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This question is probably stupid and definitely bureaucratic, but

Is writing $f\circ g$ for the composition of morphisms in the 'many hom-classes' definition of a category unambiguous?

The many hom-classes definition of a category (as given e.g. on the nlab) says that for each pair of arrows $(f,g)\in{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)$ we 'have an arrow' $f\circ g\in{\bf Hom}_\mathcal{C}(X,Z)$, but if the hom-classes may not be disjoint how do we know that the arrows we 'have' from identical composable arrow pairings in differing hom-class pairs match up?

Rephrased using the language of composition functions, the above definition is the same as a collection of functions $$\{\circ_{XYZ}:{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\to{\bf Hom}_\mathcal{C}(X,Z)\}_{X,Y,Z\in{\bf Ob}_\mathcal{C}}.$$ The axioms then specify associativity and unitarity, but how do we know that we don't have objects $X,Y,Z,X',Y',Z'\in{\bf Ob}_\mathcal{C}$ and arrows $$f\in{\bf Hom}_\mathcal{C}(Y,Z)\cap{\bf Hom}_\mathcal{C}(Y',Z'),$$ $$g\in{\bf Hom}_\mathcal{C}(X,Y)\cap{\bf Hom}_\mathcal{C}(X',Y'),$$ so $$(f,g)\in\big({\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\big)\cap\big({\bf Hom}_\mathcal{C}(Y',Z')\times{\bf Hom}_\mathcal{C}(X',Y')\big),$$ such that $$\circ_{XYZ}(f,g)\neq\circ_{X'Y'Z'}(f,g)?$$ Note that these composites live in ${\bf Hom}_\mathcal{C}(X,Z)$ and ${\bf Hom}_\mathcal{C}(X',Z')$, respectively, and as David Roberts points out in the comments these classes can be disjoint even if the original hom-classes aren't.

If it is possible to have objects and arrows as above the notation $f\circ g$ is obviously ambiguous -- also obvious is that we could fix this with an additional axiom (scheme?) if it were a problem.

Is this situation already precluded by the other axioms/data present in the many hom-classes definition of a category?

For an example where we have identical arrow pairings in differing hom-classes but composition still matches up, consider any two composable relations $R,S\in{\bf Ob_{Rel}}$. By definition $R$ and $S$ are subsets of some Cartesian squares $Y\times Z$ and $X\times Y$ (respectively), but we can take $X',Y',Z'$ to be any strict superclasses of $X,Y,Z$ (resp.) and observe that $R$ and $S$ are also subsets of $Y'\times Z'$ and $X'\times Y'$ (resp.), so $$R\in{\bf Hom_{Rel}}(Y,Z)\cap{\bf Hom_{Rel}}(Y',Z'),$$ $$S\in{\bf Hom_{Rel}}(X,Y)\cap{\bf Hom_{Rel}}(X',Y').$$ Here we obviously still have that the composition functions coincide, but this is arguably due to the fact that ${\bf Rel}$ is most naturally presented as a one hom-class category. Again, I understand that this question is very pedantic at best -- the patience involved in any clarification is greatly appreciated.

$\newcommand\bHom{\mathbf{Hom}}\newcommand\bOb{\mathbf{Ob}}\newcommand\bRel{\mathbf{Rel}}$This question is probably stupid and definitely bureaucratic, but

Is writing $f\circ g$ for the composition of morphisms in the ‘many hom-classes’ definition of a category unambiguous?

The ‘many hom-classes’ definition of a category (as given e.g. on the nlab) says that for each pair of arrows $(f,g)\in\bHom_\mathcal{C}(Y,Z)\times\bHom_\mathcal{C}(X,Y)$ we ‘have an arrow’ $f\circ g\in\bHom_\mathcal{C}(X,Z)$, but if the hom-classes may not be disjoint how do we know that the arrows we ‘have’ from identical composable arrow pairings in differing hom-class pairs match up?

