We would like to algebraize the following logic. Given a null sequence
\((m_0, m_1, \dots)\text{,}\) if we can rescale by powers of
\(g\) to get another null sequence
\((m_0, g m_1, g^2 m_2, \dots)\text{,}\) and we know that every null sequence is summable against
\(f\text{,}\) we may conclude that the original sequence is summable against
\(fg\text{.}\)
We next encounter a subtlety in the argument that can already be seen in terms of null sequences. The strategy is to invert the function
\begin{equation}
(m_0, m_1, \dots) \mapsto (m_0 - fg m_1, m_1 - fg m_2, \dots),\tag{10.5}
\end{equation}
which symbolically is achieved by
\begin{equation*}
(m_0, m_1, \dots) \mapsto \left( \sum_{n=0}^\infty (fg)^n m_n, \sum_{n=0}^\infty (fg)^n m_{n+1}, \dots \right) \text{;}
\end{equation*}
however, composing the map
\begin{equation}
(m_0, m_1, \dots) \mapsto (m_0, g m_1, g^2 m_2, \dots)\tag{10.6}
\end{equation}
with the map
\begin{equation*}
(m_0, m_1, \dots) \mapsto \left(\sum_{n=0}^\infty f^n m_n, \sum_{n=0}^\infty f^n m_{n+1}, \dots \right)
\end{equation*}
gives
\begin{equation*}
(m_0, m_1, \dots) \mapsto \left( \sum_{n=0}^\infty (fg)^n m_n, g \sum_{n=0}^\infty (fg)^n m_{n+1}, g^2 \sum_{n=0}^\infty (fg)^n m_{n+2}, \dots \right)
\end{equation*}
where the term in position
\(n\) is the term we want but multiplied by
\(g^n\text{.}\) In other words, we cannot perform the operation
(10.5) by first rescaling by
\(g\) as in
(10.6), then performing
\((m_0,m_1,\dots) \mapsto (m_0-fm_1, m_1-fm_2)\text{;}\) rather, this executes
(10.5) plus rescaling by
\(g\text{.}\)
We may resolve this by working in terms of two-dimensional arrays. To wit, to execute
(10.5), we first expand
\((m_0, m_1, \dots)\) into a two-dimensional array and rescale
columns by powers of
\(g\) to get
\begin{equation*}
\begin{pmatrix}
m_0 \amp gm_1 \amp g^2m_2 \amp \\
m_1 \amp gm_2 \amp g^2m_3 \amp \cdots \\
m_2 \amp gm_3 \amp g^2m_4 \amp \\
\amp \vdots \amp \amp \ddots
\end{pmatrix}\text{;}
\end{equation*}
difference rows by \(f\) to get
\begin{equation*}
\begin{pmatrix}
m_0-fgm_1 \amp gm_1-fg^2m_2 \amp g^2m_2-fg^3m_3 \amp \\
m_1-fgm_2 \amp gm_2-fg^2m_3 \amp g^2m_3-fg^2m-4 \amp \cdots \\
m_2-fgm_3 \amp gm_3-fg^2m_4 \amp g^2m_4-fg^2m_5 \amp \\
\amp \vdots \amp \amp \ddots
\end{pmatrix}\text{;}
\end{equation*}
and finally project onto the first column, noting that this operation is a section of the first one.
We now translate these steps into operations on condensed modules, and ultimately on condensed rings. The expansion into a two-dimensional array should give a map \(\Hom_{\CAb}(P, R) \to \Hom_{\CAb}(P \otimes P, R)\text{;}\) this is pullback by the map \(+_P\colon P \otimes P \to P\) induced by the addition map \(\NN_\infty \times \NN_\infty \to \NN_\infty\text{.}\) The rescaling by \(g\) is pullback by the given \(R\)-algebra map \(h_g\colon P \otimes R \to P \otimes R\text{,}\) or more precisely by \(\id \otimes h_g\) to indicate that we are rescaling the columns. Differencing the rows is pullback by \(\id \otimes \Delta_f\text{.}\) Finally, projecting onto the first column is pullback by the first injection \(\iota_1 \colon P = P \otimes \ZZ \to P \otimes P\text{.}\) Undoing the pullbacks (which reverses the order of composition), we end up with a putative equality
\begin{equation}
\Delta_{fg} = +_P \circ (\id \otimes (h_g \circ \Delta_f)) \circ \iota_1\tag{10.7}
\end{equation}
of morphisms \(P \to P \otimes R\text{.}\)
At this point, we still need to verify
(10.7): we can directly verify that both sides act by
\([n] \mapsto [n] - fg[n+1]\text{,}\) but this is only sufficient when
\(R\) is a topological ring. Fortunately, we may check the equality in the universal case where
\(R = P[T]\text{,}\) \(f = T\text{,}\) and
\(g = [1]\text{,}\) where an argument by elements suffices, and then specialize to the case at hand.
Now we transpose
(10.7) to get
\begin{equation*}
\Delta_{fg}^* = \iota_1^* \circ (\id \otimes (\Delta_f^* \circ h_g^*)) \circ +_P^*\text{.}
\end{equation*}
Given that \(\Delta_f^*\) is invertible on \(\Hom_{\CAb}(P, M)\text{,}\) and using the fact that the maps \(\iota_1^*\) and \((\id \otimes h_g^*) \circ +_P^*\) split each other, we obtain
\begin{equation*}
(\Delta_{fg}^*)^{-1} = \iota_1^* \circ (\id \otimes (\Delta_f^*)^{-1} \circ h_g^*)) \circ +_P^*\text{,}
\end{equation*}
which proves the claim.
