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.
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 $R \simeq 0$.)