Previous: [[Fiber (co)products in the category of sets]] Here we finally come to the reason why I've been talking about fiber coproducts. If we have two groups $G$ and $H$, we can form their coproduct $G \ast H$. In this, the elements of $G$ and the elements of $H$ are "strangers." They have no relations between than and can't move past each other. What if we *did* want to introduce some relations between them? To be more precise, what if there was a group $N$ that was a normal subgroup in *both* $G$ and $H$, and we wanted to identify its copy in $G$ with its copy in $H$? If $F$ and $G$ themselves were subgroups of a group this would be easy, but that would not preserve the rest of the elements of $G$ and $H$ remaining "strangers" - they can now interact using the ambient group multiplication! The solution is the fiber coproduct. This is much like the case of sets, where we used inclusion maps to identify elements that really should be the same. In this case, we define the "fiber product with amalgamation" with respect to $N$ via this diagram <iframe class="quiver-embed" src="https://q.uiver.app/#q=WzAsNCxbMCwwLCJOIixbMCwwLDEwMCwxXV0sWzIsMCwiSCIsWzAsMCwxMDAsMV1dLFswLDIsIkciLFswLDAsMTAwLDFdXSxbMiwyLCJHXFxhc3RfTiBIIixbMCwwLDEwMCwxXV0sWzAsMSwiaV9IIiwxLHsiY29sb3VyIjpbMCwwLDEwMF19LFswLDAsMTAwLDFdXSxbMCwyLCJpX0ciLDEseyJjb2xvdXIiOlswLDAsMTAwXX0sWzAsMCwxMDAsMV1dLFsxLDMsImpfSCIsMSx7ImNvbG91ciI6WzAsMCwxMDBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkb3R0ZWQifX19LFswLDAsMTAwLDFdXSxbMiwzLCJqX0ciLDEseyJjb2xvdXIiOlswLDAsMTAwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbMCwwLDEwMCwxXV1d&embed" width="473" height="432" style="border-radius: 8px; border: none;"></iframe> This also lets us construct it: we know that $i_H(N) = i_G(N)$ in the final, which we expect to be a free-group like thing. We do the simplest thing we can. We take the free product $H \ast G$ and manually set $i_H(n) = i_G(n)$ for any $n \in N$. We know the way to do this is by adding relations which is the same as quotienting by the smallest normal subgroup generated by the relations. Thus $G \ast_N H = G\ast H/N$ where $N$ is the smallest normal subgroup generated by all words of the form $i_G(n)i_H(n)^{-1}$ Next: [[Coproducts in the category of abelian groups]]