Does Joyal's category $\Theta$ have intersections of active monomorphisms? Recall that Joyal's category $\Theta_N$ admits at least two interesting factorization systems:

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*Split epi / Mono. This makes $\Theta_N$ an elegant Reedy category, with degree function $deg([n \mid \theta_1, \dots, \theta_n]) = n + deg(\theta_1) + \cdots + deg(\theta_n)$.


*Active / Inert. Here, a morphism $\phi = [f \mid f_1, \dots, f_m] : [m \mid \theta_1, \dots, \theta_m] \to [n \mid \zeta_1, \dots, \zeta_n]$ is active if $f : [m] \to [n]$ is an active map in $\Delta$ (meaning that $0$ and $n$ are in its image) and $f_1,\dots,f_m$ are active in $\Theta_{N-1}$.
Every object $\theta \in \Theta_N$ has a smallest active subobject $\mathbb G_n \rightarrow \theta$ (where $\mathbb G_n$ is a globe of some dimension), which picks out the pasting composite of all the atomic cells of $\theta$. A morphism $\zeta \to \theta$ is active iff its image contains this maximal pasting composite.
Let $\Theta^{act,mon}_N \subset \Theta_N$ denote the wide subcategory of active monomorphisms.
Question: Does $\Theta^{act,mon}_N$ have pullbacks?
Notes:

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*When $N = 1$, the answer is yes.


*When $N = 1$, both conditions (monomoprhism + active) are necessary: $\Delta$ fails to have the intersection of the two monomorphisms $[0] \rightrightarrows [1]$, and also fails to have the pullback of the unique map $[1] \to [0]$ along itself (and this map is an active epimorphism).


*I tried proving the answer is yes via a simple-minded induction using the wreath product description $\Theta_N = \Delta \int \Theta_{N-1}$, but this fails because there are active monomorphisms $\phi$ as above whose components $f_i$ are active non-monomorphisms (this happens often when the leading component $f$ is not the identity).


*I'm already unclear on the answer when $N = 2$. For a first example, consider the two active monomorphisms $[1 \mid 2] \rightrightarrows [2 \mid 1, 1]$. Their intersection exists and is given by the minimal active map $[1 \mid 1] \to [2 \mid 1, 1]$.
(To orient the reader who may need a refresher on the notation I'm using for objects of $\Theta_2$, note that $[2 \mid 1, 1]$ corepresents two 2-cells $\alpha,\beta$ such that $\alpha \circ_0 \beta$ is defined, while $[1 \mid 2]$ corepresents two 2-cells $a,b$ such that $a \circ_1 b$ is defined. The two maps above corepresent the facts that if $\alpha \circ_0 \beta$ is defined, then $(\alpha \circ_0 \partial^-\beta) \circ_1 (\partial^+ \alpha \circ_0 \beta)$ and $(\partial^- \alpha \circ_0 \beta) \circ_1 (\alpha \circ_0 \partial^+ \beta)$ are defined. The map from the intersection corepresents the composite cell.)
 A: As recorded in the comments, the answer is no. In the following, I write composition in diagrammatic order. Consider the object
$$X = [3 \mid 1, 1, 1] \in \Theta_2$$
which is freely generated by three 2-cells $\alpha,\beta,\gamma$ which are $\circ_0$-composable, so that $\alpha \circ_0 \beta \circ_0 \gamma$ is defined. Let
$$Y \subset X$$
be freely generated by the 2-cells $\beta' = \partial^- \alpha \circ_0 \beta$, $\alpha' = \alpha \circ_0 \partial_+ \beta$, and $\gamma$. So $(\beta' \circ_1 \alpha') \circ_0 \gamma$ is defined; we have $Y \cong [2 \mid 2, 1]$. Let
$$Z \subset X$$
be freely generated by the 2-cells $\alpha$, $\beta'' = \beta \circ_0 \partial^- \gamma$, and $\gamma'' = \partial^+ \beta \circ_0 \gamma$. So $\alpha \circ_0 (\beta'' \circ_1 \gamma'')$ is defined; we have $Z \cong [2 \mid 1, 2]$.
Here are two subobjects of $X$ which are contained in $Y$ and in $Z$. On the one hand, we have
$$A \subset Y \cap Z$$
which is generated by $\bar \beta = \partial^- \alpha \circ_0 \beta \circ_0 \partial^- \gamma$, $a = \alpha \circ_0 \partial^+ \beta \circ_0 \partial^- \gamma$, and $c = \partial^+ \alpha \circ_0 \partial^+ \beta \circ_0 \gamma$. We have that $\bar \beta \circ_1 a \circ_1 c$ is defined, so $A \cong [1 \mid 3]$. On the other hand we have
$$B \subset Y \cap Z$$
which is generated by $\bar \beta$, $c' = \partial^- \alpha \circ_0 \partial^+ \beta \circ_0 \gamma$, and $a' = \alpha \circ_0 \partial^+ \beta \circ_0 \partial^+ \gamma$. We have that $\bar \beta \circ_1 c' \circ_1 a'$ is defined, so that $B \cong [1 \mid 3]$ as well.
For degree reasons, we cannot have a pullback $A \cup B \subseteq P \subseteq Y \cap Z$: if $deg(P) = 4$, then $A \to P$ and $B \to P$ would be isomorphisms, while if $deg(P) = 5$, then $P \to Y$ and $P \to Z$ would be isomorphisms.
