A reflexive coequalizer of sets not preserved by $\mathbf{S}\mathbf{e}\mathbf{t}(\mathbb{N},-)$

Define $f=[\sigma ,1]$ and $g=[1,1]:\mathbb{N}⨿\mathbb{N}\rightrightarrows \mathbb{N}$. The second injection $i_1$ is a common section because $f{i}_1=g{i}_1=1_{\mathbb{N}}$, so the pair is reflexive by [L1].

The unique map $q:\mathbb{N}\to 1$ is the coequalizer: the relation $f(i_0(n))=n+1\sim n=g(i_0(n))$ connects every natural to $0$, and any map coequalizing $f,g$ is therefore constant and factors uniquely through $q$.

Applying $\mathbf{S}\mathbf{e}\mathbf{t}(\mathbb{N},-)$ gives a pair $(\mathbb{N}⨿\mathbb{N}{)}^{\mathbb{N}}\rightrightarrows {\mathbb{N}}^{\mathbb{N}}$. For a function $h:\mathbb{N}\to \mathbb{N}⨿\mathbb{N}$, the two resulting sequences $f\circ h$ and $g\circ h$ differ at each coordinate by either $0$ or $1$.

Every finite zigzag generated by pairs from step 2.2 has a uniform coordinatewise difference bound, namely its number of zigzag edges, by repeated use of the triangle inequality on natural-number differences.

The zero sequence $z(n)=0$ and identity sequence $d(n)=n$ have unbounded coordinatewise difference, so step 3.1 shows that no finite zigzag identifies them.

Both sequences map under $q^{\mathbb{N}}$ to the unique element of $1^{\mathbb{N}}$, yet they remain distinct in the coequalizer of the image pair. Therefore $q^{\mathbb{N}}$ is not the coequalizer required by [L2], and $\mathbf{S}\mathbf{e}\mathbf{t}(\mathbb{N},-)$ does not preserve this reflexive coequalizer.