**Proposition 6.** *Let* I = (*Sign*, *Sen*, *Mod*, |=) *be a stratified institution and let*

<sup>I</sup><sup>s</sup> = (*Sign*<sup>s</sup> , *Sen*<sup>s</sup> , *Mod*<sup>s</sup> , [[*\_*]], <sup>|</sup>=s) *be its representation as a stratified institution. Let the left hand square below represent a commutative diagram in Sign*<sup>s</sup> *, such that its projection on the first component (the right hand side square below) is a model amalgamation square in* I*.*

$$\chi\_{\left(\zeta,\eta\right)} \xrightarrow[\begin{subarray}{c} \left(\zeta,\eta\right) \\ \end{subarray}} \xrightarrow{\left(\eta,\theta\right)} \chi'\_{\left(\zeta',\eta'\right)} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\stackrel{\Sigma}{\zeta} \xrightarrow{\begin{subarray}{c} \frac{\varphi}{q} \\ \end{subarray}} \xrightarrow{\begin{subarray}{c} \Sigma'\\ \end{subarray}} \begin{subarray}{c} \Sigma'\\ \end{subarray}} \chi'\_{\left(\begin{subarray}{c} \xi\\ \end{subarray}\right)} \xrightarrow{\begin{subarray}{c} \Sigma'\\ \end{subarray}} \begin{subarray}{c} \Sigma'\\ \end{subarray}$$

*Then, the square of* <sup>I</sup>s*-signature morphisms is a stratified model amalgamation square if and only if the following lax cocone of* I*-signature morphisms has the model amalgamation property.*

**Proof.** In this proof, we rely on the fact that stratified model amalgamation in <sup>I</sup><sup>s</sup> is the same as model amalgamation in (Is) . First, we show that the model amalgamation in <sup>I</sup> implies that in (Is) .

In (Is) , we consider a *<sup>χ</sup>* -model (*M* , *N* ) and a *χ*1-model (*M*1, *N*1), such that

$$(Mod^\circ)^\sharp(\zeta,\eta)(M\_1,N\_1) = (M,N) \text{ and } (Mod^\circ)^\sharp(\zeta',\eta')(M',N') = (M,N). \tag{16}$$

Let the Σ <sup>1</sup>-model *N* <sup>1</sup> be the unique amalgamation of *N* and *N*1. Then, (*M*1, *M*, *M* , *N*1, *N*, *N* , *N* <sup>1</sup>) is a model for the *Sign*<sup>s</sup> diagram with 7 vertices and 9 (full) arrows in (15). Let *M* <sup>1</sup> be the Ω <sup>1</sup>-model that is the unique amalgamation of (*M*1, *M*, *M* , *N*1, *N*, *N* , *N* <sup>1</sup>). We have that (*M* <sup>1</sup>, *N* <sup>1</sup>) is a *χ* <sup>1</sup>-model and that:

$$(\operatorname{Mod}^\circ)^\sharp(\varrho\_1, \theta\_1)(M\_1', N\_1') = (M\_1, N\_1) \text{ and } (\operatorname{Mod}^\circ)^\sharp(\zeta', \eta')(M\_1', N\_1') = (M', N').$$

The uniqueness of the amalgamation (*M* <sup>1</sup>, *N* <sup>1</sup>) follows from the uniquenesses of the amalgamations *N* <sup>1</sup> and *M* 1.

Conversely, we now assume the amalgamation property in <sup>I</sup><sup>s</sup> and prove it at the level of I. Any model (*M*1, *M*, *M* , *N*1, *N*, *N* , *N* <sup>1</sup>) for the *Sign*<sup>s</sup> diagram with 7 vertices and 9 arrows determines two (Is) models: an *<sup>χ</sup>* -model (*M* , *N* ) and an *χ*1-model (*M*1, *N*1), such that (16) holds. By the stratified model amalgamation property in <sup>I</sup>s, interpreted in (Is) , there exists an unique amalgamation (*<sup>M</sup>* <sup>1</sup>, *N* <sup>1</sup> ) of (*M* , *N* ) and of (*M*1, *N*1). Since *Mod*(*η*)*N*<sup>1</sup> = *Mod*(*θ*)*N* , *Mod*(*θ*1)*N* <sup>1</sup> = *N*1, *Mod*(*η* )*N* <sup>1</sup> = *N* , and because the same holds for *N* <sup>1</sup> in the place of *N* <sup>1</sup> by the uniqueness property of the I-model amalgamation square (14), we have that *N* <sup>1</sup> = *N* 1.

