Lifting property

Remarks

The data $u, v$ form a commutative square, and $d$ is a diagonal filler of the square. A common notation is $i \pitchfork p$, or $i \perp p$ for orthogonality. Fix a class $M$ of morphisms. The morphisms with the left lifting property with respect to all members of $M$ contain all isomorphisms and are closed under composition.

Example: in the category of sets, let $i \colon \emptyset \to \set{*}$ and let $p \colon X \to Y$ be surjective. A square is then just a point $y = v(*)$ of $Y$, and a filler is a point $x$ of $X$ with $p(x) = y$. It exists since $p$ is surjective.

Non-example: with the same $i$, let $p \colon \emptyset \to \set{*}$. Take $u$ the empty map and $v = \id$; a filler would be a map $\set{*} \to \emptyset$, which does not exist.

In topology, a Hurewicz fibration is a map with the right lifting property with respect to all inclusions $Z \times \set{0} \to Z \times [0, 1]$; this is the homotopy lifting property.