Why are simplicial objects monadic over split (contractible) simplicial objects? Given an augmented simplicial object $d_\bullet:X_\bullet \to \Delta X_{-1}$, suppose there's a simplicial map $s_\bullet :\Delta X_{-1}\to X_\bullet$ making $d_\bullet$ a deformation retract, i.e such that $d_\bullet$ is both a retract and a homotopy-section of $s_\bullet$. This is equivalent to providing the augmented simplicial object with an "extra degeneracy" (just an alternative description of the simplicial homotopy axioms in this case). Such an extra degeneracy of an augmented simplicial object will be called a splitting.
Let $\mathrm{S}$-$s\mathsf C$ denote the category of split simplicial objects (with fixed splitting) and simplicial arrows between them respecting the simplicial homotopies. This category admits a forgetful functor to simplicial objects $s\mathsf C$.
On page 20 of Duskin's Simplicial Methods and the Interpretation of Triple Cohomology (AMS page), the author remarks this forgetful functor is a left adjoint to a shifting functor defined by deleting the top face map, viewing the top degeneracy as an extra one, and shifting to a lower index. (The simplicial homotopy is defined by $h_i=s_0^{n-i}s_{n+1}d_0^{n-i}$.)
Moreover, the author writes this shifting functor $s\mathsf C\to \mathrm{S}$-$s\mathsf C$ is monadic. In other words, simplicial objects are monadic over split simplicial objects.


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*What's the intuition behind the fact the shifting functor actually takes values in split simplicial objects? This seems strange to me - as if saying a simplicial object becomes contractible if you forget the top face map. How can that be?

*What's the intuition behind monadicity? An algebra over a split simplicial object (which is already a structured simplicial object) is an arrow to it, so how can additional structure on an already structured object yield back the original notion of object and arrow?
Added. I am looking for naive geometric intuition for these facts, namely: 1. that a simplicial object becomes contractible upon merely forgetting a face map and reindexing; 2. simplicial objects are monadic over split ones. Ideally, I would like an example of what the contraction deformation retract actually does to the Décalage of a non-contractible simplicial complex, say the boundary of a tetrahedron.
 A: I don't know about geometric intuition, but this is pretty straightforward as long as $C$ is complete and cocomplete, though you can get away with substantially weaker hypotheses.


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*Simplicial objects in $C$ form a functor category $[\Delta^\mathrm{op},C]$, while split simplicial objects also form a functor category $[\Delta_\top, C]$ where $\Delta_\top$ is the subcategory of $\Delta$ given by top-element-preserving maps (or perhaps bottom-element-preserving, depending on conventions).

*The shift and forgetful functors are both given by precomposition with functors between $\Delta^\mathrm{op}$ and $\Delta_\top$, so they each have left and right adjoints (actually, these functors are in fact adjoint to one another so the adjoints are induced functorially, not just by Kan extension, but that's beside the point).

*Moreover, the functors between $\Delta^\mathrm{op}$ and $\Delta_\top$ are both are surjective on objects, so the induced functors between functor categories are conservative and reflect (co)equalizers (in fact they reflect all limits and colimits because limits and colimits are computed objectwise in $C$ and essential surjectivity guarantees that all objects are "seen").

*So by the Beck Monadicity theorem, both of these functors are both monadic and comonadic. That is:

An essentially surjective functor $I\to J$ between small categories induces a functor $[J,C] \to [I,C]$ which is monadic and comonadic if $C$ is complete and cocomplete.

Another oft-used instance of this is when $C = Set$ and $I = Ob J$, then this says that the presheaf category $Set^I$ is both monadic and comonadic over $Set^{Ob I} = Set/ Ob I$, a slice of $Set$.
A: Coming across this old question and noticing it still unanswered, here’s an attempt at the “geometric” answer OP wanted.
First note that a split simplicial complex isn’t in general contractible — it’s just 1-truncated, i.e. each connected component is contractible.  Concretely, the augmentation indexes the connected components, and the splitting + homotopy gives a distinguished point + contraction for each component.
Write $R : \mathrm{SSet} \to \mathrm{SplSSet}$ for the desired right adjoint.  We have $(RX)_{-1} = X_0$, so $X_0$ indexes the components of $RX$, and then for each $x \in X_0$, the $n$-simplices of $RX$ in the component over $x$ are the $n+1$-simplices whose $(n+1)$th vertex is $x$ (or $0$th vertex, depending on your indexing convention).  In other words the $x$-component of $RX$ is the “out-star” (or “out-cocone”, or “out-path-space”) of $x$, consisting of all simplices whose source is $x$.  But now each out-star is contractible, contracting into the degenerate 1-simplex on $x$; this is easy to check algebraically, but also easy to see geometrically/topologically — the space of “paths in $X$ with source $x$” is contractible to the constant path on $x$.
Briefly: Under the right adjoint, a simplicial set falls apart into the collection of out-stars of its vertices; and each out-star is clearly contractible to the reflexivity path on that vertex.
