Criteria for inclusions in co-product to be monos In category of sets and functions, we observe that inclusions into the sum object (i.e. the disjoint union) are monos. I wanted to see what causes such a behaviour. Following is one criteria that I see as one of the requirements for such a behaviour to exist.
We define a binary relation $\leq$: Let $A, B$ be objects in some undetermined category $C$. $A\leq B$ if and only if $\exists i:A\hookrightarrow B$. (Please let me know if there is more appropriate symbol for this relation). 
Criterion: Let $C$ be any category. If  for all objects $A, B \in C$, ($A\leq B \ \vee B\leq A$), then inclusion maps for any sum object, if it exists, are monos.
Proof follows from the fact that for any maps $f, g, h$ such that $f=h\circ g$, if $f$ is mono, then $g$ is mono as well. More specifically, one can have 1 mono from $\leq$, say $i:A\hookrightarrow B$, and the other one will be the idenity on B, then since these maps must factor through (Please correct me if I am using non-standard or even wrong terminology) the inclusion maps, one can use the above fact to individually show that both inclusions must be monos.
My questions:


*

*Is there any other criterion which is more fundamental i.e. from which my criterion follows, assuming it is a correct criterion? 

*Are there any examples, in which inclusion maps for sum objects are not monos in the category? If yes, could you please describe it for me?


Thank you.
 A: Answer for (2): In the opposite category to the category of sets, $\mathrm{Sets}^{op}$, not all maps $A \to A \amalg B$ are monomorphisms.  For example, the canonical map $X^{op} \to X^{op} \amalg \emptyset^{op} \simeq (X \times \emptyset)^{op}$ corresponds to $\pi_1^{op}$ for the projection $\pi_1 : X \times \emptyset \to X$.  Since $\pi_1$ is not an epimorphism in the case where $X \ne \emptyset$, it follows that $\pi_1^{op}$ is not a monomorphism.
Another example for (2): In the category of commutative rings, the coproduct is the tensor product over $\mathbb{Z}$: $X \amalg Y \simeq X \otimes_{\mathbb{Z}} Y$.  However, for example $(\mathbb{Z} / 2 \mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Z} / 3 \mathbb{Z}) \simeq 0$, so neither of the maps $\mathbb{Z} / 2 \mathbb{Z} \to (\mathbb{Z} / 2 \mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Z} / 3 \mathbb{Z})$ or $\mathbb{Z} / 3 \mathbb{Z} \to (\mathbb{Z} / 2 \mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Z} / 3 \mathbb{Z})$ is a monomorphism.  (If you are not familiar with tensor products, here's an elementary way of seeing $(\mathbb{Z} / 2 \mathbb{Z}) \amalg (\mathbb{Z} / 3 \mathbb{Z}) \simeq 0$: suppose you have any ring $R$ with homomorphisms $\mathbb{Z} / 2 \mathbb{Z} \to R$ and $\mathbb{Z} / 3 \mathbb{Z} \to R$.  Then the existence of the first map implies $1_R + 1_R = 0_R$, whereas the existence of the second map implies $1_R + 1_R + 1_R = 0_R$.  Therefore, combining the two, $1_R = 0_R$ which implies $R \simeq 0$.)
