## Equality of Interleaving Distances

Finally we get to proving theorem 9. We reuse the notation and the definitions from the previous two subsections. We aim to show that the Reeb precosheaf $$\mathcal{C}$$ defines a $$1$$-homomorphism from $$\mathbf{D}$$ to $$\mathbf{C}$$. Now in order for this statement even to make any sense, the image of $$\mathbf{D}$$ under $$\mathcal{C}$$ should lie in $$\mathbf{C}$$.

1. Lemma. The image of $$\mathbf{D}$$ under $$\mathcal{C}$$ is part of the subcategory $$\mathbf{C}$$.

Now let $$f \colon X \rightarrow {\mathbb{R}}$$ be a bounded constructible $${\mathbb{R}}$$-space. To proof the previous lemma it suffices to show that $$\mathcal{C} (r + \mathcal{E} f)$$ and $$\mathcal{C} (r + \mathcal{R} \mathcal{E} f)$$ are an object of $$\mathbf{C}$$ for all $$r \in (-\infty, \infty]$$.

We consider the case $$r = 0$$ first. By lemma 19 the precosheaf $$\mathcal{C} \mathcal{E} f$$ is an object of $$\mathbf{C}$$. By lemma 30 from the last appendix the projection $$\mathcal{E} f$$ from $${\operatorname{epi}}f$$ to $$\overline{{\mathbb{R}}}$$ is a constructible $$\overline{{\mathbb{R}}}$$-space. This in conjunction with lemma 6 yields the following

1. Lemma. The homomorphism $$(\mathcal{C} \circ \pi \circ \mathcal{E})_f$$ from $$\mathcal{C} \mathcal{E} f$$ to $$\mathcal{C} \mathcal{R} \mathcal{E} f$$ is an isomorphism of precosheaves.

Now $$\mathbf{C}$$ is closed under isomorphisms and thus also $$\mathcal{C} \mathcal{R} \mathcal{E} f$$ lies in $$\mathbf{C}$$.

We continue with the case of $$r$$ not necessarily being $$0$$. To this end let $$g \colon Y \rightarrow (-\infty, \infty]$$ be a continuous function, possibly an object of $$\mathbf{D}$$. We note that for any $$r \in (-\infty, \infty]$$ we have $$r + g = \mathcal{S}((\infty, -r)) (g)$$, by definition of $$\mathcal{S}$$. We now recall that we also defined the endofunctors associated to the smoothing functor $$\mathcal{S}'$$ for $$\mathbf{D}$$ on the whole category of set-valued precosheaves on $$\overline{D}$$ and not just $$\mathbf{D}$$. With $$\mathbf{D}$$ being invariant under these endofunctors, lemma 23 follows from the following

1. Lemma. For all $$\mathbf{a} \in -D$$ we have $$(\mathcal{C} \circ \mathcal{S}(\mathbf{a}))_g = (\mathcal{S}'(\mathbf{a}) \circ \mathcal{C})_g$$.

• Proof. For $$(a, b) \in -D$$ we have $$\Delta \circ \mathcal{S}((a, b))(g) = S^{(-b, -b)} \circ \Delta \circ g$$ and this implies the claim.

Next we show that $$\mathcal{C}$$ is compatible with the natural transformations of the smoothing functors. To this end let $$\mathbf{a}, \mathbf{b} \in -D$$ with $$\mathbf{a} \preceq \mathbf{b}$$.

1. Lemma. We have $$(\overline{\mathcal{S}}(\mathbf{a} \preceq \mathbf{b}) \circ \pi^2_* \circ \mathcal{C} \circ \mathcal{E})_f = (\pi^2_* \circ \mathcal{C} \circ \mathcal{S}(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{E})_f$$.

