It is well-known that there is "no" nowhere vanishing continuous tangent vector field on $S^2$, by the so-called Hairy-ball theorem. But then, is there a continuous tangent vector field on $S^2$ which vanishes at only one point?
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2Why is the "no" in quotes? Are you sceptical ? – Henno Brandsma May 31 '20 at 13:49
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Consider the non-zero continuous tangent field to $D^1$ (the open unit disk in the plane)
$$T(r, \theta) = (1-r) \cdot \hat{x},$$
where $r$ is the distance from the origin, and $\hat{x}$ the rightward pointing unit vector.
There exists a diffeomorphism from $D^1$ to $S^2 \setminus \{ NP \}$, where NP is the north pole. Use this map to map the tangent field to a (non-zero!) tangent field $\hat{T}$ on $S^2 \setminus \{NP \}$, and complete it to a tangent field on $S^2$ by setting $\hat{T}(NP) = \vec{0}$.

Alexander Geldhof
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