Sources, targets, and why an arrow is more than a formula
Objective: Why is writing only f(x)=x² insufficient before asking whether another arrow can compose with f?
The expression x↦x² does not yet say which inputs are allowed or where outputs live. Category theory refuses to hide that missing type information. Begin with the concrete maps, equations, kernels, quotients, or lifting problem. Only after the types are visible should the same pattern be named categorically.
Read every arrow before composing it. The exact definition used here is: A morphism f:A→B consists of an arrow name f, a source object A, and a target object B inside one declared category C. The set of such arrows, when it is a set, is Hom_C(A,B). State the source and target of every morphism, the category containing it, and the direction in which composition is written.
The theorem or deliberately bounded claim is: Two morphisms can be equal only when they have the same source and target and are equal by the equality rule of the ambient category. Its complete proof route is: 1. Place f in the typed collection Hom_C(A,B) and g in Hom_C(A′,B′). 2. If f=g as morphisms, substitution into the source and target operations gives A=A′ and B=B′. 3. After the types match, apply the category-specific equality test, such as pointwise equality for functions. Equality, isomorphism, natural isomorphism, equivalence, quasi-isomorphism, and equality only after localization must never be interchanged.
Reconstruct the exact model “Let A={0,1}, B={0,1,2}, and f:A→B satisfy f(0)=0,f(1)=1. Decide whether the same value table defines an arrow A→A and whether that arrow equals f.”. The calculation is 1. Both values 0 and 1 lie in A, so the table defines g:A→A. 2. The source of f and g is A, but their targets are B and A. 3. Because typed equality includes the target, f and g are not equal morphisms even though every displayed value agrees. The checked result is The table defines g:A→A, but g≠f as typed arrows because A≠B as targets. The nearest failure boundary is The same formula can define different morphisms after the domain or codomain changes; formulas alone do not determine typed arrows.
A commutative diagram is a typed family of equations, not decoration. A finite table, computed homology group, collapsed page, or software trace proves only the declared example unless a theorem transfers it. The diagnostic question is “Why is writing only f(x)=x² insufficient before asking whether another arrow can compose with f?”; the answer is Composition is allowed by matching f’s target to the next arrow’s source; the formula alone does not reveal either object.
Two morphisms can be equal only when they have the same source and target and are equal by the equality rule of the ambient category.
Place f in the typed collection Hom_C(A,B) and g in Hom_C(A′,B′).
If f=g as morphisms, substitution into the source and target operations gives A=A′ and B=B′.
After the types match, apply the category-specific equality test, such as pointwise equality for functions.
Let A={0,1}, B={0,1,2}, and f:A→B satisfy f(0)=0,f(1)=1. Decide whether the same value table defines an arrow A→A and whether that arrow equals f.
- Both values 0 and 1 lie in A, so the table defines g:A→A.
- The source of f and g is A, but their targets are B and A.
- Because typed equality includes the target, f and g are not equal morphisms even though every displayed value agrees.
Result: The table defines g:A→A, but g≠f as typed arrows because A≠B as targets.