The term 'canonical' is very general and any attempt to create a "precise notion" may be questioned and can never be proved as such. However, there is a canonical candidate, which at least concerns canonical morphisms for constructs and other structured sets, which has a clear logical character.

First, every mathematical structure on a set is determined by relations between sets. For group structures there is a primary relation $r$ that is a function $G\times G \overset {r}\longrightarrow G$, $(x,y)\underline{r} z \Leftrightarrow z=x\centerdot y$ and some secondary relations which are conditions on $r$ (associativity, unit element and inverses). For a topological space the primary relation could be $2^X \overset{r}\longrightarrow X$, where $2^X$ is the set of all subsets of $X$ and $M\underline{r}x$ is the relation $x\in \bar{M}$ (the closure of $M$).

Whenever the main relation of a mathematical structure is given on the form $F(X)\rightarrow X$, for a functor $F$ in the category **Rel** (where sets are objects and binary relations are morphisms) which maps morphisms $X\overset {f}\longrightarrow Y$ on $F(X)\overset {F(f)}\longrightarrow F(Y)$, it is possible to define morphisms between the structures as diagrams:
$\require{AMScd}$
\begin{CD}
F(X) @>F(f)>> F(Y)\\
@VrV V @VVsV\\
X @>>f> Y
\end{CD}
such that

(1) $\rho\underline{F(f)}\sigma \Rightarrow (\rho\underline{r}x\Rightarrow\sigma\underline{s}f(x))$.

This condition on $f$ gives the canonical morphisms to every construct that provide canonical morphisms.

Example: If $F$ is the (contravariant) functor defined as $2^X\overset{2^f}\longrightarrow 2^Y$, where $M\underline{2^f}M'\Leftrightarrow M=f^{-1}(M')$ and $r,s$ are defined as
above, then due to (1):

$M=f^{-1}(M')\Rightarrow (x\in \bar{M}\Rightarrow f(x)\in \bar{M}')$, so $x\in \overline{f^{-1}(M')}\Rightarrow f(x)\in \bar{M}'$. (Continuity).

morecanonical than "pick an arbitrary element of X". – Tom Leinster Jun 5 '12 at 13:42