Consider the amalgamated sum $Q\_1 \rightarrow^{v\_1} Q\_1 \oplus\_P Q\_2 \leftarrow^{v\_2} Q\_2$ of $Q\_1 \leftarrow^{u\_1} P \rightarrow^{u\_2} Q\_2$ with $Q\_1,Q\_2,P$ being monoids.  

Why does $v:= v\_i \circ u\_i$ factor through $(Q\_1 \oplus\_P Q\_2)^{\times}$ if either $Q\_1,Q\_2$ or $P$ is a group?  

Thanks in advance for your help.



amalgamated sum of monoids
It looks like this just follows from the fact that a map of monoids $P\to Q$ where $P$ is a group factors through $Q^\times$ (the image of the inverse of some $p$ is an inverse of the image, so the image of every $p$ is invertible).


In your question, if $P$ is a group this immediately implies the conclusion, otherwise say $Q\_1$ is a group. Then $Q\_1\to (Q\_1\oplus\_P Q\_2)$ factors through $(Q\_1\oplus\_P Q\_2)^\times$, and hence also the composite $P\to Q\_1\to (Q\_1\oplus\_P Q\_2)$ does. Note also that $v\_1\circ u\_1 = v\_2 \circ u\_2$, so $v\_2 \circ u\_2$ will factor through $(Q\_1\oplus\_P Q\_2)^\times$ as well.


