モナド ── 定義と随伴により引き起こされるモナド では, 任意の随伴からモナドが構成されることを見た.
そのようなモナドの例として maybe モナドについて簡単にまとめる.
$\mathbf{Set}_*$ を点付き集合 (pointed set) のなす圏とする.
忘却関手 (forgetful functor) $U : \mathbf{Set}_* \rightarrow \mathbf{Set}$ は左随伴
\begin{equation}
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\begin{xy}
\xymatrix@=16pt {
(-)_+ \,:\, \mathbf{Set} \ar[r] & \mathbf{Set}_* \\
}
\end{xy}
\end{equation} を持つ.
関手 $(-)_+$ は各集合 $A$ に対して点付き集合
\begin{equation*}
A_+ = A \amalg \{ A \} = (A \amalg \{A\}, A)
\end{equation*} を対応させ, 関数 $f : A \rightarrow B$ に対して $f_+ : A_+ \rightarrow B_+$ を
\begin{equation*}
f_+(a) = \begin{cases}
f(a) & (a \in A) \\
B & (a = A)
\end{cases}
\end{equation*} として対応させる.
また, これに伴う単位 (unit) $\eta : \Un{\Set} \Rightarrow U(-)_+$ は, 各集合 $A$ に対して
\begin{equation*}
\begin{xy}
\xymatrix@=16pt@R=0pc {
\eta_A \,:\, A \ar[r] & A_+ \\
\hspace{2em}a \ar@{}[r]|(.6){\mapsto} & a
}
\end{xy}
\end{equation*} により, また余単位 (counit) $\epsilon : (-)_+ U \Rightarrow \Un{\Set_+}$ は, 各点付き集合 $(A,a_0)$ に対して
\begin{equation*}
\begin{xy}
\xymatrix@=16pt@R=0pc {
\epsilon_A \,:\, A_+ \ar[r] & (A,a_0) & \\
\hspace{1.6em}a \ar@{}[r]|(.6){\mapsto} & a & (a \in A) \\
\hspace{1.6em}A \ar@{}[r]|(.6){\mapsto} & a_0 & (a = A)
}
\end{xy}
\end{equation*} により与えられる.
この随伴
\begin{equation*}
\begin{xy}
\xymatrix@=32pt {
\Set_* \ar@<1ex>[r]^{(-)_+} \ar@{}[r]|{\bot} & \Set \ar@<1ex>[l]^{U}
}
\end{xy}
\end{equation*} から導かれるモナドを maybe モナドと呼ぶ.
maybe モナドを実際に構築してみる.
関手 $T : \Set \rightarrow \Set$ は $T=U(-)_+$ によって与えられる. 集合 $A$ に対して
\begin{equation*}
TA = A_+ = A \amalg \{A\}
\end{equation*} であり, 関数 $f : A \rightarrow B$ に対して
\begin{equation*}
Tf = f_+ : A_+ \rightarrow B_+
\end{equation*} である.
単位 (unit) は $\eta : \Un{\Set} \Rightarrow T$ で定まる.
積 (multiplication) は, 各集合 $A$ に対して
\begin{equation*}
(A_+)_+ = (A \amalg \{A\}) \amalg \{A \amalg \{A\}\} = (A \amalg \{A\}) \amalg \{A_+\}
\end{equation*} であるが
\begin{equation*} .
\begin{xy}
\xymatrix@=16pt@R=0pc {
\mu \,:\, (A_+)_+ \ar[r] & A_+ & \\
\hspace{2em}a \ar@{}[r]|(.6){\mapsto} & a & (a \in A) \\
\hspace{2em}A \ar@{}[r]|(.6){\mapsto} & A & (a = A) \\
\hspace{2em}A_+ \ar@{}[r]|(.6){\mapsto} & A & (a = A_+) \\
}
\end{xy}
\end{equation*} により定まる.
$(T,\eta,\mu)$ が実際にモナドになっていることを示す. 図式
\begin{equation*}
\begin{xy}
\xymatrix@=24pt {
T^3A \ar[r]^{T\mu_A} \ar[d]_{\mu_{TA}} & T^2A \ar[d]^{\mu} \\
T^2A \ar[r]_{\mu} & T
}
\qquad
\xymatrix {
TA \ar[r]^{\eta_{TA}} \ar[dr]_{\Un{TA}} & T^2A \ar[d]^{\mu} & TA \ar[l]_{T\eta_A} \ar[dl]^{\Un{TA}} \\
& TA &
}
\end{xy}
\end{equation*} を考える.
$a \in T^3A = ((A_+)_+)_+ = ((A \amalg \{A\}) \amalg \{A_+\}) \amalg \{(A_+)_+\}$ に対して
\begin{equation*}
T\mu_A(a) = \begin{cases}
a & (a \in (A_+)_+) \\
A_+ & (a = (A_+)_+),
\end{cases}
\end{equation*}
\begin{equation*}
\mu_{TA}(a) = \begin{cases}
a & (a \in A_+) \\
A_+ & (a = A_+, (A_+)_+),
\end{cases}
\end{equation*} だから
\begin{equation*}
\mu_A T\mu_A(a) = \begin{cases}
a & (a \in A) \\
A & (a=A, A_+, (A_+)_+)
\end{cases}
\end{equation*}
\begin{equation*}
\mu_A \mu_{TA}(a) = \begin{cases}
a & (a \in A) \\
A & (a=A, A_+, (A_+)_+)
\end{cases}
\end{equation*} となり, 左の図式は可換である.
また, $a \in TA$ に対して
\begin{equation*}
\eta_{TA}(a) = \begin{cases}
a & (a \in TA)
\end{cases}
\end{equation*}
\begin{equation*}
T\eta_A(a) = \begin{cases}
a & (a \in A) \\
A & (a = A)
\end{cases}
\end{equation*} だから
\begin{equation*}
\mu \eta_{TA}(a) = \begin{cases}
a & (a \in A) \\
A & (a = A)
\end{cases}
\end{equation*}
\begin{equation*}
\mu T\eta_A(a) = \begin{cases}
a & (a \in A) \\
A & (a = A)
\end{cases}
\end{equation*} となり, 右側の図式も可換である.
したがって maybe モナド $(T,\eta,\mu)$ は確かにモナドである.
プログラミングとの関係で言えば, 任意の関数 $f : A \rightarrow B$ を実装したプログラム $f_+ : A_+ \rightarrow B_+$ は, 入力 $a \in A$ に対しては出力として正常値 $f(a) \in B$ を返し, 予期せぬ入力 $a=A$ に対してはエラー値 $f_+(A) = B$ を返すということになる.
