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.

Let $S$ be a monoid and denote by $Act-S$ the category of $S$-sets. I am having a problem understanding a step in the proof of the fact that if $\left\{G_i\right\}_{i \in I}$ is a set of generators in $Act-S$, then any amalgam $\coprod_{i \in I}^U G_i$, $U \in Act-S$, is also a generator.

Since $G_i$ is a generator, there exists retraction $\pi_i:G_i \rightarrow S_S, \, \forall i \in I$ (this is a "known" fact). Why does this imply that by the universal property of the amalgam there exists a unique morphism $\pi : \coprod_{i \in I}^U G_i \rightarrow S_S$ such that $\pi_i = \pi q_i$, where $q_i : G_i \rightarrow \coprod_{i \in I}^U G_i$ is the natural embedding into the amalgam?

To be more explicit, let $j_i : U \rightarrow G_i$ be monomorphisms $\forall i \in I$, corresponding to the amalgam $\coprod_{i \in I}^U G_i$. Then for the universal property of the amalgam to come into play, thus yielding the existence of the above $\pi$, we would need that $\pi_i j_i = \pi_k j_k, \forall i, k \in I$. But i can't see why such a thing would be true.

Any alternative proof?

Thanks :-)

share|improve this question
    
This amalgam doesn't make sense if $U$ is not specified with maps to $G_i$. In my answer I assumed as if $U$ wasn't there. Perhaps you want to work with the corresponding comma category. –  Martin Brandenburg Mar 6 '12 at 14:53

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.