In what follows, we introduce the concept of 𝓠-inverse filters of a 𝓠-regular semigroup.

#### Proof

Obviously, *S*(*Q*) is a semigroup. If (*s*, *t*) ∈ *S*, where *s* ∈ *S*_{1}, *t* ∈ *S*_{2}, since *S*_{1}, *S*_{2} are all 𝓠-regular, there exists *s*′ ∈ *V*_{S1}(*s*), *t*′ ∈ *V*_{S2}(*t*) such that (*s*′, *t*′) ∈ *S*. It is easy to see that (*s*′, *t*′) ∈ *V*(*s*, *t*), and so *S* is regular. Let

$$\begin{array}{}{\displaystyle {V}_{{Q}_{1}}(s)\times {V}_{{Q}_{2}}(t)=\{({s}^{+},{t}^{+}):{s}^{+}\in {V}_{{Q}_{1}}(s),{t}^{+}\in {V}_{{Q}_{2}}(t)\}.}\end{array}$$

Clearly, it is non-empty and contained in *V*(*s*, *t*). For any (*s*^{+}, *t*^{+}) ∈ *V*_{Q1}(*s*) × *V*_{Q2}(*t*), we have

$$\begin{array}{}{\displaystyle (s,t)({s}^{+},{t}^{+})\in {V}_{{Q}_{1}}(s{s}^{+})\times {V}_{{Q}_{2}}(t{t}^{+}),}\\ ({s}^{+},{t}^{+})(s,t)\in {V}_{{Q}_{1}}({s}^{+}s)\times {V}_{{Q}_{2}}({t}^{+}t).\end{array}$$

For another, let (*s*, *t*), (*x*, *y*) ∈ *S*. For any ((*sx*)^{+}, (*ty*)^{+}) ∈ *V*_{Q1}(*sx*) × *V*_{Q2}(*ty*), since *S*_{1}, *S*_{2} are all 𝓠-regular, there exists *s*^{+} ∈ *V*_{Q1}(*s*), *x*^{+} ∈ *V*_{Q1}(*x*), *t*^{+} ∈ *V*_{Q2}(*t*), *y*^{+} ∈ *V*_{Q2}(*y*) such that

$$\begin{array}{}{\displaystyle ((sx{)}^{+},(ty{)}^{+})=({x}^{+}{s}^{+},{y}^{+}{t}^{+})}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=({x}^{+},{y}^{+})({s}^{+},{t}^{+})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\in ({V}_{{Q}_{1}}(x)\times {V}_{{Q}_{2}}(y))({V}_{{Q}_{1}}(s)\times {V}_{{Q}_{2}}(t)).\end{array}$$

Hence, *S*(*Q*) is a 𝓠-regular semigroup. □

#### Proof

If (*s*_{1}, *t*_{1}), (*s*_{2}, *t*_{2}) ∈ *π*, then *t*_{1} ∈ *s*_{1}*φ*, *t*_{2} ∈ *s*_{2}*φ*. By (2), *t*_{1}*t*_{2} ∈ (*s*_{1}*s*_{2})*φ*, and so (*s*_{1}*s*_{2}, *t*_{1}*t*_{2}) ∈ *π*. Hence *π* is a subsemigroup of *S* × *T*. If (*s*, *t*) ∈ *π*, then *t* ∈ *s**φ*. By (4), there exist *s*^{+} ∈ *V*_{Q1}(*s*) and *t*^{+} ∈ *V*_{Q2}(*t*) such that *t*^{+} ∈ *s*^{+}*φ*, and so (*s*^{+}, *t*^{+}) ∈ *π*. It is easy to see that (*s*^{+}, *t*^{+}) ∈ *V*(*s*, *t*). Thus *π* is a regular semigroup. Let

$$\begin{array}{}{\displaystyle {V}_{Q}(s,t)=\{({s}^{+},{t}^{+})\in {V}_{{Q}_{1}}(s)\times {V}_{{Q}_{2}}(t):\phantom{\rule{thinmathspace}{0ex}}{t}^{+}\in {s}^{+}\phi \}.}\end{array}$$

