to think that this is not the case and I will explain the reason, together with the motivating issue for this question, after the straight line below.
So, how can I visualise $A$? If $A$ is really homeomorphic to $S^{2}\vee S^{1}$, where does the reasoning below the straight orizontal line fail?
Following the suggestion given by G. Kelley in its Basic Concepts of Enriched Category Theory, I would like to prove that $\mathbb{Top}$ is not cartesian closed by showing that, in general, the functor $-\times Y$ does not preserve regular epimorphism, i.e. it does not commute with coequalisers.
I then tried with the simplest situation I could imagine. Take a singleton, say $\{\ast\}$, and consider the (continuous) functions $f$ and $g$ mapping $\ast$ to $0\in I$ and $1\in I$ respectively. Then the coequaliser (object) for this pair should be $S^{1}$. On the other hand, taking the product of $f$ and $g$ with (the identity map associated to) $Y=S^{1}$, gives rise to the maps $f\times id_{Y}$ and $g\times id_{Y}$ sending ${\ast}\times Y$ to $\{0\}\times Y$ and $\{1\}\times Y$ in the cylinder respectively. Now, the coequaliser of these maps is $A$ above and, if $A$ is homeomorphic to $S^{2}\vee S^{1}$, $-\times Y$ can not commmute with coequalizers, since $Coeq(f\times id_{Y},g\times id_{Y})=S^{2}\vee S^{1}$ is not homemorphic to the $2-$torus $Coeq(f,g)\times Y=S^{1}\times S^{1}$.
In the beginning, I was persuaded enough by this argument, but then I realised that it should be wrong, since $S^{1}$ is a locally compact Hausdorff space and it is known that, when $Y$ has those properties, $-\times Y$ has a right adjoint. So, in the previous case, $A$ should be homeomorphic to the $2-$torus, but I can not see it.
This is why I have asked this question.
The answer $A \cong S^2 \vee S^1$ isn't correct. Notice that if you remove the collapsed point from $A$, then you get a space homeomorphic to $(0,1) \times S^1$. There is no point of $S^2 \vee S^1$ that has this property (no matter what, you'll still be left with a part of dimension one).
Your space is actually homeomorphic to a sphere with two points identified. Indeed if you collapse separately the two subspaces $0 \times S^1$ and $1 \times S^1$ you get a sphere; when you identify the two, you get a sphere with two points identified. This space is homotopically equivalent to $S^2 \vee S^1$, but not homeomorphic to it.
As for what you wrote before, you can show that the coequalizer of $f \times id$ and $g \times id$ is the quotient $I \times S^1 / \sim$ where $(f(x), y) \sim (g(x), y)$ for all $x \in *, y \in Y$, that is the coequalizer is $I \times S^1 / (0, y) \sim (1, y) \cong T^2$, the torus. Not $A$.