You are here
Homeorthogonal morphisms
Primary tabs
orthogonal morphisms
A morphism $f:A\to B$ in a category $\mathcal{C}$ is said to be orthogonal to a morphism $g:C\to D$ in $\mathcal{C}$, written
$f\perp g$ 
if whenever we have a commutative diagram
$\xymatrix@+=3pc{A\ar[r]^{f}\ar[d]&B\ar[d]\\ C\ar[r]_{g}&D}$ 
there is a unique morphism $h:B\to C$ such that the diagram
$\xymatrix@+=3pc{A\ar[r]^{f}\ar[d]&B\ar[d]\ar@{.>}[dl]h\\ C\ar[r]_{g}&D}$ 
is commutative also. If $f\perp g$, we sometimes call the ordered pair $(f,g)$ a diagonally polar pair.
For example, in Set, the category of sets, any surjective function is orthogonal to an injective function. To see this, suppose $f:A\to B$ is surjective and $g:C\to D$ injective, with $y\circ f=g\circ x$, where $x:A\to C$ and $y:B\to D$ are functions. For any $b\in B$, there is some $a\in A$ such that $f(a)=b$ since $f$ is surjective. Define $h:B\to C$ by $h(b)=x(a)$. Now, if there is $c\in A$ such that $b=f(a)=f(c)$, then $g(x(a))=y(f(a))=y(b)=y(f(c))=g(x(c))$. Since $g$ is injective, $x(a)=x(c)$. This shows that $h$ is a welldefined function. It is clear that $h\circ f=x$ and $g\circ h=y$. Now, if $e:B\to C$ is another such a function, then $g(e(b))=y(b)=g(h(b))$, so that $e(b)=h(b)$ since $g$ is injective. This shows that $h$ is uniquely defined.
Here are some basic properties of the orthogonality relation on morphisms:

If either $f$ or $g$ is an isomorphism, then $f\perp g$.

If $f\perp f$, then $f$ is an isomorphism.

If $f\perp g$ and $f\perp h$, then $f\perp(h\circ g)$. Similarly, $g\perp f$ and $h\perp f$ imply $(h\circ g)\perp f$. Of course, both statements make sense provided that $h\circ g$ exists.
More generally, if $\mathcal{F}$ and $\mathcal{G}$ are two classes of morphisms in a category $\mathcal{C}$, we say that $\mathcal{F}$ is orthogonal to $\mathcal{G}$, or that $(\mathcal{F},\mathcal{G})$ is a diagonally polar pair, written $\mathcal{F}\perp\mathcal{G}$, if $f\perp g$ for every $f$ in $\mathcal{F}$ and every $g$ in $\mathcal{G}$.
For every class $\mathcal{X}$ of morphism, the largest class of morphisms in $\mathcal{C}$ such that $\mathcal{X}$ is orthogonal to is denoted by $\mathcal{X}_{*}$, and the largest class of morphisms that is orthogonal to $\mathcal{X}$ is denoted by $\mathcal{X}^{*}$.
Based on the properties of $\perp$ above, below are some properties of ${}^{*}$ and ${}_{*}$:

$\mathcal{X}\subseteq\mathcal{Y}_{*}$ iff $\mathcal{Y}\subseteq\mathcal{X}^{*}$. Equivalently, if $\mathscr{M}$ is the class of all subclasses of morphisms of $\mathcal{C}$, then $(^{*},_{*})$ is a Galois connection between $(\mathscr{M},\subseteq)$ and $(\mathscr{M},\supseteq)$.

A morphism is in both $\mathcal{X}^{*}$ and $\mathcal{X}_{*}$ iff it is an isomorphism.

Both $\mathcal{X}^{*}$ and $\mathcal{X}_{*}$ are closed under $\circ$.

Given that $m=m_{1}\circ m_{2}$ exists in $\mathcal{C}$ and $m_{2}\in\mathcal{X}_{*}$, then $m\in\mathcal{X}_{*}$ iff $m_{1}\in\mathcal{X}_{*}$.

If $f\in\mathcal{X}_{*}$, then the pullback of $f$ along any morphism is again in $\mathcal{X}_{*}$.
References
 1 F. Borceux Basic Category Theory, Handbook of Categorical Algebra I, Cambridge University Press, Cambridge (1994)
Mathematics Subject Classification
18A32 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections