Given a pushout square in the category of monoids
$$\begin{array}{ccc}A & \rightarrow & M \\ \downarrow && \downarrow \\ N & \rightarrow & P\end{array}$$such that $A \to M$ and $A \to N$ are injective, is it possible to deduce that $M \to P$ and $N \to P$ are injective, too?


The answer is yes if the $A$-actions on $M$ and $N$ are free. In fact, in that case there is even a normal form for the elements of $P$ (Bourbaki, Algebra, Chapter I, §7.3). This happens in particular if $A$ is a group. What happens in the general case? For example, what about $A=(\mathbb{N},+,0)$?



Pushouts of injective monoid homomorphisms
No. Mark Sapir and Marcel Jackson even showed it is undecidable if the factors embed in an amalgamated free product of finite monoids.


See the intro of [Jackson, Marcel. "The embeddability of ring and semigroup amalgams is undecidable." Journal of the Australian Mathematical Society 69.2 (2000): 272-286.](https://doi.org/10.1017/S1446788700002214)
for an easy example of finite semigroups for which the factors don't embed and add identities.


See also [Sapir, Mark V. "Algorithmic problems for amalgams of finite semigroups." Journal of Algebra 229.2 (2000): 514-531.](https://doi.org/10.1006/jabr.1999.8138).


