Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Since semigroups don't need to have an identity element, I was wondering if there's any kind of short exact sequence for semigroups.

share|improve this question
add comment

1 Answer

up vote 7 down vote accepted

Suppose that you have a homomorphism of semigroups $f\colon S\to T$. The kernel of $f$ is defined to be the congruence $\mathrm{ker} f$ on $S$ defined by: $$\mathrm{ker} f = \{ (s_1,s_1)\in S\times S\mid f(s_1)=f(s_2)\}.$$ Note that $S/\mathrm{ker} f \cong f(S)$.

We also define the image of $f$, $\mathcal{K}_{\mathrm{Im}f}$ to be the relation on $T$ defined by: $$\mathcal{K}_{\mathrm{Im} f} = f(S)\times f(S)\cup\{(t,t)\mid t\in T\}.$$ Edit: The image need not be a congruence if $f(S)$ is not an ideal in $T$, because it need not be a subsemigroup of $T\times T$. But this agrees with the situation in, say, not-necessarily-abelian groups, where the image need not be a normal subgroup (and thus, not "eligible" to be equal to a kernel).

Given homomorphism $f\colon S\to T$ and $g\colon T\to U$, we say that $$S\stackrel{f}{\longrightarrow} T \stackrel{g}{\longrightarrow} U$$ is exact (at $T$) if and only if $\mathrm{ker} g = \mathcal{K}_{\mathrm{Im} f}$.

So we say that $S\stackrel{f}{\longrightarrow} T\stackrel{g}{\longrightarrow} U$ is a short exact sequence if and only if $\mathrm{ker} f = \{(s,s)\mid s\in S\}$, $\mathcal{K}_{\mathrm{Im} f} = \mathrm{ker}g$, and $\mathcal{K}_{\mathrm{Im} g} = U\times U$.

You can also see it by extending to monoids first. Given any semigroup $S$, you can always adjoin an identity element by taking some $1\notin S$, and defining the operation on $S\cup\{1\}$ by extending the multiplication by the rule $1s=s1=s$ for all $s\in S$; this monoid is denoted $S^1$ (even if $S$ already has an identity, we adjoin a new element). If $f\colon S\to T$ is a semigroup homorphism, then this induces a monoid homomorphism $f^1\colon S^1\to T^1$ by $f^1(s)=f(s)$ for all $s\in S$, and $f(1) = 1$. If we do this, then note that $$\mathrm{ker}(f^1) = \mathrm{ker}(f)\cup\{(1,1)\}$$ is a congruence on $S^1$; and that \begin{align*} \mathcal{K}_{\mathrm{Im}f^1} &= f^1(S^1)\times f^1(S^1)\cup\{(t,t)\mid t\in T^1\}\\ &= f(S)\times f(S) \cup\{(t,t)\mid t\in T\} \cup\{(1_T,1_T)\}\\ &= \mathrm{K}_{\mathrm{Im}f}\cup\{(1_T,1_T)\}, \end{align*} so that if we have $S\stackrel{f}{\to}T\stackrel{g}{\to}U$, with corresponding $S^1\stackrel{f^1}{\to} T^1\stackrel{g^1}{\to} U^1$, then $\mathrm{ker}f = \mathcal{K}_{\mathrm{Im}g}$ if and only if $\mathrm{ker}f^1 = \mathcal{K}_{\mathrm{Im}g^1}$.

If we do this, then $$S\stackrel{f}{\longrightarrow} T \stackrel{g}{\longrightarrow} U$$ is a short exact sequence if and only if $$1 \longrightarrow S^1 \stackrel{f^1}{\longrightarrow} T^1\stackrel{g^1}{\longrightarrow} U^1 \longrightarrow 1$$ is exact at $S^1$, $T^1$, and $U^1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.