Does pushout of schemes along formal neighborhoods exist in the category of schemes? I have a question about gluing specific types of schemes which doesn't fit into any well-known gluing situation. Assume $C$ is an algebraic curve and $p$ a point on it. The formal completion of $C$ at that point is $k[[t]]$. Now imagine $\text{Spec}(k[[t]])\times C \times C$, $C\times \text{Spec}(k[[t]]) \times C$ and $C\times C \times \text{Spec}(k[[t]])$. We want to glue these in the obvious way i.e. glue the first and the second one along $\text{Spec}(k[[t]])\times \text{Spec}(k[[t]])\times C$, glue the first and the third one along $\text{Spec}(k[[t]])\times C \times \text{Spec}(k[[t]])$ and in the similar way glue the second and third one. I wonder whether the outcome is an scheme?
Intuitively the formal completion should behave like an open neighborhood but obviously it is not an open immersion.
 A: I will give a counterexample in a slightly simpler situation (two copies of $C$ instead of three) but the same argument can be made for your case.
Write $\operatorname{Sch}$ (resp. $\operatorname{Set}$) for the category of schemes (resp. sets). Pick $C=\operatorname{Spec} \mathbb C[t]$, call $p$ the origin in $C$ and put $C_p=\operatorname{Spec} \mathbb C[[t]]$. Suppose that there is a colimit $X$ in the category of schemes for the diagram $D$ formed by the two maps $C_p \times C_p \to C_p \times C$ and $C_p \times C_p \to C \times C_p$. We will derive a contradiction. There is a natural map $X \to C\times C$ by definition of the colimit. I claim that the functor of points of $X$ is the (Zariski) sheafification of the presheaf taking a scheme $T$ to the subset of $(C\times C)(T)$ consisting of morphisms $(f,g)$ such that either $f$ or $g$ factors through $C_p$. To prove this claim, call $X'$ this sheafification. Then $X'$ is the colimit of our diagram $D$ in the category of Zariski sheaves $\operatorname{Sch}^{op} \to \operatorname{Set}$. In particular, there is a natural map $X' \to X$ in this category (since $X$ is the colimit in a smaller category). Conversely, a morphism of schemes $T \to X$ gives a morphism $T \to C\times C$, which factors through $X'$ since $C_p \times C$ and $C \times C_p$ jointly surject onto $X$. So, we have $X=X'$, which proves the claim.
Now, pick an affine neighbourhood $U=\operatorname{Spec} A$ of $(p,p)$ in $X$. We can choose an affine neighbourhood $V=\operatorname{spec} B$ of $p$ in $C$ which is small enough so that $C_p \times V \to X$ and $V \times C_p \to X$ both factor through $U$. Identifying $C\times C$ with $\mathbb A^2_k=\operatorname{Spec}\mathbb C[t_1,t_2]$, we find that $A$ is a sub-$\mathbb C[t_1,t_2]$-algebra of $\mathbb C[[t_1,t_2]]$ which factors through both $\mathbb C[[t_1]]\otimes_\mathbb C B$ and $B \otimes_\mathbb C \mathbb C[[t_2]]$. Since $B$ is obtained from $\mathbb C[t]$ by inverting finitely many polynomials, $U=\operatorname{Spec} A$ is an open subscheme of $C \times C=\mathbb A^2_k$. This is incompatible with $U$ being contained in $X$, so we have our contradiction (for example because such an $U$ will contain many closed points which do not lie on one of the coordinate axes).
