%
\label{prop:decomposition-functoriality}
Let $f, g \in \mathcal{O}_\tau(\tau^3)$
be $\tau$-holomorphic,
and suppose $g \circ f$ is defined
and $\tau$-holomorphic.
Then:
\begin{enumerate}
    \item[\textup{(F1)}]
          \textbf{Channel preservation:}
          $(g \circ f)_+ = g_+ \circ f_+$
          and $(g \circ f)_- = g_- \circ f_-$.
    \item[\textup{(F2)}]
          \textbf{No channel mixing:}
          the B-channel of $g \circ f$
          depends only on the B-channels
          of $g$ and $f$;
          the C-channel of $g \circ f$
          depends only on the C-channels
          of $g$ and $f$.
    \item[\textup{(F3)}]
          \textbf{Functor property:}
          the map $f \mapsto f_+$
          is a functor from the category
          of $\tau$-holomorphic maps
          to the category of B-channel maps,
          and similarly for $f \mapsto f_-$.
\end{enumerate}

\emph{Proof of (F1).}
We compute $(g \circ f)_+$ at an arbitrary point $x \in \tau^3$:
\begin{align*}
    (g \circ f)_+(x)
    &\;=\; e_+ \cdot (g \circ f)(x) \\
    &\;=\; e_+ \cdot g(f(x)).
\end{align*}
Now write $f(x) = f_+(x) + f_-(x)$
and $g = g_+ + g_-$.
Since $g_+$ takes values in $e_+ \cdot H_\tau$
and $g_-$ takes values in $e_- \cdot H_\tau$,
and $f_+(x) \in e_+ \cdot H_\tau$,
$f_-(x) \in e_- \cdot H_\tau$:
\begin{align*}
    g(f(x))
    &= g_+(f_+(x) + f_-(x))
    + g_-(f_+(x) + f_-(x)).
\end{align*}
The key observation is that
the channels $e_+ \cdot H_\tau$ and $e_- \cdot H_\tau$
form a \emph{direct product of rings}:
\[
    H_\tau \;\cong\; e_+ H_\tau \;\times\; e_- H_\tau.
\]
Under this isomorphism,
a map $g = g_+ + g_-$
acts \emph{componentwise}:
\[
    g(z_+, z_-) = (g_+(z_+),\; g_-(z_-)),
\]
where $z_+ = e_+ z$ and $z_- = e_- z$.
This is because $g_+$ takes values in $e_+ H_\tau$
and receives inputs from $e_+ H_\tau$
(the $e_- H_\tau$ component is annihilated
by orthogonality (I2)),
and similarly for $g_-$.

Explicitly:
the evaluation $g_+(f(x))$
depends only on the $e_+$-component of $f(x)$,
which is $f_+(x)$,
because $g_+$ takes values in $e_+ H_\tau$
and $e_+ \cdot f_-(x) = 0$ by orthogonality.
Therefore $g_+(f(x)) = g_+(f_+(x))$.
Similarly $g_-(f(x)) = g_-(f_-(x))$.

Hence:
\[
    g(f(x))
    = g_+(f_+(x)) + g_-(f_-(x)),
\]
and applying $e_+$:
\[
    e_+ \cdot g(f(x))
    = e_+ g_+(f_+(x)) + e_+ g_-(f_-(x))
    = g_+(f_+(x)) + 0
    = g_+(f_+(x))
    = (g_+ \circ f_+)(x).
\]
Therefore $(g \circ f)_+ = g_+ \circ f_+$.
The argument for the C-channel is identical.

\emph{Proof of (F2).}
This is a restatement of~(F1):
$(g \circ f)_+$ depends on $g_+$ and $f_+$ only,
not on $g_-$ or $f_-$.
The channels do not communicate under composition.

\emph{Proof of (F3).}
We verify the functor axioms.
\emph{Identity:}
$(\id)_+ = e_+ \cdot \id = e_+$,
which acts as the identity on $e_+ H_\tau$.
\emph{Composition:}
$(g \circ f)_+ = g_+ \circ f_+$ by (F1).
Hence $f \mapsto f_+$ is a functor.
The argument for $f \mapsto f_-$ is the same.