= Solution
{c}
The <sphere> $S^2$ is <path-connected>. For $u,v\in S^2$ which are not an <antipodal pair>, a <continuous path> is
$$
\gamma(t)=\frac{(1-t)u+tv}{\|(1-t)u+tv\|},\qquad 0\leq t\leq1.
$$
The denominator never vanishes: a zero would make $u$ and $v$ antipodal. If $v=-u$, choose a <unit vector> $w$ perpendicular to $u$ and concatenate the paths from $u$ to $w$ and from $w$ to $v$ of the same form. Thus $S^2$ is <connected>. The <quotient map> $q_S$ is <continuous> and <surjective>, so the <continuous image of a connected space> theorem proves that \b[$\mathbb{RP}^2$ is <connected>]. In fact composing the paths above with $q_S$ also proves <path connectedness>.
Back to article page
