I realize I'm quite late for this post, but I believe it deserves an answer as the theorem is so often implicitly used when constructing maps between simplicial sets. Rather than addressing OP's original question, I try to give a complete proof here.
Denote by $X^{\text{nd}}_\bullet$ the semi-simplicial set of non-degenerate simplices of $X_\bullet$ and $\varphi: X^{\text{nd}}_\bullet \to Y_\bullet$ a map of semi-simplicial complex, that is, a graded map compatible with face maps. We extend $\varphi$ to a map of simplicial complex $\varphi: X_\bullet \to Y_\bullet$ as follows.
For each degenerate $n$-simplex $\Delta^n_\bullet \to X_\bullet$, there is a unique factorization $f\circ \alpha_*: \Delta^n_\bullet \to \Delta^k_\bullet \to X_\bullet$, where the first map is induced by a surjective map $\alpha: [n] \to [k]$ and the second map $f$ defines a non-degenerate $k$-simplex (this is proved in the subsection on skeleton filtration in Kerodon). We set $\varphi(f \circ \alpha_*) = \varphi(f) \circ \alpha_*$.
This is the only way to extend $\varphi$ so we have uniqueness. Note also that the extension respects degeneracy maps, as $f\circ (\alpha_* \circ s^i)$ is the unique factorization of $(f\circ \alpha_*) \circ s^i$. To check compatibility with face maps, we write $\alpha_*$ as compositions of (co)degeneracy maps:
$$\alpha_* = s^{i_m} \circ \cdots \circ s^{i_1}.$$
By simplicial identities,
$$\alpha_* \circ d^j = s^{i_m} \circ \cdots \circ s^{i_1} \circ d^j = d^{j'} \circ s^{i'_m} \circ \cdots \circ s^{i'_1} = d^{j'} \circ \alpha'_*,$$
where $\alpha': [n-1] \to [k-1]$ is surjective. (Reminder: This is not true. As PaulTaylors pointed out in the comment, we may encounter cases that $s^k \circ d^l = \text{id}$, resulting $\alpha_* \circ d^j = \alpha'_*$, where $\alpha': [n-1] \to [k]$ is surjective.) We thus have
\begin{align*}
\varphi(d_j(f \circ \alpha_*))
=& \varphi(f \circ \alpha_*\circ d^j) \\
=& \varphi((f \circ d^{j'})\circ \alpha'_*) \\
=&\varphi(f \circ d^{j'}) \circ \alpha'_*\\
=&\varphi(f) \circ d^{j'} \circ \alpha'_*\\
=&\varphi(f) \circ \alpha_*\circ d^j\\
=&d_j(\varphi(f) \circ \alpha_*),
\end{align*}
where in the third last equality we use that $\varphi$ on non-degenerate simplices is compatible with faces maps.
before passing to homotopy categories we should get a weak equivalence $⊗∘τ≃⊗$
Since $⊗∘τ$ and $⊗$ are both morphisms $\def\cC{{\cal C}}\cC⟨1⟩^2→\cC⟨1⟩$,
what we are looking for is a homotopy between them.
We have $$⊗=\cC φ∘m, \qquad τ=(\cC δ_1,\cC δ_2)∘\cC t∘m.$$
Now $$⊗∘τ=\cC φ∘m∘(\cC δ_1,\cC δ_2)∘\cC(t)∘m,$$
which can be simplified by observing that $m∘(\cC δ_1,\cC δ_2)$ is canonically homotopic to identity, given that $m$ has defined as the inverse of $(\cC δ_1,\cC δ_2)$.
Thus, $$⊗∘τ≃\cC φ∘\cC t∘m=\cC(φ∘t)∘m=\cC φ∘m=⊗,$$
as desired.
Best Answer
The functor that implements the equivalence between simply connected Lie groups and Lie algebras sends a simply connected Lie group to its tangent space at the origin equipped with the appropriate Lie bracket, and a morphism of simply connected Lie groups to its derivative at the origin.
Thus, the inverse functor reconstructs morphisms of Lie groups from their derivatives at the origin, so is referred to as the integration functor.
More generally, Lie integration explains how to reconstruct a smooth function X→G, where X is a connected smooth manifold and G is a Lie group, from its value at some point of X and its derivative (a differential 1-form on X valued in the Lie algebra of G). A necessary and sufficient condition for this is that the Maurer–Cartan equation holds. In this case, the original function can be uniquely reconstructed from the given data, and the process is naturally called integration, since it generalizes the usual integration on a line.