**EDIT:** On reflection, this answer is pretty incorrect. There is a point here, but it's formulated badly enough that I think this should be ignored. I wasn't sure what best practice is for a wrong answer, so I'm going to leave a comment with corrections.

Forgive both the late reply and potentially sophomoric comment, but maybe it would help to break down the Lawvere theorem and ask whether a non-diagonal undecidability proof is constructible.

An undecidability proof has the rough anatomy that we want to find some epimorphism $g: A \to Y^S = g: A \to S \to Y$ such that the image of $g$ is monomorphic. In the usual case that $A = S$ is a set and $Y$ is a finite set, $g$ picks out some $h: A \to Y$ that assigns some $y \in Y$ to each $a \in A$, or in other words, $g$ finds, for each $A$, some $h$ that decides $a$s. We then find that assuming such an epimorphism exists, we obtain a contradiction, thus $A$ is not in general decidable.

Undecidability proofs are, I think anyway, quintessentially nonconstructive, so we have to assume decidability (i.e., there exists a $g$) and show it leads to contradiction (whether explicitly or not) in order to get an undecidability proof. Assuming there exists an epimorphism $g$ is a strictly necessary element.

Then, just by uncurrying, we can show that $g$ is isomorphic to $f: A \times S \to Y$. Then, the diagonal morphism $\Delta: A \to A \times S$, which forms an identity pairing $\Delta(a) = (a, a)$ in the usual case, does not add much, as $f \circ \Delta$ is isomorphic to $g \circ id_A = g$ (more generally, $f \circ \Delta$ is isomorphic to some $g \circ i$ where $i: A \to A$ is an isomorphism). Joel's reply above suggested that the application of the diagonal morphism could be thought of as a pivotal step in singling out a diagonalization, but it seems to me to be really trivial. At best, non-diagonal proofs seen this way apply an isomorphism $A \to A$ before doing their work.

At this stage, I think most of a diagonalization proof is necessary, including, in effect, the diagonal part, in order to show undecidability. At least, if we find a proof that doesn't seem to involve diagonalization, we can be pretty well assured that it should not be too hard to transform it into an argument that uses a $\Delta$.

The final step in the diagonalization is to choose $\alpha: Y \to Y$ without any fixed points. Since we can choose such an $\alpha$, by applying $\alpha \circ f \circ \Delta: A \to Y$, we show that $f \circ \Delta$ is not unique up to isomorphism $A \to A$, and so $g$ cannot be an epimorphism. If we're looking for alternatives to diagonalization, the place might be here. There might be a nontrivially different approach to showing that $g$ cannot be an epimorphism. Though, for undecidability results, I have a hard time imagining how one proves that one cannot find an $h$ that decides arbitrary $a$ other than choosing $\alpha \circ h$ that makes some sort of inverse decision assignment.

To make a long answer short, if I haven't made some basic error, I think the answer to the question may be "no, and furthermore, there can't be".

### Corrections:

- An instance of an undecidability proof assumes the existence of a unique morphism $g: A \to Y$. Assuming $g: A \to S \to Y$ is already almost all of the heavy lifting in a diagonalization/Lawvere proof, and isn't strictly necessary to prove undecidability.
- To get from undecidability to Lawvere, we let $g = h \circ k \circ id_A: A \to A \to S \to Y \sim A \to A \times S \to Y$, where $k$ is an isomorphism. For instances where $S = A$, $k$ is trivial as the above suggests (and could just as well be the identity), but that might not always be so.
- For $g$ to be unique, $g \equiv \alpha \circ h \circ k \circ id_A \sim \alpha \circ f \circ \Delta$ for any choice of $\alpha$, but, since we may always choose $\alpha: Y \to Y$ with no fixed points (as $Y$ is nondegenerate), this cannot be.

So here is the rub. $k$ might not always be trivial (though these are cases where we want to move to $S$ because it's easier to make claims about $S \to Y$ than $A \to Y$), but if we have a proof that there does not exist a unique morphism $g$, then we can always transform this into a proof about $g \circ id_A \circ id_A \sim f \circ \Delta$. However, I can imagine cases where doing so is unnecessary and this move seems superfluous. Though it doesn't seem to me to be *whenever* $g = h$ (e.g. the halting problem), so maybe that's just a trick of the light.

I'm still not sure that I've got this right, but I think it's at least worthwhile to admit the mistake.

Oh wait.$\endgroup$13more comments