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1. • CommentRowNumber1.
   • CommentAuthorFosco
   • CommentTimeJan 30th 2017
   • (edited Jan 30th 2017)
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   Author: Fosco
   Format: MarkdownItexWhy do you call two arrows $f,g$ with mutual lifting properties "[[orthogonality|orthogonal]]"? Is there an immediate link between $\langle v,w\rangle=0$ and $f\perp g = \{*\}$, where $*$ is the unique element that lifts $g : X \to Y$ on the right of $f : A \to B$ in $$\begin{array}{ccc} A &\to& X\\ \downarrow&\nearrow&\downarrow\\ B &\to& Y \end{array} $$ ?

   Why do you call two arrows $f,g$ with mutual lifting properties “orthogonal”?

   Is there an immediate link between $\langle v,w\rangle=0$ and $f\perp g = \{*\}$, where $*$ is the unique element that lifts $g : X \to Y$ on the right of $f : A \to B$ in

   $$\begin{array}{ccc} A &\to& X\\ \downarrow&\nearrow&\downarrow\\ B &\to& Y \end{array}$$

   ?

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 30th 2017
   • (edited Jan 30th 2017)
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   Author: Todd_Trimble
   Format: MarkdownItexThat'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 $f: A \to B$. It means $X$ "thinks" $f$ is an isomorphism in the sense that $\hom(f, X): \hom(B, X) \to \hom(A, X)$ 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 $f: A \to B$ 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 $\perp$ here is a relation between tangent vectors which are infinitesimal $f$ and 1-forms $d X$, 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 $g: X \to Y$ being orthogonal to an arrow $f$ in the slice.

   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 $f: A \to B$. It means $X$ “thinks” $f$ is an isomorphism in the sense that $\hom(f, X): \hom(B, X) \to \hom(A, X)$ 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 $f: A \to B$ 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 $\perp$ here is a relation between tangent vectors which are infinitesimal $f$ and 1-forms $d X$, 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 $g: X \to Y$ being orthogonal to an arrow $f$ in the slice.

3. • CommentRowNumber3.
   • CommentAuthorFosco
   • CommentTimeJan 31st 2017
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   Author: Fosco
   Format: MarkdownItexThat's a beautiful analogy. Mine was that lifting $f$ against $g$ is like pairing two vectors: If $v\neq 0$ is a vector, then $v\cdot w = v \cdot (w_\perp+w_{//}) = v \cdot w_{//}$ where $w_\perp \in \langle v\rangle^\perp$ 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 $f : X \to Y$ is a generic arrow in $\mathcal{C}$ then it factors as $f : X \to A \to Y$, and you can "pair it" against other arrows.

   That’s a beautiful analogy. Mine was that lifting $f$ against $g$ is like pairing two vectors:

   If $v\neq 0$ is a vector, then $v\cdot w = v \cdot (w_\perp+w_{//}) = v \cdot w_{//}$ where $w_\perp \in \langle v\rangle^\perp$ 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 $f : X \to Y$ is a generic arrow in $\mathcal{C}$ then it factors as $f : X \to A \to Y$, and you can “pair it” against other arrows.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 31st 2017
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   Author: Todd_Trimble
   Format: MarkdownItexMight be worth asking at the categories list to see if anyone knows.

   Might be worth asking at the categories list to see if anyone knows.

5. • CommentRowNumber5.
   • CommentAuthorTim_Porter
   • CommentTimeJan 31st 2017
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   Author: Tim_Porter
   Format: MarkdownItexSome 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, $(\mathcal{L},\mathcal{R})$, of classes of morphisms in a category, $\mathcal{C}$, is a factorisation system if, and only if, the three conditions * every morphism, $f : X\to Y$, admits a $(\mathcal{L},\mathcal{R})$-factorisation, $$X\xrightarrow{\ell}E\xrightarrow{r}Y;$$ * the classes are closed under isomorphism; * $\mathcal{L}\perp\mathcal{R}$. are satisfied. (In fact, $\mathcal{L}=\,\!^\perp\mathcal{R}$ and $\mathcal{R}= \mathcal{L}^\perp$.) 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.)

   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, $(\mathcal{L},\mathcal{R})$, of classes of morphisms in a category, $\mathcal{C}$, is a factorisation system if, and only if, the three conditions

   • every morphism, $f : X\to Y$, admits a $(\mathcal{L},\mathcal{R})$-factorisation, $$X\xrightarrow{\ell}E\xrightarrow{r}Y;$$

   • the classes are closed under isomorphism;

   • $\mathcal{L}\perp\mathcal{R}$.

   are satisfied. (In fact, $\mathcal{L}=\,\!^\perp\mathcal{R}$ and $\mathcal{R}= \mathcal{L}^\perp$.)

   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.)

6. • CommentRowNumber6.
   • CommentAuthorKarol Szumiło
   • CommentTimeJan 31st 2017
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   Author: Karol Szumiło
   Format: MarkdownItexI 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 $0$.

   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 $0$.