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Let $G_1,\ldots,G_m$ be a sequence of graphs, all having the same number $n$ of vertices. For each pair $(G_i, G_{i+1})$ we add $n$ edges that connect the vertices of $G_i$ and $G_{i+1}$ bijectively. My question: Is there an established name for this "stack" of graphs?

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  • $\begingroup$ given @Rob Pratt's comment about this being the cartesian product whenever all the $G_i$ are equal, I wonder what the categorical formalisation of what you describe is? some sort of fibred product? I don't know $\endgroup$ – Tim Sep 2 '19 at 11:57
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On page 4 of [1 ] , you find:

[...] a multiplex network can be represented as a collection of graphs $$\mathcal{G}=\{G^{(\ell)}=(V_n,E^{(\ell)})\}_{\ell \in V_L}$$ where $V_n=\{1,\ldots,n\}$ is the set of nodes, $V_l=\{1,\ldots,L\}$ s the set of layers and $E^{(\ell)}\subset V_n\times V_n$ is the set of edges on layer $\ell$.

These structures are also sometimes called multi-layer graphs. Note that in the above formulation it is assumed that the nodes on each layer are the same (this models the bijective edges you are referring to).

[1 ] Node and layer eigenvector centralities for multiplex networks. F Tudisco, F Arrigo, A Gautier - SIAM Journal on Applied Mathematics, 2018 (arXiv)

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In the case that you only have two graphs, and they are the same, say $G$, then the edges between the two copies of $G$ can be described by a permutation, and your graphs are precisely the permutation graphs defined by Chartrand and Harary in their 1967 paper "Planar permutation graphs" [Ann. Inst. H. Poincaré Sect. B (N.S.), 3, pp. 433–438]. A cute example of such a "permutation graph" is the Petersen graph, which we can obtain with $G=C_5$.

(I cannot stress strongly enough that this is not the standard notion of permutation graphs!)

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In the special case that $G_i=G$ for all $i$, this is the Cartesian product of $G$ with the path on $n$ nodes.

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  • $\begingroup$ That's not true if the vertices are simply connected "bijectively", as the question specifies. $\endgroup$ – Vince Vatter Aug 29 '19 at 12:07
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    $\begingroup$ Yes, for the Cartesian product the bijection is the identity, not an arbitrary bijection. $\endgroup$ – Rob Pratt Aug 29 '19 at 13:15

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