The lift of a morphism $f: Y\to B$ along an epimorphism (or more general map) $p:X\to B$ is a morphism $\tilde{f}: Y\to X$ such that $f = p\circ\tilde{f}$.

The dual problem is the extension of a morphism $f: A\to Y$ along a monomorphism $i: A\hookrightarrow X$, which is a morphism $\tilde{f}:X\to Y$ such that $\tilde{f}\circ i = f$. One sometimes, extends along more general morphisms than monomorphisms.

Lifting properties

Let $K$ be a category and write $arr(K)$ for the arrow category of $K$: the category with arrows (= morphisms) $a \stackrel{f}{\to} b$ of $K$ as objects and commutative squares $g u=v f$

$\array{
a &\stackrel{u}{\to}& c
\\
\downarrow^{\mathrlap{f}}
&&
\downarrow^{\mathrlap{g}}
\\
b &\stackrel{v}{\to}& d
}$

as morphisms $(u,v) : f \rightarrow g$. We may also refer to a commutative square $g u=v f$ as a lifting problem between $f$ and $g$.

We say a morphism $f$ has the left lifting property with respect to a morphism $g$ or equivalently that $g$ has the right lifting property with respect to $f$, if for every commutative square $(u,v) :f \rightarrow g$ as above, there is an arrow $\gamma$

$\array{
a &\stackrel{u}{\to}& c
\\
\downarrow^f
&{}^{\exists \gamma}\nearrow&
\downarrow^g
\\
b &\stackrel{v}{\to}& d
}$

from the codomain $b$ of $f$ to the domain $c$ of $g$ such that both triangles commute. We call such an arrow $\gamma$ a lift or a solution to the lifting problem $(u,v)$.

(If this lift is unique, we say that $f$ is orthogonal$f \perp g$ to $g$.)