- Let $S$ be a monoid with zero and $I_{\lambda}$ be an indexed set, then $B_{\lambda}(S) = \{ (\alpha, s , \beta ) : \alpha , \beta \in I_{\lambda}, s\in S \} \cup \{0\}$ is a semigroup and $J = \{ (\alpha, 0_s , \beta ) : \alpha , \beta \in I_{\lambda}\} \cup \{0\}$ , where $0_s$ is a zero of $S$, is an ideal of $B_{\lambda}(S)$. Define $$B_{\lambda}^o(S) = B_{\lambda} \backslash J$$
is called Brandt $\lambda^o$-extension of the semigroup $S$ with zero.
Let $S$ be an monoid with zero, $\lambda \geq 1$ any cardinal and $B_{\lambda}^o(S)$ the Brandt $\lambda^o$-extension of $S$. Then every non-trivial homomorphic image of $B_{\lambda}^o$ is the Brandt $\lambda^o$-extension of some monoid with zero.
I have tried:
when $\lambda = 1$ the proof is trivial. Therefore we assume that $\lambda \geq 2.$ Let $T$ be a semigroup and $h : B_{\lambda}^o(S) \rightarrow T$ is a homomorphism . Without loss of generality we can assume that the homomorphism $h$ is surjective map. suppose $(0_s)h = 0_T$ and $k$ be in $T$, then $$ k (0_s)h = (x)h . (0_s)h = (x.0_s)h = (0_s)h$$ thus $0_T$ is zero of $T$. Also we know that the semigroup $B_{\lambda}^o(1_s)$ is a congruence free, where $1_s$ is the identity element of $S$. So any homomorphism $g : B_{\lambda}^o(1_s) \rightarrow T$ is either trivial or isomorphism.
we fix $\alpha_o \in I_{\lambda}$, for every $\alpha , \beta \in I_{\lambda}$, we denote $1_{\alpha,\beta} = (\alpha, 1_s , \beta)h$ and $T^*_{\alpha,\beta} = \{ (\alpha, s , \beta)h : s \in S\backslash \{0\} \} \backslash \{0_T\}$ and $T_o= T^*_{\alpha_o , \alpha_o}$. Firstly we shall show that for any $\alpha, \beta , \gamma , \delta \in I_{\lambda}$, we have $|T^*_{\alpha,\beta}| = |T^*_{\gamma,\delta}|$, by defining the maps $$\phi_{(\alpha,\beta)}^{(\gamma, \delta)} : T^*_{\alpha,\beta} \rightarrow T^*_{\gamma,\delta}$$ by $$(x) \phi_{(\alpha,\beta)}^{(\gamma, \delta)} = 1_{\gamma ,\alpha}.x.1_{\beta,\delta}$$
similarly we can define the maping $$\phi_{(\gamma,\delta)}^{(\alpha, \beta)} : T^*_{\gamma,\delta} \rightarrow T^*_{\alpha,\beta}$$
Composition of both map is identity map, so both are mutually invertible maps and hence we have $|T^*_{\alpha,\beta}| = |T^*_{\gamma,\delta}| = |T_o|$.
Next I want to show that $T = I_{\lambda} \times T_o \times I_{\lambda} \cup \{0_T\}$, but I am unable how to prove.