Rephrased using the language of composition functions, the above definition is the same as a collection of functions $$\{\circ_{XYZ}:\bHom_\mathcal{C}(Y,Z)\times\bHom_\mathcal{C}(X,Y)\to\bHom_\mathcal{C}(X,Z)\}_{X,Y,Z\in\bOb_\mathcal{C}}.$$ The axioms then specify associativity and unitarity, but how do we know that we don't have objects $X,Y,Z,X',Y',Z'\in\bOb_\mathcal{C}$ and arrows $$f\in\bHom_\mathcal{C}(Y,Z)\cap\bHom_\mathcal{C}(Y',Z'), \\ g\in\bHom_\mathcal{C}(X,Y)\cap\bHom_\mathcal{C}(X',Y'),$$ so $$(f,g)\in\bigl(\bHom_\mathcal{C}(Y,Z)\times\bHom_\mathcal{C}(X,Y)\bigr)\cap\bigl(\bHom_\mathcal{C}(Y',Z')\times\bHom_\mathcal{C}(X',Y')\bigr),$$ such that $$\circ_{XYZ}(f,g)\neq\circ_{X'Y'Z'}(f,g)?$$ Note that these composites live in $\bHom_\mathcal{C}(X,Z)$ and $\bHom_\mathcal{C}(X',Z')$, respectively, and as David Roberts points out in the comments these classes can be disjoint even if the original hom-classes aren't.

If it is possible to have objects and arrows as above the notation $f\circ g$ is obviously ambiguous — also obvious is that we could fix this with an additional axiom (scheme?) if it were a problem.

Is this situation already precluded by the other axioms/data present in the many hom-classes definition of a category?

For an example where we have identical arrow pairings in differing hom-classes but composition still matches up, consider any two composable relations $R,S\in\bOb_{\bRel}$. By definition $R$ and $S$ are subsets of some Cartesian squares $Y\times Z$ and $X\times Y$ (respectively), but we can take $X'$, $Y'$, $Z'$ to be any strict superclasses of $X$, $Y$, $Z$ (respectively) and observe that $R$ and $S$ are also subsets of $Y'\times Z'$ and $X'\times Y'$ (respectively), so $$R\in\bHom_{\bRel}(Y,Z)\cap\bHom_{\bRel}(Y',Z'), \\ S\in\bHom_{\bRel}(X,Y)\cap\bHom_{\bRel}(X',Y').$$ Here we obviously still have that the composition functions coincide, but this is arguably due to the fact that $\bRel$ is most naturally presented as a one hom-class category. Again, I understand that this question is very pedantic at best — the patience involved in any clarification is greatly appreciated.

This question is probably stupid and definitely bureaucratic, but

Is writing $f\circ g$ for the composition of morphisms in the 'many hom-classes' definition of a category unambiguous?

The many hom-classes definition of a category (as given e.g. on the nlab) says that for each pair of arrows $(f,g)\in{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)$ we 'have an arrow' $f\circ g\in{\bf Hom}_\mathcal{C}(X,Z)$, but if the hom-classes may not be disjoint how do we know that the arrows we 'have' from identical composable arrow pairings in differing hom-class pairs match up?

Rephrased using the language of composition functions, the above definition is the same as a collection of functions $$\{\circ_{XYZ}:{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\to{\bf Hom}_\mathcal{C}(X,Z)\}_{X,Y,Z\in{\bf Ob}_\mathcal{C}}.$$ The axioms then specify associativity and unitarity, but how do we know that we don't have objects $X,Y,Z,X',Y',Z'\in{\bf Ob}_\mathcal{C}$ and arrows $$f\in{\bf Hom}_\mathcal{C}(Y,Z)\cap{\bf Hom}_\mathcal{C}(Y',Z'),$$ $$g\in{\bf Hom}_\mathcal{C}(X,Y)\cap{\bf Hom}_\mathcal{C}(X',Y'),$$ so $$(f,g)\in\big({\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\big)\cap\big({\bf Hom}_\mathcal{C}(Y',Z')\times{\bf Hom}_\mathcal{C}(X',Y')\big),$$ such that $$\circ_{XYZ}(f,g)\neq\circ_{X'Y'Z'}(f,g)?$$ If we do the notation $f\circ g$ is obviously ambiguous -- also obvious is that we could fix this with an additional axiom (scheme?) if it were a problem.

Is this situation already precluded by the other axioms/data present in the many hom-classes definition of a category?

For an example where we have identical arrow pairings in differing hom-classes but composition still matches up, consider any two composable relations $R,S\in{\bf Ob_{Rel}}$. By definition $R$ and $S$ are subsets of some Cartesian squares $Y\times Z$ and $X\times Y$ (respectively), but we can take $X',Y',Z'$ to be any strict superclasses of $X,Y,Z$ (resp.) and observe that $R$ and $S$ are also subsets of $Y'\times Z'$ and $X'\times Y'$ (resp.), so $$R\in{\bf Hom_{Rel}}(Y,Z)\cap{\bf Hom_{Rel}}(Y',Z'),$$ $$S\in{\bf Hom_{Rel}}(X,Y)\cap{\bf Hom_{Rel}}(X',Y').$$ Here we obviously still have that the composition functions coincide, but this is arguably due to the fact that ${\bf Rel}$ is most naturally presented as a one hom-class category. Again, I understand that this question is very pedantic at best -- the patience involved in any clarification is greatly appreciated.

