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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.

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    $\begingroup$ 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 $\Bbb RP^2$.. $\endgroup$ Dec 4 '21 at 23:00
  • $\begingroup$ A continuous map can have finite fibres and still be dimension raising. Theorems exist on this. $\endgroup$ Dec 4 '21 at 23:03
  • $\begingroup$ @HennoBrandsma I probably just misunderstood him lol. Thank you, though! $\endgroup$
    – Obamafish
    Dec 4 '21 at 23:10
  • $\begingroup$ Here's a simpler example. Identifying the endpoints of $[0,1]$ gives a circle, and a circle lives in a two-dimensional space. $\endgroup$ Dec 4 '21 at 23:52
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    $\begingroup$ The instructor perhaps meant to start with the disk $D^2$ and then identify antipodal points on its boundary $S^1$. $\endgroup$ Dec 5 '21 at 1:12
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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.)

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    $\begingroup$ I probably just misunderstood him. But also, that's super interesting. Could you give an example of a space-filling curve? I tried Googling it, but just saw animations of fractals rather than explicit functions. Thanks! $\endgroup$
    – Obamafish
    Dec 4 '21 at 23:11
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    $\begingroup$ @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. $\endgroup$ Dec 4 '21 at 23:15
  • $\begingroup$ @hennobrandsma just a typo. $\endgroup$ Dec 5 '21 at 6:25

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