Let \(D\) be the open unit disk in the complex plane that is \(D:=\{z\mid |z|<1\}\). Let \(\sim\) be an equivalence relation on \(D\) defined by \(z_1\sim z_2\) if \(|z_1|=|z_2|\). Is the quotient space \(D/\sim\) Hausdorff? Prove or disprove.
Solution
Let \[ q:D\longrightarrow D/\!\sim \] be the quotient map and let \[ r:D\longrightarrow[0,1), \qquad r(z)=|z|. \]
1. The modulus map \(r\) is constant exactly on the equivalence classes of \(\sim\).
By definition, \[ z_1\sim z_2 \quad\Longleftrightarrow\quad |z_1|=|z_2| \quad\Longleftrightarrow\quad r(z_1)=r(z_2). \] Thus the fibers of \(r\) are precisely the equivalence classes.
2. There is a unique continuous map\[ \bar r:D/\!\sim\longrightarrow[0,1) \]satisfying\[ \bar r(q(z))=|z|. \]
1. By the universal property of the quotient topology, \(r\) therefore factors uniquely as\[ r=\bar r\circ q \]with \(\bar r\) continuous.
3. The map \(\bar r\) is bijective.
For surjectivity, if \(t\in[0,1)\), regard \(t\) as the point \(t+0i\in D\). Then \[ \bar r(q(t))=|t|=t. \] For injectivity, if \[ \bar r(q(z_1))=\bar r(q(z_2)), \] then \(|z_1|=|z_2|\), so \(z_1\sim z_2\) by definition and hence \(q(z_1)=q(z_2)\).
4. The inverse of \(\bar r\) is the continuous map\[ s:[0,1)\longrightarrow D/\!\sim, \qquad s(t)=q(t). \]
The inclusion \[ j:[0,1)\longrightarrow D, \qquad j(t)=t+0i, \] is continuous, and \(q\) is continuous by definition of a quotient map. Thus \[ s=q\circ j \] is continuous. Moreover, \[ (\bar r\circ s)(t)=\bar r(q(t))=t. \] Conversely, for \(q(z)\in D/\!\sim\), \[ (s\circ\bar r)(q(z)) =s(|z|) =q(|z|). \] Since \(z\sim |z|\), their quotient classes agree, so \(q(|z|)=q(z)\). Hence \(s\) and \(\bar r\) are mutual inverses.
5. Therefore\[ D/\!\sim\ \cong [0,1), \]and in particular \(D/\!\sim\) is Hausdorff.
By <1>2–<1>4, \(\bar r\) is a homeomorphism. The interval \([0,1)\) is a subspace of the Hausdorff space \(\mathbb R\), hence is Hausdorff. Hausdorffness is preserved by homeomorphism, so the quotient is Hausdorff.