If (*s*, *t*) ∈ *π*, then *t* ∈ *s**φ*. By (4), *V*_{Q}(*s*, *t*) ≠ ∅. For any (*s*, *t*) ∈ *π*, (*s*^{+}, *t*^{+}) ∈ *V*_{Q}(*s*, *t*),

$$\begin{array}{}{\displaystyle (s,t)({s}^{+},{t}^{+})=(s{s}^{+},t{t}^{+}),}\end{array}$$

where *s*^{+} ∈ *V*_{Q1}(*s*), *t*^{+} ∈ *V*_{Q2}(*t*), *t*^{+} ∈ *s*^{+}*φ*. Since *S*(*Q*_{1}), *T*(*Q*_{2}) are 𝓠-regular, it follows that *ss*^{+} ∈ *V*_{Q1}(*ss*^{+}), *tt*^{+} ∈ *V*_{Q2}(*tt*^{+}). And *tt*^{+} ∈ (*ss*^{+})*φ*, hence

$$\begin{array}{}{\displaystyle (s,t)({s}^{+},{t}^{+})\in {V}_{Q}(s{s}^{+},t{t}^{+})={V}_{Q}((s,t)({s}^{+},{t}^{+})).}\end{array}$$

Similarly, (*s*^{+}, *t*^{+})(*s*, *t*) ∈ *V*_{Q}((*s*^{+}, *t*^{+})(*s*, *t*)).

On the other hand, let (*s*_{1}, *t*_{1}), (*s*_{2}, *t*_{2}) ∈ *π*. For any

$$\begin{array}{}{\displaystyle (({s}_{1}{s}_{2}{)}^{+},({t}_{1}{t}_{2}{)}^{+})\in {V}_{Q}({s}_{1}{s}_{2},{t}_{1}{t}_{2}),}\end{array}$$

since *S*(*Q*_{1}) and *T*(*Q*_{2}) are all 𝓠-regular semigroups, there exist $\begin{array}{}{\displaystyle {s}_{1}^{+}}\end{array}$ ∈ *V*_{Q1}(*s*_{1}), $\begin{array}{}{\displaystyle {s}_{2}^{+}}\end{array}$ ∈ *V*_{Q1}(*s*_{2}), $\begin{array}{}{\displaystyle {t}_{1}^{+}}\end{array}$ ∈ *V*_{Q2}(*t*_{1}), $\begin{array}{}{\displaystyle {t}_{2}^{+}}\end{array}$ ∈ *V*_{Q2}(*t*_{2}) such that

$$\begin{array}{}{\displaystyle (({s}_{1}{s}_{2}{)}^{+},({t}_{1}{t}_{2}{)}^{+})=({s}_{2}^{+}{s}_{1}^{+},{t}_{2}^{+}{t}_{1}^{+})=({s}_{2}^{+},{t}_{2}^{+})({s}_{1}^{+},{t}_{1}^{+}).}\end{array}$$

Since $\begin{array}{}{t}_{2}^{+}{t}_{1}^{+}\in ({s}_{2}^{+}{s}_{1}^{+})\phi ,\end{array}$ by (6), $\begin{array}{}{t}_{2}^{+}\in {s}_{2}^{+}\phi ,\phantom{\rule{thinmathspace}{0ex}}{t}_{1}^{+}\in {s}_{1}^{+}\phi .\end{array}$ And so

$$\begin{array}{}{\displaystyle ({s}_{2}^{+},{t}_{2}^{+})\in {V}_{Q}({s}_{2},{t}_{2}),\phantom{\rule{2em}{0ex}}({s}_{1}^{+},{t}_{1}^{+})\in {V}_{Q}({s}_{1},{t}_{1}).}\end{array}$$

Hence,

$$\begin{array}{}{\displaystyle {V}_{Q}({s}_{1}{s}_{2},{t}_{1}{t}_{2})\subseteq {V}_{Q}({s}_{2},{t}_{2}){V}_{Q}({s}_{1},{t}_{1}).}\end{array}$$

that is, *π*(*Q*) is a 𝓠-regular semigroup.

