amalgamated sum of monoids

Question

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.

Answer 1

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.