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Let $S$ be a scheme and $G$ be a sheaf in groups on the big étale site over $S$. Let $e:S\rightarrow G$ be the unit section. Is it true that given an algebraic space in groups $H$, étale over $S$, and a homomorphism of sheaves in groups $f:H\rightarrow G$ the fiber product $H\times_{f, G,e}S$ is representable by an algebraic space?

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No. For one thing, it would imply that every group subsheaf $K$ of an (abelian) algebraic space in groups $H$ is an algebraic space (just take $G=H/K$). Now let me give a "concrete" example.

For any field $k$, take $S=\mathrm{Spec}(k[t])$ and $H=(\mathbb{Z}/2\mathbb{Z})_S$. Put $U=\mathrm{Spec}(k[t,t^{-1}])$, and let $j:U\to S$ be the inclusion. Finally, take for $f$ the canonical morphism $H\to G:=j_*j^*H$. Let $K:=H\times_{f,G,e}S$ be the kernel of $f$.

Writing $H$ as the sum $S_0\coprod S_1$ of two copies of $S$ (with $S_0$ as the unit section), we find $K=S_0\coprod K_1$ where $K_1\subset S_1\cong S$ is the "locus" where $t$ is locally nilpotent. This is not an algebraic space.

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