= Solution
{c}
Let $q_S:S^2\to S^2/{\sim_S}$ and $q_D:D^2\to D^2/{\sim_D}$ be the <quotient maps>. The <continuous map>
$$
h:D^2\longrightarrow S^2,\qquad h(x,y)=\left(x,y,\sqrt{1-x^2-y^2}\right)
$$
is a <homeomorphism> onto the upper <hemisphere>; its <inverse function> is the <restriction> of the <projection map> $(x,y,z)\mapsto(x,y)$. Every <antipodal pair> on $S^2$ meets that <hemisphere>. Moreover, two points of this <hemisphere> are antipodal only if both lie on the <equator>. Thus
$$
q_S(h(u))=q_S(h(v))\quad\Longleftrightarrow\quad u\sim_D v.
$$
By the <universal property of the quotient topology>, $q_S\circ h$ induces a <continuous bijection>
$$
H:D^2/{\sim_D}\longrightarrow S^2/{\sim_S},\qquad H(q_D(u))=q_S(h(u)).
$$
To justify <continuity> of the <inverse function>, the source is <compact>, being a <continuous image of a compact space> $D^2$; the <Heine-Borel theorem> gives <compactness> of $D^2$. The target is <Hausdorff>, as follows. For distinct <equivalence classes> $[u]\neq[v]$, put
$$
\delta=\min(\|u-v\|,\|u+v\|)>0.
$$
In $S^2$, take the <union> $U$ of the two relative <open balls> of radius $\delta/3$ about $u,-u$, and similarly take $V$ about $v,-v$. Every distance between their respective centres is at least $\delta$, so the <triangle inequality> gives $U\cap V=\varnothing$. These <open sets> contain entire <antipodal pairs>, hence $q_S^{-1}(q_S(U))=U$ and similarly for $V$. The <quotient topology> therefore makes $q_S(U)$ and $q_S(V)$ disjoint <open> <neighbourhoods> of the two classes.
The <compact-to-Hausdorff continuous bijection theorem> now gives \b[the required <homeomorphism> $H$].
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