• Proof. First we set $$(a', a) := \mathbf{a}$$ and $$(b', b) := \mathbf{b}$$. Now let $$r \in {\mathbb{R}}$$. If we unravel the definitions we obtain $(\overline{\mathcal{S}}(\mathbf{a}) \pi^2_* \mathcal{C} \mathcal{E} f)( [-\infty, r) ) = \Lambda({\operatorname{epi}}f \cap X \times [-\infty, r + a))$ and $(\overline{\mathcal{S}}(\mathbf{b}) \pi^2_* \mathcal{C} \mathcal{E} f)( [-\infty, r) ) = \Lambda({\operatorname{epi}}f \cap X \times [-\infty, r + b)) .$ Let $$i$$ be the inclusion of $${\operatorname{epi}}f \cap X \times [-\infty, r + a)$$ into $${\operatorname{epi}}f \cap X \times [-\infty, r + b)$$ and let $\tau \colon {\operatorname{epi}}f \cap X \times [-\infty, r + a) \rightarrow {\operatorname{epi}}f \cap X \times [-\infty, r + b), (p, t) \mapsto (p, t - a + b) ,$ then $(\overline{\mathcal{S}}(\mathbf{a} \preceq \mathbf{b}) \circ \pi^2_* \circ \mathcal{C} \circ \mathcal{E})_{f ~ [-\infty, r)} = \Lambda(i)$ and $(\pi^2_* \circ \mathcal{C} \circ \mathcal{S}(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{E})_{f ~ [-\infty, r)} = \Lambda(\tau) .$ Now let $$(p, t) \in {\operatorname{epi}}f \cap X \times [-\infty, r + a)$$, then $$\{p\} \times [t, t - a + b]$$ is contained in $${\operatorname{epi}}f \cap X \times [-\infty, r + b)$$. Moreover we have $$(p, t), (p, t - a + b) \in \{p\} \times [t, t - a + b]$$, hence $$\Lambda(i) = \Lambda(\tau)$$.

• Corollary. We have $$(\mathcal{S}'(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{C} \circ \mathcal{E})_f = (\mathcal{C} \circ \mathcal{S}(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{E})_f$$.

• Proof. This follows from $$\pi^2_*$$ being a $$1$$-homomorphism and being full and faithful on $$\mathbf{C}$$.

• Corollary. We have $$(\mathcal{S}'(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{C} \circ \mathcal{R} \circ \mathcal{E})_f = (\mathcal{C} \circ \mathcal{S}(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{R} \circ \mathcal{E})_f$$.

• Proof. By the lemmata 22 and 25 and the previous corollary we have the commutative diagram $\xymatrix@R+=4pc@C+=6pc{ \mathcal{S}'(\mathbf{a}) \mathcal{C} \mathcal{E} f \ar[r]^{ (\mathcal{S}'(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{C} \circ \mathcal{E})_f } \ar[d]|-{ (\mathcal{S}'(\mathbf{a}) \circ \mathcal{C} \circ \pi \circ \mathcal{E})_f } & \mathcal{S}'(\mathbf{b}) \mathcal{C} \mathcal{E} f \ar[d]|-{ (\mathcal{S}'(\mathbf{b}) \circ \mathcal{C} \circ \pi \circ \mathcal{E})_f } \\ \mathcal{S}'(\mathbf{a}) \mathcal{C} \mathcal{R} \mathcal{E} f \ar[r]_{ (\mathcal{C} \circ \mathcal{S}(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{R} \circ \mathcal{E})_f } & \mathcal{S}'(\mathbf{b}) \mathcal{C} \mathcal{R} \mathcal{E} f . }$ Now by lemma 24 the vertical arrows are isomorphisms and thus the naturality of $$\mathcal{S}'(\mathbf{a} \preceq \mathbf{b})$$ implies that $$(\mathcal{C} \circ \mathcal{S}(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{R} \circ \mathcal{E})_f = (\mathcal{S}'(\mathbf{a} \preceq \mathbf{b}) \circ \mathcal{C} \circ \mathcal{R} \circ \mathcal{E})_f$$.

Eventually the previous two corollaries and lemma 25 imply the

• Proposition. The functor $$\mathcal{C}$$ is a $$1$$-homomorphism from $$\mathbf{D}$$ to $$\mathbf{C}$$.

Now let $$g \colon Y \rightarrow {\mathbb{R}}$$ be another bounded constructible $${\mathbb{R}}$$-space. Then we have the following

• Corollary. The interleavings of $$\mathcal{R} \mathcal{E} f$$ and $$\mathcal{R} \mathcal{E} g$$ with respect to $$\mathcal{S}$$ are in bijection with the interleavings of $$\mathcal{C} \mathcal{E} f$$ and $$\mathcal{C} \mathcal{E} g$$ with respect to $$\mathcal{S}'$$.

