Let $\Sigma_n$ be the permutation group of order $n$. Then the regular representation of $\Sigma_n$ gives an injective homomorphism $f:\Sigma_n\to O(n)$. Why $f$ induces a map between their classifying spaces $$ F: B\Sigma_n\to BO(n)?$$
In general, let $\phi: G\to H$ be a homomorphism of Lie groups. Does $\phi$ always induce a map between their classifying spaces $$ \Phi: B G\to B H?$$
To produce a map $B G \to B H$ it suffices to define a principal $H$-bundle on $B G$. So consider $E G \times_G H$. Assuming the topology works out nicely (which is the case when one of $G$ or $H$ is discrete), $E G \times_G H \to B G$ is a principal $H$-bundle, hence induces a map $B G \to B H$, as required.
To illustrate what's really going on more clearly, let me just work with discrete (but possibly infinite) groups. The point is that if you have a group homomorphism $\phi : G \to H$, then you get a functor $(-) \times_G H$ that sends principal $G$-bundles to principal $H$-bundles and is compatible with the pullback operation. In particular, we get a natural map $$\phi_! : \mathrm{Prin}_G (X) \to \mathrm{Prin}_H (X)$$ between the sets of isomorphism classes of principal bundles for each space $X$, which implies we have natural maps $$\phi_! : [X, B G] \to [X, B H]$$ for all sufficiently nice spaces $X$, where $[-, -]$ denotes the set of homotopy classes of maps; so by taking $X = B G$ and considering the image of $\mathrm{id} : B G \to B G$, we get a map $B \phi : B G \to B H$, well defined up to homotopy.