Does $e^{\frac{1}{z}}$ belongs to $W^{1,p}(B(0,1),\mathbb{CP}^1)$? Some background for this question. Given a Riemannian manifold $(M,g)$ and a compact Riemannian manifold $(N,h)$ together with an isometric embedding $(N,h) \hookrightarrow (\mathbb{R}^N, g_{\mathrm{euc}})$, one can define the Sobolev spaces of mappings between manifolds as
$$W^{k,p}(M,N) = \{ u \in W^{k,p}(M,\mathbb{R}^n) \,\, | \,\, u \in N \, \mathrm{a.e} \}.$$
This definition depends on the embedding of $N$, but one can show that different embeddings result in equivalent definitions in an appropriate sense.
Now, I'm trying to understand whether the function $u(z) = e^{1/z} : B(0,1) \rightarrow \mathbb{CP}^1$ is in $W^{1,p}(B(0,1), \mathbb{CP}^1)$ for some $1 \leq p < 2$, where we put the Fubini-Study metric on $\mathbb{CP}^1$ and use the Euclidean metric on the unit disk $B(0,1) \subset \mathbb{C}$. Writing explicitly the embedding of $\mathbb{CP}^1$ into $\mathbb{R}^3$ and choosing one of the coordinates, the question is translated, up to some constants, into whether the following function
$$f(x,y) = \frac{2e^{\frac{x}{x^2+y^2}} \cos \left( \frac{y}{x^2+y^2} \right)}{1+e^{\frac{2x}{x^2+y^2}}} = \mathrm{sech}\left(\frac{x}{x^2+y^2}\right)\cos\left(\frac{y}{x^2+y^2}\right)$$
belongs to $W^{1,p}(B)$. I'm somewhat lost with the calculations here. Can this be answered without much technical work?
 A: The function 
$$f(x,y)=\operatorname{sech} \left(\frac{x}{x^2+y^2}\right)\cos\left(\frac{y}{x^2+y^2}\right)$$
is in $W^{1,p}(B)$ for $1\le p<3/2$. Indeed, the function is ACL (since it's locally Lipschitz on the punctured disk), so the integrability of $|\nabla f|^p$ is the only thing to worry about. 
It helps to unwind the inversion $1/z$ and work with the coordinates $u=x/(x^2+y^2)$ and $v=y/(x^2+y^2)$. (Chances are it would be easier for you to work this way from the beginning.) Letting $g(u,v)=f(x,y)$ we find that $|\nabla f(x,y)|^p=(u^2+v^2)^p |\nabla g(u,v)|^p$ because conformal transformation simply multiplies the gradient by derivative of the map. Taking the Jacobian into account, we obtain 
$$\iint_B |\nabla f|^p \,dxdy = \iint_{B^c} |\nabla g|^p (u^2+v^2)^{p-2}\,dudv$$
Both partial derivatives of $g(u,v)=\operatorname{sech} u\cos v$ are majorized by $\operatorname{sech} u$, which decays exponentially. Thus, the convergence of integral over $B^c$ is determined by its convergence over the set $|u|\le 1\le |v|$. (Contribution of the strips $1\le |u|\le 2$, $2\le |u|\le 3$, etc is a negligible exponential tail.) Since
$$
\iint_{|u|\le 1\le |v|} |\operatorname{sech} u|^p (u^2+v^2)^{p-2}\,dudv 
\approx \int_{1}^\infty v^{2(p-2)}\,dv 
$$
we have convergence when $p<3/2$. And only then, because none of the estimates were too wasteful.

Answer with a different function in it, might be of some $\epsilon$ value
Every Sobolev function $f$ has an ACL representative. Namely, after redefining $f$ on a set of measure zero, we can ensure that for almost every $y$ the one-variable function $f(\cdot,y)$ is absolutely continuous (on the segment on which it's defined). Of course, the same holds for the other coordinate. The proof is based on Fubini's theorem: since the weak derivative $\partial f/\partial x$ is integrable with respect to planar measure, its restriction to almost every segment is integrable with respect to linear measure. Integrability of one-dimensional weak derivative gives absolute continuity. 
The ACL property allows us to quickly check that 
$$f(x,y)=\sec \left(\frac{x}{x^2+y^2}\right)\cos\left(\frac{y}{x^2+y^2}\right)$$
is not a Sobolev function on $B$. Indeed, the secant blows up along the circle $x=\frac{\pi}{2}(x^2+y^2)$. Every horizontal segment passing near the origin will cross the circle. So, no matter which representative of $f$ we take, the restriction to almost every horizontal segment near the origin will be unbounded, hence not continuous, and a fortiori not absolutely continuous.
