3 ms·
No, the identity id:A->A is characterized by the property that for all arrows f:B->A and g:A->C (P1) id . f = f and (P2) g . id = g, in particular it is uniq
by orbifold 12y ago
No, the identity id:A->A is characterized by the property that for all arrows
f:B->A and g:A->C
(P1) id . f = f and (P2) g . id = g, in particular it is unique, since if id':A->A had the same property, then
id' = id . id' = id,
where the first equality follows by (P1) and the second by (P2').
- tfgg 12y agoOk, I think I got the wrong idea that the identity was any arrow pointing back on to an object. Thanks.