I am trying to wrap my head around the construction of the simplicial replacement of a diagram, e.g. following Rodríguez-González (Realizable Homotopy Colimits, 1.5, https://arxiv.org/pdf/1104.0646.pdf):
Given a category $\mathcal{D}$ with coproducts and an index category $I$, denote the functor category from $I$ to $\mathcal{D}$ by $\mathcal{D}^I$. The simplicial replacement $\amalg^I X$ of $X: I \rightarrow \mathcal{D}$ is the simplicial object given in degree $n$ by: $$\amalg_n^I X = \coprod_{(i_0 \rightarrow \cdots \rightarrow i_n)\in N_n(I)} X(i_0),$$ with the coproduct indexed by the simplicial nerve of $I$. The face maps $d_k$ of this simplicial object are defined as follows (denote with $d_k^I$ the face maps of the nerve $N(I)$). For $0 < k\leq n$, $d_k : \amalg_n^I X \rightarrow \amalg_{n-1}^I X$ maps the term $X(i_0)$ indexed by $(i_0 \rightarrow \cdots \rightarrow i_n)$ to the term $X(i_0)$ indexed by $d_k^I((i_0 \rightarrow \cdots \rightarrow i_n)) \in N_{n-1}(I)$, while $d_0$ sends this term via $X(i_0 \rightarrow i_1)$ to the term $X(i_1)$ indexed by $d_0^I((i_0 \rightarrow \cdots \rightarrow i_n)) = (i_1 \rightarrow \cdots \rightarrow i_n) \in N_{n-1}(I)$.
My question is related to these face maps. In general, we should have several different simplices $(i_0 \rightarrow \cdots \rightarrow i_n) \in N_n(I)$ for which $d_0^I((i_0 \rightarrow \cdots \rightarrow i_n)) \in N_{n-1}(I)$ coincides (e.g. by having different entries $i_0$ or different maps $i_0 \rightarrow i_1$). In my (mis-)understanding, all the terms $X(i_0)$ indexed by these simplices are mapped by $d_0$ to the same entry indexed by $(i_1 \rightarrow \cdots \rightarrow i_n)$, hence the images of all the terms $X(i_0)$ have to be "cramped" into the same entry. I would be happy about either an explanation on how this looks, or about a clarification about where in the above I am getting something wrong.
Maybe to clarify my question with a simple example: let $I$ be the category $\{0 \rightarrow 1\}$, $\mathcal{D}$ the category of groups, $X \in \mathcal{D}^I$. Is $$d_0 : X(0)\oplus X(0) \oplus X(1) \rightarrow X(0) \oplus X(1)$$ then mapping the $X(0)$-term indexed by $0 \rightarrow 1$ via $X(0 \rightarrow 1)$ and $X(1)$ (indexed by $1\rightarrow 1$) via the identity both to the $X(1)$-term? (and if so, how is this map looking like?)