Now, we show that *M* <sup>1</sup> is the amalgamation of (*M*1, *M*, *M* , *N*1, *N*, *N* , *N* 1).


$$\begin{aligned} N' &= \text{Mod}(\zeta\_{\sharp}') N\_1' & \text{ definition of } N\_1' \\ &\in \text{Mod}(\zeta\_{\sharp}') (\text{Mod}(\chi\_1') M\_1') & N\_1' \in \text{Mod}(\chi\_1') M\_1' \\ &\subseteq \text{Mod}(\zeta\_{\sharp}'; \chi\_1') M\_1' & & \text{Mod} \text{ kax.} \end{aligned}$$

• By a similar argument to the above one, we establish that *N*<sup>1</sup> ∈ *Mod*(*ϕ*1; *χ* 1)*M* 1. • Finally,


The uniqueness of *M* <sup>1</sup> follows from the uniqueness of the amalgamation (*M* <sup>1</sup>, *N* <sup>1</sup>) by relying on the first implication of the proposition.

In the context of proposition 6, the following general result provides a sufficient condition for the lax cocone of I-signature morphisms to have the model amalgamation property. Then, by the conclusion of proposition 6, this leads to the left-hand side square of diagrams (14) to be a stratified model amalgamation square. We need to recall from [13] two concepts as follows:

• In any 3/2-category, a strict commutative square

is a *3/2-pushout* when for any strict cocones (*θ* <sup>1</sup>, *θ* <sup>2</sup>) and (*θ* <sup>1</sup> , *θ* <sup>2</sup> ) over the span (*ϕ*1, *ϕ*2), if *θ <sup>k</sup>* ≤ *θ <sup>k</sup>* , *k* = 1, 2, there exists unique mediating arrows *μ* ≤ *μ* , such that *θk*; *μ* = *θ k* and *θ <sup>k</sup>* ; *μ* = *θ <sup>k</sup>* , *k* = 1, 2. Note that 3/2-pushouts are stronger than ordinary pushouts.

• *3/2-institutional seeds* were mentioned above when we discussed examples of 3/2 institutions. For the full definition, see [13]. For the purpose of proposition 7 below, we only need the property that there exists a signature Π, such that for each signature Σ

$$|\operatorname{Mod}(\Sigma)| = \{ \mathcal{M} \, : \, \Sigma \to \Pi \mid \operatorname{Sen}(\mathcal{M}) \text{ total} \}$$

and for each signature morphism *ϕ* and for each *ϕ*✷-model *M* ,

$$\mathcal{M}\text{od}(\mathfrak{q})\mathcal{M}' = \{ \mathcal{M} \in |\mathcal{M}\text{od}(\Box \mathfrak{q})| \mid \mathfrak{q} ; \mathcal{M}' \le \mathcal{M} \}.$$

**Proposition 7.** *Let us assume a 3/2-institution* I*, such that when we remove its model homomorphisms it is generated by a 3/2-institutional seed. Let us consider a lax cocone of* I *signature morphisms such as in diagrams* (15) *with the following properties:*


*Then, the lax cocone of diagram* (15) *has the model amalgamation property.*

**Proof.** Let us consider (*M*1, *M*, *M* , *N*1, *N*, *N* , *N* <sup>1</sup>), a model of the diagram of 7 vertices and 9 arrows of diagrams (14). Let *M* <sup>1</sup> be the amalgamation of *M* and *M*<sup>1</sup> by using the model amalgamation property of the outer square (*η*, *θ*, *η* , *θ*1). We show that *M* <sup>1</sup> is also the amalgamation of (*M*1, *M*, *M* , *N*1, *N*, *N* , *N* <sup>1</sup>), and its uniqueness follows from the uniqueness as amalgamation of *M* and *M*<sup>1</sup> only.

