Monomorphisms in operad algebras Setup: Let $\mathcal{O}$ be an operad in the category of sets, and let $\mathcal{O}\text{-Alg}$ denote the category of algebras on it (i.e., operad functors $\mathcal{O}\to\mathbf{Set}$. This category is cocomplete; let $\kappa$ denote an initial object. Call an algebra $X$ initially monic if the unique map $!_X\colon\kappa\to X$ is monic. 
Question 1: If $X$ and $Y$ are initially monic, is their coproduct $X\sqcup Y$ initially monic as well?
Remark: If, in the setup, we replace "operad" by "algebraic theory", then the answer to the question becomes no, as shown here by Zhen Lin in the case of $R$-algebras. I'm wondering if the issue might somehow come down to the diagonal maps.
Question 2: Can one characterize algebraic theories $\mathcal{T}$ for which the coproduct of initially monic algebras is initially monic?
 A: Even for single-sorted operads, the coproduct of initially monic algebras need not be initially monic. 
First a general construction. Let $A$ be a commutative monoid. Then the comma category $A \downarrow \mathrm{CMon}$ (aka the undercategory or co-slice under $A$) is the category of algebras of an operad $\mathcal{O}_A$. I'll indicate a proof of that later, but it allows us to relativize the question to consideration of pushouts of monos in $\mathrm{CMon}$, where we ask: if $i_B: A \to B$ and $i_C: A \to C$ are monos in $\mathrm{CMon}$ and we take the pushout $P = B \oplus_A C$, is it true that the canonical map $A \to P$ is monic? This is equivalent to the question of initial monicity for the operad $\mathcal{O}_A$. 
The answer is 'no'. The following is a modification of an example that appears in 


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*Kimura, On semigroups, PhD dissertation, Tulane University 1957. 


Take $A = \{u, v, 1, 0\}$ to be a commutative monoid with identity element $1$ and null element $0$, with all products $x y = 0$ for $x \neq 1, y \neq 1$. Define $B = A \sqcup \{b\}$, adjoining a new element $b$ to $A$, and extending the multiplication on $A$ by $b u = u b = v$ and $b \cdot 1 = 1 \cdot b = b$ and $b x = x b = 0$ for any $x \notin \{u, 1\}$. Similarly define $C = A \sqcup \{c\}$ with $c v = v c = u$ and $c \cdot 1 = 1 \cdot c = c$ and $c x = x c = 0$ for $x \notin \{v, 1\}$. Let $i_B: A \to B$ and $i_C: A \to C$ be the inclusion maps, and let $P$ be the pushout of $i_B$ and $i_A$ with coproduct coprojections $j: B \to P$, $k: C \to P$. I claim $k \circ i_C: A \to P$ is not monic. The short version of the calculation is that in $P$ we have 
$$u = v c = b u c = b \cdot 0 = 0$$ 
and the long version (if anyone really needs it) is that $(k \circ i_C)(u) = (k \circ i_C)(0)$ according to the string of equations 
$$k(i_C u) = k(i_C v \cdot c) = k(i_C v) k(c) = j(i_B v) k(c) = j(b i_B u) k(c) = j(b)j(i_B u)k(c) = j(b)k(i_C u)k(c) = j(b)k(i_C u \cdot c) = j(b)k(i_C 0) = j(b)j(i_B 0) = j(b \cdot i_B 0) = j(i_B 0) = k(i_C 0).$$ 
Now for the claim of an operad $\mathcal{O}_A$ whose algebras are $A \downarrow \mathrm{CMon}$. We have pairs of adjoint functors 
$$F_A \dashv U_A: A\downarrow \mathrm{CMon} \to \mathrm{CMon}, \qquad F \dashv U: \mathrm{CMon} \to \mathrm{Set}$$ 
where $U F$ is the monad attached to the commutative monoid operad and the forgetful functor $U_A: A \downarrow \mathrm{CMon} \to \mathrm{CMon}$ is monadic. The composite $U \circ U_A$ is also monadic; this follows for example from the crude monadicity theorem (both $U$ and $U_A$ reflect isomorphisms and preserve reflexive coequalizers). What remains to be seen is that the monad attached to $U \circ U_A$ is an analytic monad in the sense of the theory of Joyal species. This has two parts: that the underlying functor of the monad $U U_A F_A F$ is an analytic functor, and that the monad unit and multiplication are cartesian natural transformations. 
One way to characterize analytic functors $\mathrm{Set} \to \mathrm{Set}$ is that they are functors which preserve filtered colimits and weak wide pullbacks (see for example this paper), so we check that the three functors $U, F,$ and $U_A F_A = A \oplus -: \mathrm{CMon} \to \mathrm{CMon}$ have these properties. That $U$ and $F$ do follows from the fact that $U F$ does (being an analytic functor) plus the fact that $U$ preserves and reflects limits and even weak limits, plus the fact that $U$ preserves filtered colimits. Meanwhile, the underlying functor of the monad $A \oplus -: \mathrm{CMon} \to \mathrm{CMon}$ preserves filtered colimits (this is true for the functor $A \oplus - \cong A + -$, seen as taking the coproduct with $A$) and preserves wide pullbacks (this is true for the functor $A \oplus - \cong A \times -$, seen as taking the product with $A$). 
That the monad data for the monadic functor $A \downarrow \mathrm{CMon} \to \mathrm{Set}$ are cartesian follows from (1) the observation that the unit and multiplication of the monad $A \oplus -: \mathrm{CMon} \to \mathrm{CMon}$ are cartesian; this is most easily seen by considering $A \oplus -$ in the guise of $A \times -$, where the cartesianness boils down to consideration of products of pullback squares, and (2) the fact that cartesian natural transformations are closed under pasting, e.g., given that the units $u: 1_{\mathrm{Set}} \to U F$ and $\eta: 1_{\mathrm{CMon}} \to A \oplus -$ are cartesian, so is the unit of the monad formed by pasting 
$$1 \stackrel{u}{\to} U F \stackrel{1 \eta 1}{\to} U(A \oplus -)F$$ 
(using the fact that $U$ preserves pullbacks). 
