Is this possible to define the continuous map from equivalent class?

Question

Let $M$ be a compact metric space and $f: M \rightarrow M$ be a quotient map.

Consider the following equivalent relation $\sim$ such that $x_1\sim x_2 \Leftrightarrow f(x_1)=f(x_2)$ for a given quotient map $f$.

We know that $f$ is closed map. Hence, there exists a homeomorphism $\tilde{f}:M/\sim \rightarrow M$ by $\tilde{f}([x])=f(x)$, where $[x]=\{x' \mid f(x)=f(x')\}$ is equivalence class.

Now I am curious about this part.

Can we define a continuous function $\alpha:M/\sim \rightarrow M$ such that $\alpha([x])=x' \in [x]$ ?

Answer 1

If $[x]$ has at least two elements then $\alpha([x])=x'\in[x]$ would depend on the choice of $x'$. In other words $\alpha$ is not well defined in general.

But the definition $\tilde f([x]) = f(x')$ for $x'\in[x]$ is correct. It doesn't depend on the choice of $x'\in[x]$, because for any $x_1, x_2\in[x]$ we have $f(x_1)=f(x_2)$ by the definition of $\sim$.

The question about the continuity of $\tilde f$ is a whole different story though and requires a proof. But it's a more general fact:

Let $X, Y$ be topological spaces and $f:X\to Y$ a quotient map. Define a relation on $X$ by $x\sim y$ iff $f(x)=f(y)$. Then there's a well defined homeomorphism $F:X/\sim \to Y$ given by $F([x])=f(x)$.

Note that $X, Y$ don't have to be compact, metric or equal and $f$ does not have to be closed.