Let $G$ be a multiplicative group. A function $\text{sqrt}: G \to G$ is a square root function iff, for every $x \in G$, $(\text{sqrt}(x))^2 = x$.
The unit real numbers, namely $\pm 1$, doesn't even have a well-defined square root function. The unit complex numbers $S^1 \subset \mathbb{C}$ has a well-defined square root function, but no branch cut is continuous. But what about the unit quaternions $S^3 \subset \mathbb{H}$?
I think the answer is negative. Here's my attempt. Suppose a continuous $\text{sqrt}: S^3 \to S^3$ exists. Since $S^3$ is a topological group (and thus an H-space), the homotopy set $[S^3, S^3]$ has a group structure (denoted by $+$), defined as pointwise multiplication. Now consider the homotopy class $[\text{sqrt}]$. By the definition of $\text{sqrt}$, and by the group structure of $[S^3, S^3]$, $[\text{sqrt}] + [\text{sqrt}] = [(\cdot)^2 \circ \text{sqrt}] = [\text{id}]$. This contradicts against that $[\text{id}]$ is a generator of the infinite cyclic group $[S^3, S^3] = \mathbb{Z}$.
By a similar proof, the unit octonions $S^7 \subset \mathbb{O}$ doesn't admit a continuous square root function either. Is my proof correct?