For any

$$\begin{array}{}{\displaystyle ({s}_{1}^{+},{t}_{1}^{+})\in {V}_{{Q}_{1}}({s}_{1})\times {V}_{{Q}_{2}}({t}_{1}),\phantom{\rule{2em}{0ex}}({s}_{2}^{+},{t}_{2}^{+})\in {V}_{{Q}_{1}}({s}_{2})\times {V}_{{Q}_{2}}({t}_{2}).}\end{array}$$

If $\begin{array}{}({s}_{1}^{+},{t}_{1}^{+})({s}_{2}^{+},{t}_{2}^{+})\end{array}$ ∈ *π*, namely $\begin{array}{}({s}_{1}^{+}{s}_{2}^{+},{t}_{1}^{+}{t}_{2}^{+})\end{array}$ ∈ *π*, then $\begin{array}{}{t}_{1}^{+}{t}_{2}^{+}\in ({s}_{1}^{+}{s}_{2}^{+})\phi .\end{array}$ By (6), $\begin{array}{}{t}_{1}^{+}\in {s}_{1}^{+}\phi ,\phantom{\rule{thinmathspace}{0ex}}{t}_{2}^{+}\in {s}_{2}^{+}\phi ,\end{array}$ and so $\begin{array}{}({s}_{1}^{+},{t}_{1}^{+}),\phantom{\rule{thinmathspace}{0ex}}({s}_{2}^{+},{t}_{2}^{+})\end{array}$ ∈ *π*. Hence *π*(*Q*) is a 𝓠-inverse filter of *S*(*Q*_{1}) × *T*(*Q*_{2}).

By (1) and (5), the projection *f*_{1}: *π* → S, (*s*, *t*) ↦ *s* is a surjective 𝓠-homomorphism. By (3) and (5), the projection *f*_{2}: *π* → T, (*s*, *t*) ↦ *t* is a surjection 𝓠-homomorphism. Therefore, *π*(*Q*) is a 𝓠-subdirect product of *S*(*Q*_{1}) and *T*(*Q*_{2}).

Conversely, let *H*(*Q*) be a 𝓠-subdirect product of *S*(*Q*_{1}) and *T*(*Q*_{2}). Let

$$\begin{array}{}{\displaystyle \phi :\phantom{\rule{thinmathspace}{0ex}}S\to {2}^{T},\phantom{\rule{thinmathspace}{0ex}}s\mapsto \{t\in T:\phantom{\rule{thinmathspace}{0ex}}(s,t)\in H\}.}\end{array}$$

Since *H*(*Q*) is a 𝓠-subdirect product of *S*(*Q*_{1}) and *T*(*Q*_{2}), there exists *t* ∈ *T* such that (*s*, *t*) ∈ *H* for any *s* ∈ *S*. Thus *t* ∈ *s**φ*. Hence (1) holds. Since *H* is a subsemigroup of *S*(*Q*_{1}) × *T*(*Q*_{2}), (2) holds. Since the projection *f*_{2}: *H* → *T* is surjective, (3) holds. If *t* ∈ *s**φ* (*s* ∈ *S*, *t* ∈ *T*), then (*s*, *t*) ∈ *H*. Since *H*(*Q*) is 𝓠-regular, there exists

$$\begin{array}{}{\displaystyle ({s}^{+},{t}^{+})\in ({V}_{{Q}_{1}}(s)\times {V}_{{Q}_{2}}(t))\cap H.}\end{array}$$

Hence *s*^{+} ∈ *V*_{Q1}(*s*), *t*^{+} ∈ *V*_{Q2}(*t*), and *t*^{+} ∈ *s*^{+}*φ*, so (4) holds. For any *p*_{1} ∈ *Q*_{1}, there exists (*p*_{1}, *p*_{2}) ∈ *Q* = *H* ∩ (*Q*_{1} × *Q*_{2}), and so *p*_{2} ∈ *Q*_{2} and *p*_{2} ∈ *p*_{1}*φ*. Additionally, since *f*_{2} is a surjective 𝓠-homomorphism, there exists

$$\begin{array}{}{\displaystyle ({p}_{1},{p}_{2})\in Q=H\cap ({Q}_{1}\times {Q}_{2})}\end{array}$$

for any *p*_{2} ∈ *Q*_{2}. Hence *p*_{1} ∈ *Q*_{1} and *p*_{2} ∈ *p*_{1}*φ*. Thus (5) holds. Since *H*(*Q*) is a 𝓠-inverse filter of *S*(*Q*_{1}) × *T*(*Q*_{2}), (6) holds. By the proof above, *φ* is a surjective 𝓠-subhomomorphism of *S*(*Q*_{1}) onto *T*(*Q*_{2}). Obviously, *H* = *π*(*S*, *T*, *φ*). □

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