= Solution
{c}
We first show that a <continuous injection> $j:S^1\to S^1$ must be <surjective>. Otherwise its <image> misses a point $q$. <Stereographic projection> identifies $S^1\setminus\{q\}$ with $\mathbb R$, giving a <continuous injection> $u:S^1\to\mathbb R$. The <extreme value theorem> supplies distinct points $p_-,p_+$ where $u$ attains its minimum $m$ and maximum $M$, with $m<M$ because $u$ is <injective>.
The two <circular arcs> from $p_-$ to $p_+$ each contain a point at which $u=(m+M)/2$, by the <intermediate value theorem>. Those points are distinct because they lie in the disjoint interiors of the two arcs. This contradicts <injectivity>, proving the claim about $j$.
Now suppose $F:S^2\to S^1$ were a <continuous injection>. Its restriction to the equatorial <circle> $E=\{(x,y,0):x^2+y^2=1\}$ would be a <continuous injection> from $S^1$ to itself, hence $F(E)=S^1$. The north pole $N=(0,0,1)$ is outside $E$, but $F(N)$ must equal $F(e)$ for some $e\in E$. This contradicts <injectivity>. Therefore
$$
\boxed{\text{there is no continuous injection }S^2\longrightarrow S^1.}
$$
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