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There are a couple questions on this site that seem to be asking "the same" question, but I'm asking about the category theory version.

Let $\mathcal{C}$ be a lextensive category. A subobject $R \hookrightarrow Y$ is complemented if there is a subobject $R' \hookrightarrow Y$ such that the induced map $R+R' \rightarrow Y$ is an isomorphism. Moreover, a subobject of $X \times X$ is called a relation, and a relation $r : R \hookrightarrow X \times X$ is symmetric if the arrow $$ R\overset{r}{\rightarrow} X \times X \overset{\sigma}{\rightarrow} X \times X $$ factors through $r$ (where $\sigma =\langle \pi_2, \pi_1 \rangle$ is the arrow swapping the components).

It seems reasonable that if $R$ is symmetric, then $R'$ should be too. But I can't seem to work out a proof (I don't see what you could use to build a map $R' \rightarrow R'$), and I can't find a mention of this in my references.

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  • $\begingroup$ Are you missing a prime from $R + R \rightarrow Y$ or am I missing something? $\endgroup$ Commented May 22 at 16:57
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    $\begingroup$ @Z.A.K. Fixed it, thanks! $\endgroup$ Commented May 22 at 17:03

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As a complement (no pun intended) to Zhen Lin's excellent answer, here is a more direct, diagrammatic way to see that $R'$ is symmetric. Consider the following diagram:

commutative diagram

Both $s$ and $\delta$ are involutions, so the bottom left square is a pullback; the top left square is a pullback by disjointness of coproducts and the right rectangle is a pullback by construction. Hence by stability of coproducts under pullback, the top row is a coproduct diagram, which means that $\pi_1$ is invertible, which in turn means that $\delta \circ r'$ factors through $r'$ via $\pi_2$.

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  • $\begingroup$ While I do appreciate Zhen Lin's answer (notably since it gives a technique that can be used to solve other such problems), this solution feels the most satisfying to me. Thanks! (Also, nice pun!) $\endgroup$ Commented May 23 at 20:48
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It is useful to use the language of generalised elements.

For any object $T$ and any morphisms $x_0, x_1 : T \to X$, the induced morphism $\langle x_0, x_1 \rangle : T \to X \times X$ factors through $R' \hookrightarrow X \times X$ if and only if, for any object $S$ and any morphism $t : S \to T$ such that $\langle x_0, x_1 \rangle \circ t : S \to X \times X$ factors through $R \hookrightarrow X \times X$, $S$ is an initial object. Indeed, this is just a paraphrase of the defining properties of a binary coproduct in a lextensive category: the "only if" direction is simply saying that in any commutative square of the form below, $$\require{AMScd} \begin{CD} S @>>> R' \\ @VVV @VVV \\ R @>>> X \times X \end{CD}$$ the object $S$ must be a (strict) initial object, whereas the "if" direction means that if we form pullback squares as below, $$\begin{CD} S @>>> T @<<< S' \\ @VVV @V{\langle x_0, x_1 \rangle}VV @VVV \\ R @>>> X \times X @<<< R' \end{CD}$$ and $S$ is a (strict) initial object then $S' \to T$ must be an isomorphism.

With the above characterisation, it is easy to see $R$ is symmetric if and only if $R'$ is symmetric: after all, symmetry of $R$ means $\langle x_0, x_1 \rangle : T \to X \times X$ factors through $R \hookrightarrow X \times X$ if and only if $\langle x_1, x_0 \rangle : T \to X \times X$ factors through the same.

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  • $\begingroup$ I'm having some trouble following the "only if" direction (and also applying this equivalence to the symmetry case). I don't think it should be true that in any such commutative square, $S$ must be initial; rather, we should demand that the square be a pullback. Should that be required in the condition? I.e. for any $S$, $t$ such that $\langle x_0, x_1 \rangle \circ t$ factors through $R$ in a way that makes a pullback square? $\endgroup$ Commented May 23 at 16:07
  • $\begingroup$ Also, I would really appreciate if you could provide a source for this argument. Can it be found in a standard category theory text? $\endgroup$ Commented May 23 at 16:08
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    $\begingroup$ @Sambo if the square is not a pullback, there is a "gap map" from $S$ to the actual pullback, which is a strict initial object, hence $S$ is also a (strict) initial object. $\endgroup$ Commented May 23 at 17:14
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    $\begingroup$ @Sambo This is a special case of internal logic in a (sheaf) topos. $\endgroup$ Commented May 23 at 23:29

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