I am in the beginning of Category Theory.

Let $\theta\colon G\rightarrow K$ be a group homomorphism, and $\ker\theta$ denote the kernel of this homomorphism. Let $\varepsilon\colon G\rightarrow K$ denote the trivial homomorphism and $i\colon \ker\theta\rightarrow G$ denote the inclusion, i.e. $i(x)=x$. Then the pair $(\ker\theta,i)$ satisfies the property that $$\varepsilon\circ i=\theta\circ i.$$ Question(Exercise) What is the universal property of $(\ker\theta,i)$?

I didn't get any direction towards the solution. But, I tried to move in the following direction: by definition, $\ker\theta$ is the largest subset (subgroup) of $G$ whose elements go to identity via $\theta$. This "largest" and "..go to identity via $\theta$" I tried to put in categorical language. The answer I came up then is as follows:

Given a subset $H$ of $G$ and $j\colon H\rightarrow G$, the inclusion map, if $j$ and $\theta$ (or $\ker\theta$) satisfy
$$\varepsilon\circ j=\theta\circ j,$$ then there is a unique (inclusion?) map $f\colon H\rightarrow \ker\theta$ such that $j=i\circ f$.

Is this answer (universal proerty of $(\ker\theta,i)$) correct?

(I don't know how to draw commutative diagram in the latex, here; one may edit to draw the diagram here.)

  • 3
    $\begingroup$ "Subset" is not a categorical notion. $\endgroup$ – Alex G. Sep 5 '15 at 12:28

You're close, but not quite there. As was mentioned already in the comments, "subset" is not a categorical notion.

Because we are dealing with the category $\sf Grp$ of groups and morphisms, all our objects should, in fact, be groups.

So we might contemplate replacing "subset" by "subgroup". However, the notion of "subgroup" is not preserved under group isomorphism because it relies on the set-theoretical nature of the groups.

However, we know that for every $j: G' \to G$, $\operatorname{im} j$ is a subgroup of $j$, and indeed the inclusion homomorphisms show that every subgroup $H$ of $G$ arises in this way.

We can thus phrase the UMP as follows:

Given a group homomorphism $j: H \to G$, if $j$ and $\theta$ satisfy $$\varepsilon j = \theta j,$$ then there is a unique $f: H \to \ker\theta$ such that $j = if$.

In categorical lingo, we say that $(\ker\theta, i)$ is the equalizer of $\varepsilon$ and $\theta$. See also ProofWiki and WikiPedia.

  • $\begingroup$ Note that in terms of generalized elements, the kernel of an arrow can be described as the subobject of the domain consisting precisely of those generalized elements, which are killed by that arrow. Of course, one must be aware of the correct meaning of "subobject" and "element", here. $\endgroup$ – Jakob Werner Sep 5 '15 at 14:28
  • $\begingroup$ as you said, that we are in the category of groups, the unique map f should be homomorphism? $\endgroup$ – Groups Sep 6 '15 at 3:43
  • 1
    $\begingroup$ @Groups Naturally. $\endgroup$ – Lord_Farin Sep 6 '15 at 10:17

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.