ag.algebraic geometry - Can a coequalizer of schemes fail to be surjective?

This might lead to a pedestrian example: Let $R$ be a local ring with maximal ideal $m$, $X$ a scheme, and $f:Spec(R) to X$ a map. If $U$ is an open subscheme of $X$ containing $f(m)$ then $f$ factors through a map $Spec(R) to U$. Thus if we have two arrows $g,h:Y to Spec(R)$, the coequalizer $c: Spec(R)to C$ must be an affine scheme (else $c$ would factor strictly through an affine neighborhood $U$ of $c(m)$, and $U$ would be a "better coequalizer" than $C$). So If $Y=Spec(B)$ is also affine, then the coequalizer of $g,h$ is just $Spec(A)$ where $A$ is the equalizer of $g^sharp,h^sharp$.

let R be $k[x_i]_{(x_i)}$ and $S'=k[y_i,z_i]_{(y_i,z_i)}.$ there are two maps $g',h':R to S$ given by

$$g'(x_i) = y_i$$

$$h'(x_i) = z_i$$

suppose that $I$ is an ideal of $R$. let $I_y, I_z$ be the ideals generated by $g(I)$ and $h(I)$ respectively and let $S = S'/(I_y + I_z)$. write $g$ and $h$ for the induced maps $R to S$. the equalizer of $g,h$ is just $A = k + I subset R$. it seems unlikely to me that $spec R to spec A$ is surjective for all choices of $I$.