Definition of category: composition of incompatible morphisms

Question

The definition of a category specifies that if two morphisms are compatible (i.e. output of one morphism is the input of the other) then we can compose them. In other words, $ f \in C(y, z), g\in C(x, y) \implies \exists h \in C(x, z) : f\circ g = h$

However, there seem to be no requirement to have the equality only for the compatible morphisms. In other words, this is not true: $f\circ g = h \implies source(f) = target(g)$.

1. Is my understanding correct?
2. If so, we could have more structure on top of the category definition. For example, we could extend the composition to functions that satisfy some weaker form of equality between input and output. Are there such objects?
3. If we require the composition defined on compatible morphisms and only on those, this will give a different "strict" definition of category. Is there a name for such a thing?

Answer 1

I don't believe this changes the definition of category. That said, there is not only one way to define a category.

$\newcommand{\C}{\mathsf{C}}$In the set-theoretic definitions of category, composition is a function $\circ_{x,y,z}:\C(y,z)\times\C(x,y)\to\C(x,z)$ and we assert some axioms of these functions. $f\circ g=h$ is technically an abuse of notation. What we really mean is that $\circ_{x,y,z}(f,g)=h$ for some $x,y,z$. Oh, but that is only defined if $f\in\C(y,z)$ and $g\in\C(x,y)$, so the morphisms have to be compatible; composition wouldn't exist otherwise. $\circ$ is a meaningless symbol on its own, we must supply the context of a (large) family of given functions. Similarly, $f\circ g$ is a meaningless symbol without the morphisms being compatible.

^ N.B. if you allow your hom-sets to be non-disjoint there is a small wrinkle in that the same object $f$ of your universe could lie in some intersection $\C(a,b)\cap\C(y,z)$ for $(a,b)\neq(y,z)$. However, for $f\circ g$ to be well-defined we still must have $f\in\C(y,z)$ and $g\in\C(x,y)$ (for some $x,y,z$)... while we can have situations where, depending on how you "look at it" (we have lost a well-defined global concept of "source" and "target") $f,g$ are both compatible and incompatible but they must at least be compatible. They cannot be exclusively incompatible and still have $f\circ g$ make sense. This situation is not completely stupid, since if we want ordinary categories to be the same thing as $\mathsf{Set}$-enriched categories - which we might do - then we must allow for it to be possible that the hom-sets are nondisjoint. Some authors will demand disjoint hom-sets anyway especially in texts that don't cover enrichment. Many authors don't care at all, since it's implicit we should work only with what the axioms of a category give us and not bother with specific implementations of these axioms.

Answer 2

If the source of $f$ does not equal the target of $g$, then $f\circ g$ is not even defined. Therefore, it doesn't make sense to assert that for all morphisms $f$ and $g$, we have$$f\circ g = h\implies\operatorname{source}(f)=\operatorname{target}(g).$$

Since the definition of a category does not mention $f\circ g$ when the source of $f$ is unequal to the target of $g$, it is implicit that $f\circ g$ is not defined whenever this situation occurs. This is because definitions are supposed to be exhaustive. An analogy: when I write "an integer $n$ is even if there is an integer $k$ such that $n=2k$", then it is implict that the only objects which can be considered even are integers.