We first prove that *N* <sup>1</sup> ∈ *Mod*(*χ* 1)*M* <sup>1</sup>. This goes as follows. Since *N* ∈ *Mod*(*χ* )*M* , this means that *χ* ; *M* ≤ *N* . It follows that:

$$\mathbb{Z}';\chi\_1';M\_1'=\chi';\eta';M\_1'=\chi';M'\leq N'=\text{Mod}(\mathbb{Z}')N\_1'=\mathbb{Z}';N\_1'.\tag{17}$$

Similarly,

$$
\varphi\_1; \chi\_1'; \mathcal{M}\_1' \le \varphi\_1; \mathcal{N}\_1'. \tag{18}
$$

Because the inner square of diagram (15), i.e., the square (*ϕ*, *ζ*, *ζ* , *ϕ*1), is a 3/2-pushout square, from (17) and (18), we obtain that *χ* <sup>1</sup>; *M* <sup>1</sup> ≤ *N* <sup>1</sup>, which means *N* <sup>1</sup> ∈ *Mod*(*χ* 1)*M* 1. From here, the proof that *M* <sup>1</sup> is an amalgamation of (*M*1, *M*, *M* , *N*1, *N*, *N* , *N* <sup>1</sup>) follows the same steps as in the second part of the proof of proposition 6. In particular, this means that the three specific conditions of the current proposition are not used anymore.

In concrete situations, it is quite common that the 3/2-pushout condition on the inner square in proposition 7 implies the model amalgamation condition on the same square in proposition 6, a fact that enhances the applicability of proposition 7 within the context of the equivalence established by proposition 6.

**Proposition 8.** *In the context of a 3/2-institution generated by a 3/2-institutional seed, let us assume that Sen preserves and reflects maximality (i.e., ϕ is maximal if and only if Sen*(*ϕ*) *is total). Then, any pushout cocone of signature morphisms determines a model amalgamation square.*

**Proof.** Let (*ϕ*1, *ζ* ) be a pushout cocone for a span (*ϕ*, *ζ*) of signature morphisms (such as in the diagram below).

Given a Σ -model *N* and a Σ1-model *N*1, such that *Mod*(*ϕ*)*N* = *Mod*(*ζ*)*N*1, we obtain a strict cocone (*N* , *N*1) for the span (*ϕ*, *ζ*). By the pushout property of (*ϕ*1, *ζ* ), there exists an unique *N* <sup>1</sup> : Σ <sup>1</sup> → Π, such that *N* = *ζ* ; *N* <sup>1</sup> and *N*<sup>1</sup> = *ϕ*1; *N* <sup>1</sup>. It remains to prove that *N* <sup>1</sup> qualifies as a Σ <sup>1</sup>-model, i.e., that *Sen*(*N* <sup>1</sup>) is total.

By the hypothesis on *Sen* that it preserves maximality, it is enough to prove that *N* 1 is maximal. Consider any *x* : Σ <sup>1</sup> → Π, such that *N* <sup>1</sup> ≤ *x*. It follows that *N* ≤ *ζ* ; *x* and *N*<sup>1</sup> ≤ *ϕ*1; *x*. By the hypothesis on *Sen* that it reflects maximality, we have that both *N* and *N*<sup>1</sup> are maximal, hence *ζ* ; *x* = *N* and *ϕ*1; *x* = *N*1. By the pushout hypothesis, it follows that *x* = *N* 1.

