Is this square diagram cocartesian for every regular local ring? Let $K$ be a field and $R=\{f\in K[X]\mid f(0)=f(1)\}$ the $K$-algebra obtained by pulling back $K[X]\to K\times K$, $X\mapsto (0,1)$ along the diagonal.
Is the induced square
\begin{eqnarray}
Hom(K\times K,L)& \to & Hom(K,L)\\
\downarrow && \downarrow \\
Hom(K[X],L) &\to & Hom(R,L)
\end{eqnarray}
of sets cocartesian for every regular local ring $L$? The $Hom$ sets are the sets of $K$-algebra maps.
 A: This is true for any integrally closed $K$-algebra $L$. 
Let $p_1: Hom(K[X], L)\to S$, $p_2: Hom(K[X], L)\to S$ be two maps making commutative what has to be commutative. We want to complete with a map $p: Hom(R, L)\to S$. 
Denote by $U=X(X-1)$ and $V=X^2(X-1)$. Then it is easy to see that $R=K[U,V]$ and $V^2=U^3+UV$. 
Let $\phi: R\to L$ be an element of $Hom(R, L)$. We first lift $\phi$ to $\phi'\in Hom(K[X], L)$: if $\phi(U)=0$, then $\phi(V)^2=0$, hence $\phi(V)=0$ because $L$ is reduced. This $\phi$ lifts to $\rho: X\mapsto 0$ of $Hom(K[X], L)$ (it also lifts to $\rho': X\mapsto 1$, but $\rho$ and $\rho'$ have the same image in $S$ by the initial commutative diagram). If $\phi(U)\ne 0$. Then $\phi(V)/\phi(U)\in \mathrm{Frac}(L)$ is integral over $L$ because
$$ (\phi(V)/\phi(U))^2=\phi(U)+(\phi(V)/\phi(U)).$$
So $\phi(V)/\phi(U)=a\in L$. Then $\phi$ lift to $\rho : X\mapsto a$ of $Hom(K[X], L)$. In this case the lifting is unique. 
Now define $p : Hom(R, L)\to S$ by $p(\phi)=p_1(\rho)$. It is straightforward to check that $p$ is well defined and is compatible with $p_1$ and $p_2$ (the latter is constant as $Hom(K, L)$ is just one point).  In fact, the proof above says that $Hom(R, L)$ is the quotient of $Hom(K[X], L)$ by the equivalence relation which just identifies $X\mapsto 0$ and $X\mapsto 1$. 
