Quick question: functions to spaces with equivalence relations

Question

So I'm a little confused about sending functions from spaces without equivalence relations to a space with equivalence relations. For example, I'm trying to define a function $f : S^{n} \rightarrow D^{n} / ({x \sim -x, x \in S^{n-1})}$by taking points (x,y,z,...) to (x,y,z,...0) (taking away the last coordinate). Does this make sense? Will the equivalence relation kick in after the fact? I.e. if I just map $S^{n}$ to $D^{n}$ in an onto manner will this end up being a valid function from S^{n} $\rightarrow D^{n} / ({x \sim -x, x \in S^{n-1})}$?

Answer 1

Any continuous function valued in $D^n$ extends to your quotient space. This is actually simply because the composition of continuous functions is continuous: you have a quotient map $D^n\to D^n/\sim$, and so you get a continuous map $S^n \to D^n/\sim$ as the composition $S^n\to D^n\to D^n/\sim$.

Answer 2

In general, you can define a map from $X$ to $A/B$ by defining a map from $X$ to $A$, and then taking the equivalence class of the resulting element of $A$. But not every map from $X$ to $A/B$ can be obtained this way -- indeed, many of the interesting ones turn out not to be. (To be more exact: if you see a continuous map from $X$ to $A/B$, it may not arise from a continuous map from $X$ to $A$.)

As an example: The map from $S^1$ to $[0, 2] / (0\sim 2)$ defined by $\theta \mapsto \theta/2 \bmod 2$ is not the quotient of any continuous map from $S^1$ to the interval $[0, 2]$.