Finn Lawler product-absolute pullback

In category with finite products, the following squares must be pullbacks, for any morphisms $f,g$:

$$\array{ X & \xrightarrow{(1, f)} & X \times Y \\ \mathllap{f} \downarrow & (A) & \downarrow \mathrlap{f \times Y} \\ Y & \xrightarrow{d} & Y \times Y } \qquad \qquad \qquad \array{ X & \xrightarrow{1} & X \\ \mathllap{1} \downarrow & (B) & \downarrow \mathrlap{d} \\ X & \xrightarrow{d} & X \times X } \qquad \qquad \qquad \array{ X \times X' & \xrightarrow{1 \times g} & X \times Y' \\ \mathllap{f \times 1} \downarrow & (C) & \downarrow \mathrlap{f \times 1} \\ Y \times X' & \xrightarrow{1 \times g} & Y \times Y' }$$

as must the naturality square for the symmetry $\sigma \colon X \times Y \cong Y \times X$, the product on one side or the other with an identity morphism of a pullback square (which we will call type D), and the pasting of two pullback squares side by side. These pullbacks must be preserved by any product-preserving functor, so we call them product-absolute.

This fact was already noted for squares of types A and C by Lawvere; the others are given by Seely. See also Todd Trimble’s exposition, noting in particular that the squares expressing coassociativity of diagonal maps $d \colon X \to X \times X$ are product-absolute pullbacks, which we call type E.