Changing $[0,2\pi)$ with $S^1$ such that a map defined on $[0,2\pi)$ stays unchanged *
Consider the following procedure of changing the domain of a map, but the map remaining essentially the same - illustrated, for concreteness, in case of the polar-coordinates map.*
Let
\begin{gather*}
\varphi:(0,\infty)\times[0,2\pi)\rightarrow\mathbb{R}^{2}\\
\varphi\binom{r}{\theta}:=\binom{r\sin\theta}{r\cos\theta}.
\end{gather*}
be the polar-coordinates map. Now I'm replacing one part of the cartesian product of which the domain
of $\varphi$ consist, namely I replace $[0,2\pi)$ with some manifold
parametrized by $[0,2\pi)$, for example the sphere $S^{1}$.
 Therefore
I also have to change the definition of $\varphi$ slightly to an
analoguous map $\tilde{\varphi}$ that makes sense on $S^{1}$ instead
of $[0,2\pi)$:
\begin{gather*}
\tilde{\varphi}:(0,\infty)\times S^{1}\rightarrow\mathbb{R}^{2}\\
\tilde{\varphi}\binom{r}{u}:=\varphi\binom{r}{p^{-1}(u)},
\end{gather*}
where $p:[0,2\pi)\rightarrow S^{1}$ parametrizes $S^{1}$. The thing is that the
map $\tilde{\varphi}$ now agrees with $\varphi$ in the following sense:
$$
\varphi\binom{r}{\theta}=\tilde{\varphi}\binom{r}{p(\theta)},\quad\quad\text{for all}\ \theta\in[0,2\pi).
$$
(The parametrization $p$ is of course a bit problematic, as $[0,2\pi)$
isn't an open set, as required in the definition of a manifold by
local parametrizations, but I just wanted to convey the idea of changing
one object, $[0,2\pi)$ with another $S^{1}$ such that the map stays
the same.) 
Does this procedure have a name ? Can it be done in a more general
fashion ? (If yes, what is the abstract algebraic/geometric mechanism
underlying it ?)
 A: I think what you're looking for is the idea of passing to the quotient. 
Theorem. Suppose $X$, $Y$, and $Z$ are topological spaces, and $q\colon X\to Y$ is a quotient map. If $\varphi\colon X\to Z$ is a continuous map that is constant on the fibers of $q$ (meaning that $q(x)=q(y)\implies \varphi(x)=\varphi(y)$), then there is a unique continuous map $\widetilde \varphi\colon Y\to Z$ satisfying $\widetilde \varphi\circ q =\varphi$. 
This is proved, for example, in my book Introduction to Topological Manifolds (Theorem 3.73).
This theorem doesn't apply directly in your situation, however, because the map $\operatorname{id} \times p\colon (0,\infty)\times [0,2\pi) \to (0,\infty)\times S^1$ is not a quotient map. It's easy to see what goes wrong in that case: a map $\varphi\colon  (0,\infty) \times [0,2\pi) \to \mathbb R^2$ will not descend to a continuous map from $(0,\infty)\times S^1$ unless $\varphi(r,\theta)$ approaches $\varphi(r,0)$ as $(r,\theta)\to (r,2\pi)$. If that's the case, $\phi$ extends to a continuous map defined on $(0,\infty)\times [0,2\pi]$, and the natural extension of $\operatorname{id} \times p$ to $(0,\infty)\times [0,2\pi]$ is a quotient map.
By the way, there's also a smooth version of this theorem: If $X$, $Y$, and $Z$ are smooth manifolds, $q$ is a smooth surjective submersion, and $\varphi$ is a smooth map, then the same conclusion holds with the additional property that $\widetilde\varphi$ is smooth. (See my Introduction to Smooth Manifolds, Theorem 4.30.)