This question is probably stupid and definitely bureaucratic, but

Is writing $f\circ g$ for the composition of morphisms in the 'many hom-classes' definition of a category unambiguous?

The many hom-classes definition of a category (as given e.g. on the nlab) says that for each pair of arrows $(f,g)\in{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)$ we 'have an arrow' $f\circ g\in{\bf Hom}_\mathcal{C}(X,Z)$, but if the hom-classes may not be disjoint how do we know that the arrows we 'have' from identical composable arrow pairings in differing hom-class pairs match up?

Rephrased using the language of composition functions, the above definition is the same as a collection of functions $$\{\circ_{XYZ}:{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\to{\bf Hom}_\mathcal{C}(X,Z)\}_{X,Y,Z\in{\bf Ob}_\mathcal{C}}.$$ The axioms then specify associativity and unitarity, but how do we know that we don't have objects $X,Y,Z,X',Y',Z'\in{\bf Ob}_\mathcal{C}$ and arrows $$f\in{\bf Hom}_\mathcal{C}(Y,Z)\cap{\bf Hom}_\mathcal{C}(Y',Z'),$$ $$g\in{\bf Hom}_\mathcal{C}(X,Y)\cap{\bf Hom}_\mathcal{C}(X',Y'),$$ so $$(f,g)\in\big({\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\big)\cap\big({\bf Hom}_\mathcal{C}(Y',Z')\times{\bf Hom}_\mathcal{C}(X',Y')\big),$$ such that $$\circ_{XYZ}(f,g)\neq\circ_{X'Y'Z'}(f,g)?$$ Note that these composites live in ${\bf Hom}_\mathcal{C}(X,Z)$ and ${\bf Hom}_\mathcal{C}(X',Z')$, respectively, and as David Roberts points out in the comments these classes can be disjoint even if the original hom-classes aren't.

If it is possible to have objects and arrows as above the notation $f\circ g$ is obviously ambiguous -- also obvious is that we could fix this with an additional axiom (scheme?) if it were a problem.

Is this situation already precluded by the other axioms/data present in the many hom-classes definition of a category?

For an example where we have identical arrow pairings in differing hom-classes but composition still matches up, consider any two composable relations $R,S\in{\bf Ob_{Rel}}$. By definition $R$ and $S$ are subsets of some Cartesian squares $Y\times Z$ and $X\times Y$ (respectively), but we can take $X',Y',Z'$ to be any strict superclasses of $X,Y,Z$ (resp.) and observe that $R$ and $S$ are also subsets of $Y'\times Z'$ and $X'\times Y'$ (resp.), so $$R\in{\bf Hom_{Rel}}(Y,Z)\cap{\bf Hom_{Rel}}(Y',Z'),$$ $$S\in{\bf Hom_{Rel}}(X,Y)\cap{\bf Hom_{Rel}}(X',Y').$$ Here we obviously still have that the composition functions coincide, but this is arguably due to the fact that ${\bf Rel}$ is most naturally presented as a one hom-class category. Again, I understand that this question is very pedantic at best -- the patience involved in any clarification is greatly appreciated.

This question is probably stupid and definitely bureaucratic, but

Is writing $f\circ g$ for the composition of morphisms in the 'many hom-classes' definition of a category unambiguous?

The many hom-classes definition of a category (as given e.g. on the nlab) says that for each pair of arrows $(f,g)\in{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)$ we 'have an arrow' $f\circ g\in{\bf Hom}_\mathcal{C}(X,Z)$, but if the hom-classes may not be disjoint how do we know that the arrows we 'have' from identical composable arrow pairings in differing hom-class pairs match up?

Rephrased using the language of composition functions, the above definition is the same as a collection of functions $$\{\circ_{XYZ}:{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\to{\bf Hom}_\mathcal{C}(X,Z)\}_{X,Y,Z\in{\bf Ob}_\mathcal{C}}.$$ The axioms then specify associativity and unitarity, but how do we know what we don't have objects $X,Y,Z,X',Y',Z'\in{\bf Ob}_\mathcal{C}$ and arrows $$f\in{\bf Hom}_\mathcal{C}(Y,Z)\cap{\bf Hom}_\mathcal{C}(Y',Z'),$$ $$g\in{\bf Hom}_\mathcal{C}(X,Y)\cap{\bf Hom}_\mathcal{C}(X',Y'),$$ so $$(f,g)\in\big({\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\big)\cap\big({\bf Hom}_\mathcal{C}(Y',Z')\times{\bf Hom}_\mathcal{C}(X',Y')\big),$$ such that $$\circ_{XYZ}(f,g)\neq\circ_{X'Y'Z'}(f,g)?$$ If we do the notation $f\circ g$ is obviously ambiguous -- also obvious is that we could fix this with an additional axiom (scheme?) if it were a problem.

