# Defining a functor $H : \mathbf{Grp} \to \mathbf{Set}$ by $H(G') := \{f \in \mathrm{Hom}(G,G') : \mathrm{ker}(f) = N\}$

I am currently working through the book Categories for the Working Mathematician by Mac Lane. I am reading page 57 which gives some examples on universal elements, i.e. "universal properties". We have:

Let $G$ be a group and $N \unlhd G$. ...is a universal element for that functor $H : \mathbf{Grp} \to \mathbf{Set}$ defined by $$H(G') := \{f \in \mathrm{Hom}(G,G') : \mathrm{ker}(f) = N\}.$$ for $G'$ in $\mathbf{Grp}$.

As an exercise, I would like to verify that this is indeed a functor. So it is clear, that $H(G')$ is in $\mathbf{Set}$. Thus we have to consider $H(\varphi) : H(A) \to H(B)$ for a group homomorphism $\varphi : A \to B$. However, it is nowhere explictely stated how this map should look. Since I do not quite see a "natural" way defining it, I ask kindly for a hint (maybe I misunderstood something).

Edit.

• I think $H(G')=\{f\colon G\to G':\ker f\supseteq N\}$, so you only require the kernel contains $N$, not equal $N$, reflecting the fact that group morphisms factor though $G/N$ if their kernel contains $N$, even when this containment is proper. I haven't looked closely at it, but I suspect that if $\varphi\colon A\to B$, you can define $H(\varphi)=\varphi\circ -$ (that is, postcomposition by $\varphi$), so that $H(\varphi)\colon H(A)\to H(B)$, and everything should work out. – BW. Jul 31 '17 at 23:22

You have misread Mac Lane. Note that he says $fN=1$, which means that $N$ is contained in the kernel of $f$ but the kernel of $f$ could also be larger than $N$. So the correct definition is $$H(G') := \{f \in \mathrm{Hom}(G,G') : \mathrm{ker}(f) \color{red}\supseteq N\}.$$ There is now an obvious way to define $H$ on morphisms: given $\varphi:A\to B$ and $f\in H(A)$, define $(H\varphi)(f)=\varphi\circ f$. I'll leave it to you to verify that this gives an element of $H(B)$ and defines a functor.