Observe that $H_n = \beta^{-1}(I^nM)/\Im(\alpha) + I^nL$. For $n \geq 2$ we have $\beta^{-1}(I^nM) \subset \beta^{-1}(I^{n - 1}M)$ and $\Im(\alpha) + I^nL \subset \Im(\alpha) + I^{n - 1}L$. Thus we obtain our canonical map $H_n \to H_{n - 1}$. Similarly, we have $\Ker(\beta) \subset \beta^{-1}(I^nM)$ and $\Im(\alpha) \subset \Im(\alpha) + I^nL$ which produces the canonical map $H \to H_n$. We omit the verification that the diagram commutes. \medskip\noindent By Artin-Rees we may choose $c_1, c_2 \geq 0$ such that $\beta^{-1}(I^nM) \subset \Ker(\beta) + I^{n - c_1}L$ for $n \geq c_1$ and $\Ker(\beta) \cap I^nL \subset I^{n - c_2}\Ker(\beta)$ for $n \geq c_2$, see Algebra, Lemmas \ref{algebra-lemma-map-AR} and \ref{algebra-lemma-Artin-Rees}. Set $c = c_1 + c_2$. \medskip\noindent Let $n \geq c$. We define $\psi_n : H_n \to H/I^{n - c}H$ as follows. Say $x \in H_n$. Choose $y \in \beta^{-1}(I^nM)$ representing $x$. Write $y = z + w$ with $z \in \Ker(\beta)$ and $w \in I^{n - c_1}L$ (this is possible by our choice of $c_1$). We set $\psi_n(x)$ equal to the class of $z$ in $H/I^{n - c}H$. To see this is well defined, suppose we have a second set of choices $y', z', w'$ as above for $x$ with obvious notation. Then $y' - y \in \Im(\alpha) + I^nL$, say $y' - y = \alpha(v) + u$ with $v \in K$ and $u \in I^nL$. Thus $$ y' = z' + w' = \alpha(v) + u + z + w \Rightarrow z' = z + \alpha(v) + u + w - w' $$ Since $\beta(z' - z - \alpha(v)) = 0$ we find that $u + w - w' \in \Ker(\beta) \cap I^{n - c_1}L$ which is contained in $I^{n - c_1 - c_2}\Ker(\beta) = I^{n - c}\Ker(\beta)$ by our choice of $c_2$. Thus $z'$ and $z$ have the same image in $H/I^{n - c}H$ as desired. \medskip\noindent The composition $H/I^n H \to H_n \to  H/I^{n - c}H$ is the canonical map because if $z \in \Ker(\beta)$ represents an element $x$ in $H/I^nH = \Ker(\beta)/\Im(\alpha) + I^n\Ker(\beta)$ then it is clear from the above that $x$ maps to the class of $z$ in $H/I^{n - c}H$ under the maps constructed above. \medskip\noindent Let us consider the composition $H_n \to H/I^{n - c}H \to H_{n - c}$. Given $x, y, z, w$ as in the construction of $\psi_n$ above, we see that $x$ is mapped to the cass of $z$ in $H_{n - c}$. On the other hand, the canonical map $H_n \to H_{n - c}$ from the first paragraph of the proof sends $x$ to the class of $y$. Thus we have to show that $y - z \in \Im(\alpha) + I^{n - c}L$ which is the case because $y - z = w \in I^{n - c_1}L \subset I^{n - c}L$. \medskip\noindent Statements (1) -- (4) are formal consequences of what we just proved. Namely, (1) follows from the existence of the maps and the definition of morphisms of pro-objects in Categories, Remark \ref{categories-remark-pro-category}. Part (2) holds because isomorphic pro-objects have isomorphic limits. Part (3) is immediate from part (4). Part (4) follows from the factorization $H_{n + c} \to H/I^nH \to H_n$ of the canonical map $H_{n + c} \to H_n$. \medskip\noindent Proof of part (5). Let $x \in I^cH_n$. Write $x = \sum f_i x_i$ with $x_i \in H_n$ and $f_i \in I^c$. Choose $y_i, z_i, w_i$ as in the construction of $\psi_n$ for $x_i$. Then for the computation of $\psi_n$ of $x$ we may choose $y = \sum f_iy_i$, $z = \sum f_i z_i$ and $w = \sum f_i w_i$ and we see that $\psi_n(x)$ is given by the class of $z$. The image of this in $H_n/I^{n - c}H_n$ is equal to the class of $y$ as $w = \sum f_i w_i$ is in $I^nL$. This proves (5). \medskip\noindent Proof of part (6). Let $y \in \Ker(\beta)$ whose class is $x$ in $H$. If $x$ maps to zero in $H_n$, then $y \in I^nL + \Im(\alpha)$. Hence $y - \alpha(v) \in \Ker(\beta) \cap I^nL$ for some $v \in K$. Then $y - \alpha(v) \in I^{n - c_2}\Ker(\beta)$ and hence the class of $y$ in $H/I^nH$ is annihilated by $I^{c_2}$. Finally, let $x \in H_n$ be the class of $y \in \beta^{-1}(I^nM)$. Then we write $y = z + w$ with $z \in \Ker(\beta)$ and $w \in I^{n - c_1}L$ as above. Clearly, if $f \in I^{c_1}$ then $fx$ is the class of $fy + fw \equiv fy$ modulo $\Im(\alpha) + I^nL$ and hence $fx$ is the image of the class of $fy$ in $H$ as desired.