How can there exist a quotient map to a space of higher dimension than the domain?

I don't need a completely formal explanation, just some intuition. My professor stated identifying antipodal points on $$S^1$$, yields the projective plane $$P^2$$. That means there exists a quotient map (and thus a surjective map) $$q: S^1\to P^2$$.

Intuitively, this confuses me. How can identifying points together raise the dimension of the resulting space? Lowering or keeping the dimension both make sense to me, but I can't wrap my mind around it raising the dimension. Like, I knew that surjective linear maps $$T: X\to Y$$ require that $$\dim Y \leq \dim X$$, but this clearly doesn't hold for all continuous maps, just linear ones.

Could someone offer some intuition behind this, or how they think about it? I can wrap my head around the codomain being "larger" than the domain (like a continuous bijection $$f: [0, 1]\to\mathbb{R}$$, even though since there's a bijection $$\mathbb{R}$$ technically isn't larger) but not the codomain having higher dimension.

• Your professor is wrong (or you quoted him wrongly). Identifying anipodal points on $\Bbb S^1$ gives the real projective line $\Bbb RP^1$ which is still one-dimensional and in fact homeomorphic to $\Bbb S^1$...The two-sphere $\Bbb S^2$ will give the projective plane $\Bbb RP^2$.. Dec 4 '21 at 23:00
• A continuous map can have finite fibres and still be dimension raising. Theorems exist on this. Dec 4 '21 at 23:03
• @HennoBrandsma I probably just misunderstood him lol. Thank you, though! Dec 4 '21 at 23:10
• Here's a simpler example. Identifying the endpoints of $[0,1]$ gives a circle, and a circle lives in a two-dimensional space. Dec 4 '21 at 23:52
• The instructor perhaps meant to start with the disk $D^2$ and then identify antipodal points on its boundary $S^1$. Dec 5 '21 at 1:12

It’s just an error, by you or by your professor. The projective plane results from identifying antipodal points on $$S^2.$$
However, there do exist counterintuitive quotient maps that increase dimension. A space-filling curve is a continuous surjection $$[0.1]\to [0,1]^2$$; since the domain is compact and the codomain is Hausdorff, this is also a closed map, thus in particular, a quotient map. (This cannot happen with differentiable functions, or with injective continuous functions, happily.)
• @Obamafish Take the Cantor function $f: C \to [0,1]$ (see Wikipedia) and use that $C \simeq C^2$ for the Cantor set and extend linearly (or by Tietze) the resulting map $f \times f: C \to [0,1]^2$. Bit of trickery, OK, but quite standard. So we can go from zero-dimensional $C$ onto two-dimensional $[0,1]^2$ too. Or any dimension really. Any compact metric space is a quotient image of a zero-dimensional space. Dec 4 '21 at 23:15