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 ...

  • 3
    $\begingroup$ The category of sets contains all set functions. The category of groups only contains those functions which are group homomorphisms. $\endgroup$
    – John Douma
    Jan 16, 2021 at 7:04
  • $\begingroup$ @JohnDouma, thank you ! if you write that down as an answer, I'll accept it $\endgroup$
    – Link L
    Jan 16, 2021 at 7:05
  • 1
    $\begingroup$ I appreciate that but I only confirmed what you already knew. $\endgroup$
    – John Douma
    Jan 16, 2021 at 7:06
  • $\begingroup$ Looking up the definition of Set should have clarified this $\endgroup$ Jan 16, 2021 at 7:06
  • 2
    $\begingroup$ @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). $\endgroup$ Jan 16, 2021 at 11:12

1 Answer 1


I am not sure if "by default" is a right term. It may be more precise to say that in the category of sets, or groups, or pointed sets, and many others, the maps are precisely all possible maps that "preserve the structure" of the objects in the category. (Group homomorphism being a concrete example of this idea).

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.