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An object $S$ is subterminal when its unique morphism $S\to 1$ to the terminal object is a monomorphism. Equivalently, every object has at most one morphism to $S$. Subterminal objects form the preorder $\mathrm{Sub}(1)$.

In a Cartesian closed category, exponentiation satisfies $(-{)}^{A\times B}\cong ((-{)}^B{)}^A$. If the two exponential functors have right adjoints, their composite has the composite of those right adjoints in reverse order. Therefore a product of tiny objects is tiny. The terminal object is tiny since exponentiation by it is the identity, covering the empty product.

A nondegenerate elementary topos is two-valued when its terminal object has exactly the empty and whole subobjects. This is a global-truth property, distinct from being a Boolean topos, which requires complements for all subobjects of all objects.

An object $A$ in a category with a terminal object is well-pointed when every monomorphism $m:B\to A$ through which all points $1\to A$ factor is invertible. This definition concerns detection of proper subobjects by points, and does not designate a chosen basepoint. If there are no points, it asserts that every subobject is the whole object. In particular, the empty set is well-pointed.

An object of a regular category is well-supported when its support of an object in a regular category is the terminal object, equivalently when $A\to 1$ is a regular epimorphism. Finite products of well-supported objects are well-supported, by pullback and composition stability of regular epimorphisms. A regular epimorphism has the same support as its codomain, and an object with a point $1\to A$ is well-supported.

A zero object is both an initial object and a terminal object. In a pointed category, the composite through the zero object is the zero morphism $0_{A,B}:A\to B$.

If the terminal object is also initial and $E$ is exponentiable, the left adjoint $-\times E$ preserves the initial object. But $0\times E\cong E$ because $0$ is terminal, while preservation of initiality gives $0\times E\cong 0$. Hence $E\cong 0$.