Given a commutative ring $k$ (not necessarily a field), do the regular monomorphisms in the category of (coassociative and counital as usual) $k$-coalgebras (henceforth denoted $k-\mathrm{Coalg}$) coincide with the injective ones?
If not, then one must find a equalizer in $k-\mathrm{Coalg}$ that is not injective and/or an injective $k$-coalgebra homomorphism that is not an equalizer (of its cokernel pair).
If $k$ is a field, then the question holds true. Indeed, the functor $V \mapsto V \otimes_{k} V$ on the category of $k$-vector spaces preserves (regular) monomorphisms, and from this, one can easily show that the image of any $k$-coalgebra homomorphism is a subcoalgebra of the codomain, and the inclusion map into the codomain is then the equalizer of the cokernel pair in $k-\mathrm{Coalg}$.
If $k$ is not a field, then let $C$ be a $k$-coalgebra and $D$ be a subcoalgebra of $C$ (i.e. a $k$-submodule together with a $k$-coalgebra structure on it making the inclusion map into a $k$-coalgebra homomorphism, which in general, might not be unique). Then, the obstacle in proving that the inclusion map $D \hookrightarrow C$ is the equalizer of the two maps $C \to (C \oplus C)/\{(x,-x) \mid x \in D\}$ (which form the cokernel pair of the inclusion map) in $k-\mathrm{Coalg}$ is that the map $D \otimes_{k} D \to C \otimes_{k} C$ might not be injective, and so there could be a $k$-module homomorphism $E \to D$ (where $E$ is a third $k$-coalgebra) that is not a $k$-coalgebra homomorphism but whose composition with the inclusion map $D \hookrightarrow C$ is a $k$-coalgebra homomorphism.