Let $f\in Hom(b,a)$, then $f^\star:Hom(a,-)\to Hom(b,-)$ is monic iff $f$ is epi.

100 Views Asked by At

Let $f\in Hom(b,a)\in Set$ with $a,b\in Obj(C)$ for some $C$. I will assume $Hom(-,-):C\times C\to Set$ or I have to assume $Hom(-,-):C\times C\to D$ with $D$ equipped with pullback and pushout.

Consider 2 natural transformations.

  1. $f^\star:Hom(a,-)\to Hom(b,-)$ is mono/epi in $Set^C$ iff $f$ is epi/split mono.(Split mono means it has left inverse.)

2.$f_\star:Hom(-,a)\to Hom(-,b)$ is mono/epi in $Set^C$ iff $f$ is mono/split epi.(Split epi means it has right inverse.)

$\textbf{Q:}$ I am done with the proof but I do not find those facts totally intuitive. It is clear that $2$ and $1$ are related by $(Set^{op})^{C^{op}})$ duality.(i.e. For $S,T:C\to Set$, this induces $S,T:C^{op}\to Set^{op}$.) For 2, it seems reasonable to expect so but $f_\star$ is natural transformation between contravariant functors. Why do I expect epi/mono natural transformation between contravariant functors inducing epi/mono maps?(I will ignore split epi part.) I am not asking for the proof but asking for a reason why or how I can find this natural and obvious.

$\textbf{Q':}$ Why there is split epi and split mono showing up in above consequences? (The proof provides $1\in Hom(b,b)$ in the image for either mono or epi case of $f^\star$ or $f_\star$ respectively.) In module category, this indicates splitting from left or splitting from the right for a short exact sequence. Does this mean set category has more fine structure than functor category?