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Why do you call two arrows f,g with mutual lifting properties “orthogonal”?
Is there an immediate link between ⟨v,w⟩=0 and f⊥g={*}, where * is the unique element that lifts g:X→Y on the right of f:A→B in
A→X↓↗↓B→Y?
That’s a good question. I don’t know an “official” answer, but here’s a thought: consider what we mean when we say an object X is orthogonal to an arrow f:A→B. It means X “thinks” f is an isomorphism in the sense that hom(f,X):hom(B,X)→hom(A,X) is an isomorphism (i.e., as far as X-probes or coprobes are concerned, f is an iso). If we think of an isomorphism as a notion of sameness, then f makes A and B look just the same as gauged by the instrument X.
Now in elementary physics, say if we are measuring temperature by a thermometer X, there are certain trajectories f:A→B between points A,B in space which exhibit sameness as gauged by X, namely by traveling along isothermal lines. These are orthogonal to the “direction” of X given by gradients (really the ⊥ here is a relation between tangent vectors which are infinitesimal f and 1-forms dX, but hopefully you get the idea).
When you transport all this to the slice category over Y, you get the standard notion of an object g:X→Y being orthogonal to an arrow f in the slice.
That’s a beautiful analogy. Mine was that lifting f against g is like pairing two vectors:
If v≠0 is a vector, then v⋅w=v⋅(w⊥+w//)=v⋅w// where w⊥∈⟨v⟩⊥ and w// is the projection of w along v. So you factor every vector as something in the direction of v and something “orthogonal”.
If f:X→Y is a generic arrow in 𝒞 then it factors as f:X→A→Y, and you can “pair it” against other arrows.
Might be worth asking at the categories list to see if anyone knows.
Some time ago when I was working on a section of some monograph i asked myself exactly that question. The well known key result is:
A pair, (ℒ,ℛ), of classes of morphisms in a category, 𝒞, is a factorisation system if, and only if, the three conditions
every morphism, f:X→Y, admits a (ℒ,ℛ)-factorisation, $Xℓ→Er→Y;$
the classes are closed under isomorphism;
ℒ⊥ℛ.
are satisfied. (In fact, ℒ=⊥ℛ and ℛ=ℒ⊥.)
Looking at the last statement is reminiscent of the ‘orthogonal complement’ relationship between subspaces of an inner product space, and I think that that was how it was explained in some source that I was consulting. (I may even have found this on an nLab page or on Joyal’s pages.)
I don’t know the history, but I guess I always interpreted it as Tim suggests. The analogy with orthogonal complements in a inner product space can be seen in some other ways too. For example, the factorization axiom is reminiscent of the fact the a subspace and its orthogonal complement together generate the ambient inner product space. Also, two classes of a factorization system intersect in isomorphisms, similarly to orthogonal complements in an inner product space intersecting at 0.
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