We know that the classifying space construction $G \mapsto BG$ gives a functor from topological groups to spaces.
I was wondering if the whole construction of the universal bundle is functorial in $G$? I.e., given a topological group homomorphism $f : G\rightarrow G'$ we get a commutative diagram $$\require{AMScd} \begin{CD} EG @>{Ef}>> EG' \\ @VVV @VVV \\ BG @>{Bf}>> BG' \end{CD}$$ where the vertical maps are the universal bundles?
To a $G$-set $X$, one can define a category with objects the elements of $X$ and $\operatorname{Hom}(x,y)=\{g \in G | gx=y\}$. This construction has the property that it is functorial with respect to maps of $G$-sets. When $X=G$ we call this category $EG$, and when $X=*$ we call this category $BG$. There is a unique map of $G$-sets $G \rightarrow *$, and so it induces a map of categories $EG \rightarrow BG$.
The realization of this map $EG \rightarrow BG$ is a fibration with fiber $G$. Since $EG$ has a terminal object, that makes its realization $|EG|$ contractible, implying that $|BG|$ is a $BG$. This construction clearly is functorial with respect to group homomorphisms $G \rightarrow H$