# Question on forgetful functor from group to set is not full

I read that the forgetful functor $$U: \textbf{Grp} \rightarrow \textbf{Set}$$ is not full, i.e. given two objects $$X$$ and $$Y$$ in $$\textbf{Grp}$$, there are some arrows in the $$\textbf{Set}$$ category between $$U(X)$$ and $$U(Y)$$ that are not mapped to corresponding arrows in $$\textbf{Grp}$$ between $$X$$ and $$Y$$...

But does this mean that the category $$\textbf{Set}$$ includes all possible maps among its objects "by default" ? I tried looking for confirmation on this point in the text am reading but could not find any ...

• The category of sets contains all set functions. The category of groups only contains those functions which are group homomorphisms. Jan 16, 2021 at 7:04
• @JohnDouma, thank you ! if you write that down as an answer, I'll accept it Jan 16, 2021 at 7:05
• I appreciate that but I only confirmed what you already knew. Jan 16, 2021 at 7:06
• Looking up the definition of Set should have clarified this Jan 16, 2021 at 7:06
• @LinkL Maybe it would be nice to find a concrete example to fully convince yourself. That shouldn't be hard now that you know where to look (and you can answer your own question then). Jan 16, 2021 at 11:12

So, in coming to your specific example, what structure a map in the category of sets has to satisfy? I think it would make the situation clear when you ask the same question in the category of groups. This is gives you a proof of statement that the forgetful functor $$U: \textbf{Grp} \rightarrow \textbf{Set}$$ is not full.