I need some clarification about the reason why we have a sphere bubbling off in the situation described by Seidel in [The Symplectic Floer Homology of a Dehn Twist](https://pdfs.semanticscholar.org/64f9/126064e02574bbc9de79608304264b09fbfb.pdf).


I’ll try to summarize to the best of my abilities the situation I’m interested in:


We are stretching the neck along a circle in $\Sigma$. Let us fix an identification of the tubular neighborhood of such circle with $[-1,1]\times S^1$. Let $R>0$, by $\Sigma^R$ I’ll denote the surface diffeomorphic to $\Sigma$ but with neck $[-R,R]\times S^1$. 


Assume that for $R\_i\to \infty$, we have a sequence of $J^{R\_i}$-holomorphic strips $\{u^i\}\_i$, with $u^i\in \mathcal{M}^{R\_i}(x\_-,x\_+)$ (for a definition see bottom of page 832 of the paper) with uniform bounded energy. 


Notice that we **can’t** have uniform bound on $|du^i|\_{L^{\infty}}$, in particular we can find a sequence of points $z\_i \in [-R\_i,R\_i]\times S^1$ such that $|du^i|$ has a maximum there, with value $C\_i$, and $C\_i\to \infty$. Let $\Sigma’=\Sigma \setminus [-1,1]\times S^1$, (the definition of $\Sigma’$ is independent of the length of the neck!) now assume that, for $i$ big enough, $$d(u^i(z\_i),\Sigma’)\leq k < \infty$$


(I.e. no matter the length of the neck that is stretching to infinity, my Maximums stays within finite distance to some edge of the neck)



> 
> From there the author immediately concludes that after some reparametrization, the $u^i$’s converged to a $J$-holomorphic sphere in the surface obtained from $\Sigma’$ by attaching a semi-infinite cylinder to each boundary component. I can show that, after reparametrization my curves converges to a non-constant $J$-hol map defined on $\Bbb C$, but I don’t see why this map should extend to the sphere. My understanding is that in order to use the removable of singularity theorem we need a compact image, but $\Sigma’$ With the cylindrical ends is non compact.
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Most likely there is some clever way to reparametrize this curves (i used the standard conformal map $z\mapsto z/C\_i+z\_i$ that centers $u^i$ at $0$ and normalize the norm of its differential at $0$), or to infer something about the image of the limit, but I’m unable to see it.


Any hint is really appreciated, since this is bugging me. Thanks in advance!


**Update**


I agree with the given answer that *monotonicity lemma* could be a way to establish compactness of the image of the limiting function. What I don't understand are the following point:


1) Assuming that for small enough balls we can ensure that the topological boundary of our curve is outside such ball (as required by the *monotonicity lemma*) how do we deal with the fact that the target manifold is not compact? That's the key assumption in the lemma. As I wrote in the comments I though that we could just restrict our attention to some compact sub manifolds of $\Sigma' \cup \partial \Sigma' \times [0,\infty)$ but the lower bound provided by the lemma would depend on such a choice. Hence it's not clear to me how to obtain a uniform lower bound on the energy of the curve passing through a given ball.


2)If we work with a vertical almost complex structure on the neck, then I think the lower bound provided by the monotonicity lemma is the same in every sector $[a-n,a+n]\times \partial \Sigma'$ (since they all are isometric and with the same a.c.s.) In that case I can see how to get a uniform lower bound. But what worries me is that I don't see any problem in applying this reasoning to case #1, but as far as I understood, in that case we can't rule out non-constant $j$-hol planes that easily and we must use a different strategy (namely integrability of the a.c.s.) 



Bubbling off of a pseudo holomorphic sphere on surface with cylindrical ends
It turns out that the Removable Singularities Theorem and the Monotonicity Lemma do not require compactness, but that the target manifold should have bounded geometry, such as in our case of inserting a cylindrical piece to the compact surface. See for example, Lemma 5.11 in *"Compactness results in symplectic field theory"* by Bourgeois-Eliashberg-Hofer-Wysocki-Zehnder and Proposition 2.71 of Abbas' book *"An introduction to compactness results in symplectic field theory"*. Seidel brought to my attention Theorem 4.5.1 in Sikorav's paper *"Some properties of holomorphic curves in almost complex manifolds"*, which only needs the (possibly noncompact) target to have bounded geometry and does not need the disc to have bounded image, for removing singularities.


By assumption our points $u^i(z\_i)$ don't run away along the semi-infinite ends attached to $\Sigma'$, i.e. our bubble points are stuck in $\Sigma'$ union a collar $\partial\Sigma'\times[0,k]$ and cannot escape along $\partial\Sigma'\times[0,\infty)$, otherwise we're in Case#1 (in the paper) which has the points running away along the neck of $\Sigma^{R\_i}$. We now zoom in our around these bubble points (hence away from the boundary of the holomorphic strips), forming a rescaled sequence of holomorphic discs. Note that in both the case at hand (Case#2 in the paper) and Case#1 in the paper, our sequence of rescaled maps have domains being open disks centered around the bubble points. It can happen in Case#1 that these disks just shift along with the bubble points down the infinite neck region, which doesn't happen in Case#2.


*Aside:* I hoped that the monotonicity lemma would prevent the sequence of rescaled holomorphic discs from "thinning" out and having unbounded image, that is, our maps must use up at least a certain amount of energy for every ball whose center it passes through. We couldn't a priori apply the usual monotonicity lemma (for compact target manifolds) to each compact set in a compact exhaustion of the noncompact surface, because the resulting (lower) energy bound may vary (hence decrease) with each set. But due to the cylindrical end there is a uniform lower bound. In any case, the a priori issue is that the (image of the) boundary of the discs might squeeze into the balls with which we are trying to compute the area bounds. 


