Let $f : X \to Y$ be a D1519: Constant map from $X$ to $Y$ such that

(i)	$M_Y = (Y, \mathcal{F}_Y)$ is a D1108: Measurable space
(ii)	$\sigma_{\text{pullback}} \langle f \rangle$ is a D1730: Pullback sigma-algebra on $X$ under $f$ with respect to $M_Y$

By definition of a sigma-algebra pullback, we have $$\begin{equation} \sigma_{\text{pullback}} \langle f \rangle = \{ f^{-1}(E) : E \in \mathcal{F} \} \end{equation}$$ Result R1176: Inverse image of a constant map is either empty or equal to the domain set shows that $\sigma_{\text{pullback}} \langle f \rangle \subseteq \{ \emptyset, X \}$. It remains to establish the converse inclusion. By definition of a sigma-algebra, we have we have $\emptyset \in \mathcal{F}_Y$. Additionally, result R2076: Inverse image of empty set shows that $f^{-1}(\emptyset) = \emptyset$, so that $\emptyset \in \sigma_{\text{pullback}} \langle f \rangle$. If $X$ is empty, then we are done since in that case $\{ \emptyset, X \} = \{ \emptyset, \emptyset \} = \{ \emptyset \}$. Suppose thus that $X \neq \emptyset$ and fix $x \in X$. Since a map is, by definition, a D359: Left-total binary relation, then there exists $y \in Y$ such that $f(x) = y$. A sigma-algebra is closed under complements, so $Y \setminus \emptyset = Y \in \mathcal{F}_Y$. Since $y \in Y$ and since $f$ is constant, then $f^{-1}(Y) = X$, whence $X \in \sigma_{\text{pullback}} \langle f \rangle$ and $\{ \emptyset, X \} \subseteq \sigma_{\text{pullback}} \langle f \rangle$. $\square$