Is this situation already precluded by the other axioms/data present in the many hom-classes definition of a category?

For an example where we have identical arrow pairings in differing hom-classes but composition still matches up, consider any two composable relations $R,S\in{\bf Ob_{Rel}}$. By definition $R$ and $S$ are subsets of some Cartesian squares $Y\times Z$ and $X\times Y$ (respectively), but we can take $X',Y',Z'$ to be any strict superclasses of $X,Y,Z$ (resp.) and observe that $R$ and $S$ are also subsets of $Y'\times Z'$ and $X'\times Y'$ (resp.), so $$R\in{\bf Hom_{Rel}}(Y,Z)\cap{\bf Hom_{Rel}}(Y',Z'),$$ $$S\in{\bf Hom_{Rel}}(X,Y)\cap{\bf Hom_{Rel}}(X',Y').$$ Here we obviously still have that the composition functions coincide, but this is arguably due to the fact that ${\bf Rel}$ is most naturally presented as a one hom-class category.

  Again, I understand that this question is very pedantic at best -- the patience involved in any clarification is greatly appreciated.

This question is probably stupid and definitely bureaucratic, but

Is writing $f\circ g$ for the composition of morphisms in the 'many hom-classes' definition of a category unambiguous?

The many hom-classes definition of a category (as given e.g. on the nlab) says that for each pair of arrows $(f,g)\in{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)$ we 'have an arrow' $f\circ g\in{\bf Hom}_\mathcal{C}(X,Z)$, but if the hom-classes may not be disjoint how do we know that the arrows we 'have' from identical composable arrow pairings in differing hom-class pairs match up?

Rephrased using the language of composition functions, the above definition is the same as a collection of functions $$\{\circ_{XYZ}:{\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\to{\bf Hom}_\mathcal{C}(X,Z)\}_{X,Y,Z\in{\bf Ob}_\mathcal{C}}.$$ The axioms then specify associativity and unitarity, but how do we know that we don't have objects $X,Y,Z,X',Y',Z'\in{\bf Ob}_\mathcal{C}$ and arrows $$f\in{\bf Hom}_\mathcal{C}(Y,Z)\cap{\bf Hom}_\mathcal{C}(Y',Z'),$$ $$g\in{\bf Hom}_\mathcal{C}(X,Y)\cap{\bf Hom}_\mathcal{C}(X',Y'),$$ so $$(f,g)\in\big({\bf Hom}_\mathcal{C}(Y,Z)\times{\bf Hom}_\mathcal{C}(X,Y)\big)\cap\big({\bf Hom}_\mathcal{C}(Y',Z')\times{\bf Hom}_\mathcal{C}(X',Y')\big),$$ such that $$\circ_{XYZ}(f,g)\neq\circ_{X'Y'Z'}(f,g)?$$ If we do the notation $f\circ g$ is obviously ambiguous -- also obvious is that we could fix this with an additional axiom (scheme?) if it were a problem.

Is this situation already precluded by the other axioms/data present in the many hom-classes definition of a category?

For an example where we have identical arrow pairings in differing hom-classes but composition still matches up, consider any two composable relations $R,S\in{\bf Ob_{Rel}}$. By definition $R$ and $S$ are subsets of some Cartesian squares $Y\times Z$ and $X\times Y$ (respectively), but we can take $X',Y',Z'$ to be any strict superclasses of $X,Y,Z$ (resp.) and observe that $R$ and $S$ are also subsets of $Y'\times Z'$ and $X'\times Y'$ (resp.), so $$R\in{\bf Hom_{Rel}}(Y,Z)\cap{\bf Hom_{Rel}}(Y',Z'),$$ $$S\in{\bf Hom_{Rel}}(X,Y)\cap{\bf Hom_{Rel}}(X',Y').$$ Here we obviously still have that the composition functions coincide, but this is arguably due to the fact that ${\bf Rel}$ is most naturally presented as a one hom-class category. Again, I understand that this question is very pedantic at best -- the patience involved in any clarification is greatly appreciated.