By abelianization I mean, for any group $G$, its commutator subgroup is the subgroup $[G,G]$ generated by elements of the form $ghg^{-1}h^{-1}$ for $g,h\in G$.
Then the abelianization of $G$ is $G^{\text{ab}} := G/[G,G]$.
For any two groups $X,Y$, by functoriality we have a map $$\text{Hom}(X,Y)\rightarrow\text{Hom}(X^{\text{ab}},Y^{\text{ab}})$$
My question is, is this map always surjective for any two groups $X,Y$?
No. Take $X=C_2=\{1,x\}$ and $Y=Q_8=\{1,i,j,k,-1,-i,-j,-k\}$. Then $\operatorname{Hom}(X,Y) = \{ x\mapsto 1, x\mapsto -1\}$ where $x$ is the unique element of order 2 in $X$ and $-1$ is the unique element of order 2 in $Y$. However $X^{ab}=C_2=\{\{1\},\{x\}\}$ and $Y^{ab}=C_2 \times C_2=\{\{\pm1\},\{\pm i\},\{\pm j\},\{\pm k\}\}$ so that $\operatorname{Hom}(X^{ab},Y^{ab}) = \{ \{x\} \mapsto \{ \pm 1 \}, \{x\} \mapsto \{\pm i\}, \{x\} \mapsto \{\pm j\}, \{x\}\mapsto \{\pm k\}\}$ has 4 elements. The image of $\operatorname{Hom}(X,Y)$ is only the single map $\{x\} \mapsto \{\pm1\}$.