The topological cone is defined as $(S^1 \times [0,1]) / (S^1 \times \{1\})$. This is a quotient group that which elements are the cosets of $S^1 \times \{1\}$. Now I'm trying to understand how do these form the cone. The cosets are of the form $(1,0) + S^1 \times \{1\} , (0,1) + S^1 \times \{1\}$ e.g $(x,y) + S^1 \times \{1\}$, where $(x,y) \in S^1$ right?
So if $S^1 \times \{1\} = \{(x,y) \mid \|(x,y)\| = 1 \} \times \{1\} = \{(x,y,1) \mid \|(x,y,1)\| = 1 \}$, then is for example $(1,0) + S^1 \times \{1\} = \{(x,y,1) \mid \|(1+x,y,1) = 1\} \in (S^1 \times [0,1]) / (S^1 \times \{1\})$?
The other way to think about the set $(S^1 \times [0,1]) / (S^1 \times \{1\})$ seems to be that all elements of $ (S^1 \times \{1\})$ are being "collapsed" to zero? which would make sense since $ (S^1 \times \{1\})$ defines just the tip of the cone at $z=1$.
I have hard time with the second interpreation since I cannot formalize it in any way. It seems to be just one of those things that you just have to accept... How is this formalized?
Welcome to MSE. Because there is no (natural) group structure on the cylinder $S^1 \times [0, 1]$, there is no real way of getting your proposed group-theoretic definition off the ground.
On the other hand, here is how we formalize the second notion of “collapsing”. The idea is to impose a certain equivalence relation on $S^1 \times [0, 1]$. The denominator of the expression $S^1 \times [0, 1] / S^1 \times \{1\}$ specifies the equivalence classes of the relation: $(x, 1) \sim (y, 1)$ for all $x, y \in S^1$, and all other points $(x, t)$ for $x \in S^1$ and $t \in [0, 1)$ are inequivalent to each other (i.e. their equivalence classes are singletons). This set of equivalences is then topologized with the quotient topology to give the set of equivalence classes the shape of a cone.
Finally, let me mention that when you are working with an object that is both a group and a space (see under “quotients and normal subgroups” here) the group-theoretic/“cosets” and topological/ “equivalence classes” notions of quotient object do coincide.