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Write products of endomorphisms as composition, with the rightmost map applied first. Because $1$ is terminal, the only hom-sets that can contain multiple arrows are $\mathrm{Hom}(1,D)$ and $\mathrm{End}(D)$. If the latter contained only the identity, any point $p:1\to D$ would have $p!={\mathrm{id}}_D$, where $!:D\to 1$. Then any other point $q$ would satisfy $q=p!q=p$. Thus the category would be a preorder. The hypothesis therefore implies $|\mathrm{End}(D)|>1$.

If the product object $D\times D$ were $1$, its universal property applied to arrows from $D$ would give $\mathrm{End}(D{)}^2\cong \mathrm{Hom}(D,1)$, a singleton, which is impossible. Since there are only two objects, $D\times D=D$. Similarly, if $D^D=1$, the exponential adjunction at $1$ would giveagain impossible. Hence $D\times D=D^D=D$ as chosen objects.

Let $M=\mathrm{End}(D)$, let $\pi ,{\pi }^{\prime }$ be the product projections, and let $\epsilon :D^D\times D\to D$ be evaluation, all now endomorphisms. Pairing is $⟨-,-⟩$, and for $x:D\times D\to D$ write $x^{\ast }$ for its currying. The product identities and the two inverse exponential identities giveThis is the structure of a cartesian closed monoid.

For the converse, start only from the monoid and these identities. The first three make pairing a bijection from $M^2$ to $M$, inverse to $z\mapsto (\pi z,{\pi }^{\prime }z)$. In particularas follows by applying both projections and using uniqueness. Define $U(y)=\epsilon ⟨y\pi ,{\pi }^{\prime }⟩$. The last two identities say that $U$ and $(-{)}^{\ast }$ are inverse bijections. These facts also derive, rather than assume, the naturality lawIndeed $U(x^{\ast }z)=\epsilon ⟨x^{\ast }z\pi ,{\pi }^{\prime }⟩=U(x^{\ast })⟨z\pi ,{\pi }^{\prime }⟩=x⟨z\pi ,{\pi }^{\prime }⟩$; applying the inverse bijection proves the law.

Put $e=({\pi }^{\prime }{)}^{\ast }$. The law applied to $x={\pi }^{\prime }$ gives, for every $z\in M$,Taking $z=e$ proves $e^2=e$. This stronger absorption identity is what makes the split object terminal.

Assume first that $M$ is nontrivial; the degenerate case is handled below. Construct the required category using the two idempotents $d=1_M$ and $e$, as a two-object part of the Karoubi envelope. Call their objects $D$ and $1$. A morphism from the object with idempotent $a$ to the object with idempotent $b$ is an element $f\in M$ satisfying $bfa=f$. Composition is monoid multiplication, and the identity on each object is its idempotent. These laws are well-defined and associative. Its hom-sets are concretelyFor example, $ef=e$ for every $f$, so the condition for $D\to 1$ forces $f=e$. Thus $1$ is terminal, and the original monoid appears unchanged at $D$.

Set $D\times D=D$ with projections $\pi ,{\pi }^{\prime }$ and the given pairing. The product property works for arrows from either object: for global arrows $x,y$ satisfying $xe=x$, $ye=y$, pairing also satisfies $⟨x,y⟩e=⟨x,y⟩$. The original projection identities give existence and uniqueness. Products with $1$ are the other object, with its identity and the unique terminal arrow as projections; $1\times 1=1$.

For the exponential set $D^D=D$ with evaluation $\epsilon$. For a parameter object $D$, the given $(-{)}^{\ast }$ and $U$ already establish the universal property. We must additionally verify it for parameter object $1$. An arrow $f:1\times D=D\to D$ is an arbitrary element of $M$. Define its transpose byNaturality gives $\kappa (f)z=\kappa (f)$ for every $z$, because $f{\pi }^{\prime }⟨z\pi ,{\pi }^{\prime }⟩=f{\pi }^{\prime }$. In particular $\kappa (f)e=\kappa (f)$, so it really is an arrow $1\to D$.

Its uncurrying is $\epsilon ⟨\kappa (f),1_M⟩=f$. To see this, start from $U(\kappa (f))=f{\pi }^{\prime }$ and compose with $⟨e,1_M⟩$. Pairing naturality, ${\pi }^{\prime }⟨e,1_M⟩=1_M$, and the absorption property of $\kappa (f)$ give precisely the claimed equality. Conversely, if $g:1\to D$ satisfies $ge=g$, then $gz=gez=ge=g$ for every $z$. ThereforeThus transpose and uncurrying are inverse for the terminal parameter object too. This verifies the full exponential property, rather than only its endomorphism instance.

The remaining exponentials are $D^1=D$, $1^D=1$, and $1^1=1$, by the unit product law and terminality. Hence the constructed two-object category is cartesian closed and $\mathrm{End}(D)=M$.

For nontrivial $M$, the constructed $D$ is nonterminal because it has more than one endomorphism, and $e\ne 1_M$ by the absorption identity. This is the case corresponding to the forward non-preorder hypothesis.

If $M$ has one element, all the algebraic identities still hold, but splitting $e=1_M$ alone would make the original object terminal. To satisfy the literal converse's requirement of a nonterminal object, handle this case separately: take the two-object poset category $D<1$. Products are meets, $D^D=1$, $1^D=1$, $D^1=D$, and $1^1=1$, which verify the exponential adjunction on this two-element order. It is cartesian closed, $D$ is nonterminal, and its one-element endomorphism monoid is $M$. It is a preorder, so it does not recover the forward non-preorder setting or the self-exponential equality. Thus the converse's stated endomorphism realization works also in the degenerate case, while the stronger non-preorder reconstruction uses nontrivial $M$.