Let $X\rightarrow Y$ be etale. Let $R$ be a strict henselian ring with residue field $k$. Say that we have a map $Spec(k)\rightarrow X$ and so also $Spec(k)\rightarrow Y$. Assume that the map $Spec(k)\rightarrow Y$ factors thusly: $Spec(k)\rightarrow Spec(R)\rightarrow Y$. Then the question is: would there be (a unique?) map $Spec(R)\rightarrow X$ that would make this commute?

This would be a little cleaner if I knew how to do commutative diagrams in mathoverflow, but hopefully you get the picture. Intuitively, this means that if you have a point on $Y$ and a point above it on $X$, and if you have a "path" (I use this word very loosely here) near that point on $Y$ then it extends to a "path" near the respective point on $X$. When trying to prove this, the first thing I thought about is that every etale map is formally etale, but in the definition of formally etale they only talk about lifting $1^{st}$ order deformations.

Do you know how to prove (or god forbid disprove) this?

Let $X\rightarrow Y$ be a finite etale map. Let $R$ be a strict henselian ring with residue field $k$. Say that we have a map $Spec(k)\rightarrow X$ and so also $Spec(k)\rightarrow Y$. Assume that the map $Spec(k)\rightarrow Y$ factors thusly: $Spec(k)\rightarrow Spec(R)\rightarrow Y$. Then the question is: would there be (a unique?) map $Spec(R)\rightarrow X$ that would make this commute?

This would be a little cleaner if I knew how to do commutative diagrams in mathoverflow, but hopefully you get the picture. Intuitively, this means that if you have a point on $Y$ and a point above it on $X$, and if you have a "path" (I use this word very loosely here) near that point on $Y$ then it extends to a "path" near the respective point on $X$. When trying to prove this, the first thing I thought about is that every etale map is formally etale, but in the definition of formally etale they only talk about lifting $1^{st}$ order deformations.

Do you know how to prove (or god forbid disprove) this?