• Proof. We observe that all join trees of bounded constructible $${\mathbb{R}}$$-spaces lie in a $$-D$$-subcategory of $$\mathbf{D}$$. By the previous proposition $$\mathcal{C}$$ yields a $$1$$-homomorphism from this $$-D$$-category to $$\mathbf{C}$$. Moreover corollary 8 implies that $$\mathcal{C}$$ is full and faithful on this category and thus the interleavings of $$\mathcal{R} \mathcal{E} f$$ and $$\mathcal{R} \mathcal{E} g$$ are in bijection with those of $$\mathcal{C} \mathcal{R} \mathcal{E} f$$ and $$\mathcal{C} \mathcal{R} \mathcal{E} g$$. Now by lemma 24 these are isomorphic to $$\mathcal{C} \mathcal{E} f$$ and $$\mathcal{C} \mathcal{E} g$$ and this implies the claim.

Now we can finally proof theorem 9. To this end let $$f \colon X \rightarrow {\mathbb{R}}$$ and $$g \colon Y \rightarrow {\mathbb{R}}$$ be continuous maps with $$X$$ and $$Y$$ smooth and compact manifolds.

• Proof (Theorem 9). Let $$\delta > 0$$, by Bröcker and Jänich (1973 Satz 14.8) there are smooth functions $$f' \colon X \rightarrow {\mathbb{R}}$$ and $$g' \colon Y \rightarrow {\mathbb{R}}$$ with $${\lVert f - f' \rVert}_{\infty} \leq \frac{\delta}{8} \geq {\lVert g - g' \rVert}_{\infty}$$. Further there are Morse functions $$f'' \colon X \rightarrow {\mathbb{R}}$$ and $$g'' \colon Y \rightarrow {\mathbb{R}}$$ with $${\lVert f' - f'' \rVert}_{\infty} \leq \frac{\delta}{8} \geq {\lVert g' - g'' \rVert}_{\infty}$$ by Milnor (1963 corollary 6.8). These two results together yield $${\lVert f - f'' \rVert}_{\infty} \leq \frac{\delta}{4} \geq {\lVert g - g'' \rVert}_{\infty}$$. By corollary 14, corollary 4, proposition 16, proposition 21, and the previous corollary, we have $\begin{split} \mu(\mathcal{C} \mathcal{E} f, \mathcal{C} \mathcal{E} g) & \leq \mu(\mathcal{C} \mathcal{E} f, \mathcal{C} \mathcal{E} f'') + \mu(\mathcal{C} \mathcal{E} f'', \mathcal{C} \mathcal{E} g'') + \mu(\mathcal{C} \mathcal{E} g'', \mathcal{C} \mathcal{E} g) \\ & \leq {\lVert f - f'' \rVert}_{\infty} + \mu_J (\mathcal{R} \mathcal{E} f'', \mathcal{R} \mathcal{E} g'') + {\lVert g'' - g \rVert}_{\infty} \\ & \leq \mu_J (\mathcal{R} \mathcal{E} f'', \mathcal{R} \mathcal{E} f) + \mu_J (\mathcal{R} \mathcal{E} f, \mathcal{R} \mathcal{E} g) + \mu_J (\mathcal{R} \mathcal{E} g, \mathcal{R} \mathcal{E} g'') + \frac{\delta}{2} \\ & \leq {\lVert f'' - f \rVert}_{\infty} + \mu_J (\mathcal{R} \mathcal{E} f, \mathcal{R} \mathcal{E} g) + {\lVert g - g'' \rVert}_{\infty} + \frac{\delta}{2} \\ & \leq \mu_J (\mathcal{R} \mathcal{E} f, \mathcal{R} \mathcal{E} g) + \delta . \end{split}$ Similarly we have $$\mu_J (\mathcal{R} \mathcal{E} f, \mathcal{R} \mathcal{E} g) \leq \mu(\mathcal{C} \mathcal{E} f, \mathcal{C} \mathcal{E} g) + \delta$$. Since $$\delta > 0$$ was arbitrary, we have $$\mu_J (\mathcal{R} \mathcal{E} f, \mathcal{R} \mathcal{E} g) = \mu(\mathcal{C} \mathcal{E} f, \mathcal{C} \mathcal{E} g)$$.

The proof for $$M$$ and $$M_J$$ is completely analogous.