Let $k=\mathbb F\_q\left(\left(\frac1T\right)\right)$, $\overline k$ be an algebraic closure of $k$ and $K$ be the completion of $\overline k$ for the $\frac1T$ valuation. Consider the morphism $\sigma:k\to L=K\left(\left(\frac1Z\right)\right)$ defined by $\sigma(\sum\_{n\ge m}a\_n\left(\frac1T\right)^n)=\sum\_{n\ge m}a\_n\left(\frac1Z\right)^n$. Let $\alpha\in \overline k$. Does there exist a continuous morphism $\tilde\sigma: k(\alpha)\to L$ ($L$ is endowed with the topology induced by the $\frac1Z$-valuation) whose restriction to $k$ is $\sigma$?


Thanks in advance for any answer.



Extension of morphisms in function fields
Not if $\alpha = \sqrt{T}$ since $\alpha^2 = T$ is sent to $Z$ which does not have a square root in $L$.


