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If M a monoid and $Set^M$ a topos associate to M, I found $Set^M$ have (epi, strong,mono) factorization system(https://math.stackexchange.com/questions/541300/epi-mono-factorization-in-presentable-categories), I think $Set^M$ has (epi,mono source) factorization [http://katmat.math.uni-bremen.de/acc/acc.pdf, pag. 257] but I can not found the proof of that, I would like see some reference of the proof such that don't use the fact $Set^M$ is a locally presentable category or If I suppose it's wrong a liked see the counterexample.

Note: If $(f_i, X)$ is a source and $f: X \rightarrow X/ \sim$ where $a \sim b$ iif for all $f_i : X \rightarrow Y_i$, $f_i(a) = f_i(b)$, the first I think was associate the action of $X$ to $X / \sim$ so $\lambda: X \times M \rightarrow X$, we have $f(\lambda(x,m))$ and $\lambda(f(x),m)$ but I think, I can´t write $f(\lambda(x,m)) = \lambda(f(x),m)$ and finished the proof, because I can't justify the equality.

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  • $\begingroup$ Look at the Example 15.3(2) (page 259); it is done for sets there but can be repeated verbatim for $M$-sets. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 13, 2017 at 18:06
  • $\begingroup$ I writed a note whit the reason why I think can´t repeated verbatim the proof $\endgroup$
    – Wilson Forero
    Commented Feb 14, 2017 at 1:11
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    $\begingroup$ Use the fact that $f_i(\lambda(x,m))=\lambda_i(f_i(x),m)$ (where $\lambda_i$ is the action of $M$ on $Y_i$) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 14, 2017 at 7:43

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Based of the idea of მამუკა ჯიბლაძე and $Set$ has (epi, mono source) factorization, if $(f_i, (X, \lambda))$ is a source, the (epi, initial source factorization) is $(((X/ \sim), \lambda), f_i)$ where $a \sim b$ iff $f_i(a)=f_i(b)$ for all $i \in I$ and $f: (X, \lambda) \rightarrow (X/\sim, \lambda)$ preserve the action because: $$f_i(\lambda(f(x),m)) = \mu( f_i(f(x)),m)$$ $$= \mu( f_i(x)),m)$$ $$= f_i(\lambda(x,m))$$ $$= f_i(f(\lambda(x,m)))$$ $$\lambda(f(x),m)) = f(\lambda(x,m))$$ and $f_i: (X/\sim, \lambda) \rightarrow (Y_i, \mu_i)$.

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