# Is the inverse of a quotient map continuous?

I tried my best looking this up, but could not find anything. Very sorry if this is a duplicate.

We can define a quotient map $$q:S^\infty \to \mathbb{R}P^\infty$$ by identifying all antipodal points. We can possibly define an inverse to this map by choosing one of the two points. Can we choose the points in such a way that this inverse map is continuous, or does the structure of $$\mathbb{R} P^\infty$$ holds us back in creating such a continuous inverse?

Thank you!

• Do you mean right or left inverse? Right doesn't exist by looking at homology, $S^\infty$ is contractible while $\mathbb{R}P^\infty$ has a non-trivial homology. I'm not sure about left though. – freakish Jun 14 at 19:02
• I mean a map $i:\mathbb{R}P^\infty \to S^\infty$ such that $q\circ i = id$. Currently forgot if thats called a right or a left inverse. – EBP Jun 14 at 19:04
• That's a right inverse (it's on the right side of $q$). – freakish Jun 14 at 19:07
• @freakish A left inverse can’t exist since it isn’t injective. Unless you mean up to homotopy in which case any map will do. – Connor Malin Jun 15 at 1:45

No, there are no continuous $$\mathbb{RP}^\infty\to S^\infty$$ that is right-inverse to the quotient $$S^\infty\to\mathbb{RP}^\infty$$.
Indeed, it fails at the $$1$$-skeleton level already: there are no continuous $$\sqrt{}$$ on the unit circle $$S^1\subset\mathbb{C}$$.
• How $S^1$ generalizes to whole $S^\infty$ exactly? – freakish Jun 14 at 18:59
• If you can't find a right inverse for $\mathbb{RP}^1\subset\mathbb{RP}^\infty$, you can't do it for $\mathbb{RP}^\infty$. – user10354138 Jun 14 at 19:05
• In fact a right inverse $r : \mathbb{RP}^\infty\to S^\infty$ has the property $r(\mathbb{RP}^1) = r(q(S^1) \subset S^1$. Thus $r$ restrics to a right inverse for $q_1 : S^1 \to \mathbb{RP}^1$. – Paul Frost Jul 17 at 16:01