With respect to the conditions underlying proposition 8, note that:


Often, in concrete situations that are related to the basic context of proposition 6, the pushout squares of signature morphisms are already 3/2-pushout squares. The following result illustrates such a case that is emblematic for the concrete applications not only because it is sometimes involved as such (e.g., in 3/2PL) but also because when it is not the case then the respective category of signature morphisms can be often treated in a similar way.

#### **Proposition 9.** *Any pushout square in* **Set** *is a 3/2-pushout square in* **Pfn***.*

**Proof.** For this proof, it is convenient to use the representation of partial functions as homomorphisms between *pointed sets*. A pointed set *A* is a set with a universally designated element ⊥. A homomorphism *f* : *A* → *B* of pointed sets is a function that preserves the designated element ⊥, i.e., *<sup>f</sup>*⊥ = ⊥. This yields a category **Set**<sup>⊥</sup> and a canonical isomorphism **Pfn** ∼= **Set**<sup>⊥</sup> that:


$$f\_{\perp}x = \begin{cases} fx, & fx \text{ defined} \\ \bot, & fx \text{undefined.} \end{cases}$$

Now, let us consider a pushout square in **Set** as follows.

$$\begin{array}{c} \Sigma \xrightarrow{\begin{array}{c} \varphi\\ \varphi \end{array}} \begin{array}{c} \Sigma'\\ \varphi'\\ \Sigma\_1 \xrightarrow{\begin{array}{c} \varphi\\ \end{array}} \Sigma'\_1 \end{array} \tag{19}$$

By mapping the pushout square (19) to **Set**⊥, we obtain the commutative square

$$\begin{array}{c} \Sigma\_{\perp} \xrightarrow{\overline{\varphi\_{\perp}}} \begin{array}{c} \Sigma'\_{\perp} \\ \downarrow \\ (\Sigma\_{1})\_{\perp} \xrightarrow{} \begin{array}{c} \Sigma'\_{\perp} \\ (\Sigma'\_{1})\_{\perp} \end{array} \end{array} \begin{array}{c} \Sigma'\_{\perp} \\ \downarrow \\ (\Sigma'\_{1})\_{\perp} \end{array} \end{array}$$

This is a pushout square because (\_)<sup>⊥</sup> : **Set** → **Set**<sup>⊥</sup> is a left adjoint to the forgetful functor **Set**<sup>⊥</sup> → **Set** and left adjoint functors preserve all colimits [15]. This left adjoint property can be either checked directly, or else, it can be established by noticing that it is a special case of a free algebra construction corresponding to a reduct functor of categories of algebras for the signature inclusion of the signature consisting of one sort into the signature consisting of one sort and one constant.

This showed that (19) is a pushout square in **Pfn**. In order to prove that this is a 3/2-pushout square in **Pfn**, we let (*a* , *a*1) and (*b* , *b*1) be strict cocones for the span (*ϕ*, *ζ*), such that *a* ⊆ *b* and *a*<sup>1</sup> ⊆ *b*1. Let *α* and *β* be the unique mediating partial functions for (*a* , *a*1) and (*b* , *b*1), respectively. We have to show that *α* ⊆ *β*. Note that *α* ⊆ *β* means that for each *x* ∈ Σ <sup>1</sup>, *<sup>α</sup>*⊥*<sup>x</sup>* = ⊥ implies *<sup>α</sup>*⊥*<sup>x</sup>* = *<sup>β</sup>*⊥*x*. However, *<sup>α</sup>*<sup>⊥</sup> = ⊥ implies *<sup>x</sup>* = ⊥. By the pushout property of (19), it follows that there exists *y* in Σ or in Σ1, such that *ζ y* = *x* or *ϕ*1*x* = *y*. By symmetry, without any loss of generality, we may assume that *y* ∈ Σ and *ζ y* = *x*. We have that:

$$\alpha\_\perp \mathbf{x} = \alpha\_\perp(\zeta' \mathbf{y}) = a\mathbf{y} = b\mathbf{y} \text{ (since } a \subseteq b\text{) } = \beta\_\perp(\zeta' \mathbf{y}) = \beta\_\perp \mathbf{x}.$$
