Why is there no functor $\mathsf{Group}\to\mathsf{AbGroup}$ sending groups to their centers? The category $\mathbf{Set}$ contains as its objects all small sets and arrows all functions between them.  A set is "small" if it belongs to a larger set $U$, the universe.  
Let $\mathbf{Grp}$ be the category of small groups and morphisms between them, and $\mathbf{Abs}$ be the category of small abelian groups and its morphisms.  
I don't see what it means to say there is no functor $f: \mathbf{Grp} \to \mathbf{Abs}$ that sends each group to its center, when $U$ isn't even specified.  Can anybody explain?
 A: This is very similar to Arturo Magidin's answer, but offers another point of view.
Consider the dihedral group $D_n=\mathbb Z_n\rtimes \mathbb Z_2$ with $2\nmid n$ (so the $Z(D_n)=1$). From the splitting lemma we get a short exact sequence
$$1\to\mathbb Z_n\rightarrow D_n\xrightarrow{\pi} \mathbb Z_2\to 1$$
and an arrow $\iota\colon \mathbb Z_2\to D_n$ such that $\pi\circ \iota=1_{\mathbb Z_2}$.
Hence the composite morphism
$$\mathbb Z_2\xrightarrow{\iota} D_n\xrightarrow{\pi}\mathbb Z_2$$
is an iso and would be mapped by the centre to an iso
$$\mathbb Z_2\to 1\to \mathbb Z_2$$
what is impossible. (One can also recognize a split mono and split epi above and analyze how they behave under an arbitrary functor).
Therefore the centre can't be functorial.
A: The problem with such a functor is group theoretical, not categorical. The problem arises because morphisms between groups need not map centers to centers. It doesn't have anything to do with universes, smallness, or foundational issues.
Consider for example $G=C_2$, $H=S_3$, $K=C_2$, and the maps $f\colon G\to H$ sending the nontrivial element of $G$ to $(1,2)$, and $g\colon H\to K$ by viewing $S_3/A_3$ as the cyclic group of order $2$.
Since $Z(G) = Z(K) = C_2$, and $Z(H) = \{1\}$, such a putative functor $\mathcal{F}$ would give that  $\mathcal{F}(f)\colon C_2\to\{1\}$ is the zero map $\mathbf{z}$, and $\mathcal{F}(g)\colon \{1\}\to C_2$ is the inclusion of the trivial group into $C_2$. But $g\circ f=\mathrm{id}_{C_2}$, so 
$$\mathrm{id}_{C_2} = \mathcal{F}(\mathrm{id}_{C_2}) = \mathcal{F}(gf) = \mathcal{F}(g)\mathcal{F}(f) = \mathbf{z}$$
where $\mathbf{z}\colon C_2\to C_2$ is the zero map. 
Thus, no such functor $\mathcal{F}$ can exist. 
