# Why do these geometric assumptions imply these statements about relative homology?

I'm reading the paper Coverage in sensor networks via persistent homology.

As in the paper, let $\mathcal{D}$ be a bounded domain in $\mathbf{R}^d$. We make the following assumptions:

A5 The restricted domain $\mathcal{D} - N_{\hat{r}}(\partial \mathcal{D})$ is connected, where $$N_{\hat{r}}(\partial \mathcal{D}) = \{x \in \mathcal{D} : \|x - \mathcal{D}\| \leq \hat{r}\}, \text{ where } \hat{r} = r_f + r_s/\sqrt{2}.$$ A6 The fence detection hypersurface $\Sigma = \{x \in \mathcal{D}: \|x-\partial \mathcal{D}\| = r_f\}$ has internal injectivity radius at least $r_s/\sqrt{2}$ and external injectivity radius at least $r_s$.

Here, $r_f$ and $r_s$ are real numbers.

In section 3.1 on page 8 they say:

With assumptions A5 and A6, the top dimensional relative homology $H_d(\mathcal{D}, N_{\hat{r}}(\partial \mathcal{D}))$ has rank one. Furthermore, $H_d(\mathcal{U} \cup N_{\hat{r}}(\partial \mathcal{D}), N_{\hat{r}}(\partial \mathcal{D}))$ is nonzero if and only if $\mathcal{U}$ contains $\mathcal{D} - N_{\hat{r}}(\partial \mathcal{D})$.

(By the way, $\mathcal{U}$ is a union of open balls centered at points in $\mathcal{D}$.)

Why do these things hold?

Edit. More detail about the assumptions:

Assumption A5 is needed to prevent the domain from being too 'pinched'.

(...)

Assumption A6 means that the outermost boundary cannot exhibit large-scale 'wrinkling'.

(...)

Let $\mathcal{S}$ denote the set of points in $\mathbf{R}^d$ within distance $r_s\sqrt{(d-1)/2d}$ of $\Sigma$ on the interior side (i.e. the side of $\Sigma$ corresponding to the interior of $\mathcal{D}$), and within distance $r_s$ of $\Sigma$ on the exterior side.

(...)

From A6, we know that $\mathcal{S}$ is a disjoint collection of thickened $(d-1)$-dimensional surfaces in $\mathbf{R}^d$ each of thickness at most $$\Delta = r_s\left(1+\sqrt{\frac{d-1}{2d}}\right).$$

(...)

Remark 4.5. The precise statement of A6 in terms of injectivity radii requires the curve to be smooth. From the proof of theorem 3.4, it is clear that the crucial condition is to have the shell $\mathcal{S}$ represent annular domains of thickness bounded by $\frac{3}{2}r_s$. In practice, having $\partial\mathcal{D}$ piecewise-linear is admissible: even though the injectivity radii degenerate to zero, the set $\mathcal{S}$ is still an annular region(s) of width bounded by some larger length, depending on the sharpness of the curves. For a piecewise-linear $\partial\mathcal{D}$, an increase in $r_w$ based on the angle of the sharpest corner in the outermost boundary component makes the criterion rigorous.

This may help clarify the meaning of A6, though most of this (everything except the first two statements) comes after the part that I'm asking about.

• What are the internal and external injectivity radii? I have my guesses, but they might not (and, probably, do not) agree with the definitions you are using. You also need to add more assumptions on the domain $D$, say, contractibility, otherwise, I am pretty sure the statements will fail. – Moishe Kohan May 23 '14 at 22:03
• There aren't any more assumptions on the domain $\mathcal{D}$ in the paper as far as I can see. They don't define the injectivity radii, but I've added some more detail to the question that may help. – Ricardo Buring May 24 '14 at 8:49
• In your edit, you still did not define what $r_f$ and $r_s$ are. Without this definition, your question is meaningless as homology depends on these radii. If $r_f$ and $r_s$ are not defined in the paper you are reading, you should look at other papers by its authors and see if they have a precise definition there. If they do not, write to the authors. – Moishe Kohan May 24 '14 at 17:46

In addition to the Assumption A5, I will make another one, namely, that the complement $D\setminus N_{r}(\partial D)$ is nonempty (without this assumption the claim is clearly false). Here and below I set $r=\hat{r}$.
I will also assume that the boundary of $N_r(\partial D)$ is a topological submanifold in $R^d$, in order to simplify the calculation (this assumption can be avoided). Then, by excision: $$H_*(D, N_r(\partial D))\cong H_*(D_r, \partial D_r),$$ where $D_r$ is the complement in $D$ of the interior of $N_r(\partial D)$. Thus, $D_r$ is a $d$-dimensional compact connected manifold with nonempty boundary. By the Lefschetz-Poincare duality: $$H_d(D_r, \partial D_r)\cong H^0(D_r)\cong {\mathbb Z},$$ by the connectivity assumption. This answers the first question.
Consider now, some open subset $U\subset D$ and $U_r:= U\cap D_r, V_r:= U_r\cap \partial D_r$. Again, by Lefschetz-Poincare duality: $$H_d(U_r, V_r)\cong H^0_c(U_r).$$ The latter is zero precisely when $U_r$ has only noncompact components, i.e., $U$ does not contain $D_r$.