# Question on the definition of the category of elements of a presheaf.

I am new in Topos theory. I have actually just started learning. I am reading MacLane-Moerdijk's book, as it was suggested to me as the best introduction to the subject. Unfortunately I can not make sense of the following.

In section 5 of Chapter I, (page 41) they build what is needed to prove that in the presheaf category $[\mathcal{C}^{\textrm{op}},\mathbf{Set}]$, any object is the colimit of a diagram of representable objects. They define the category of elements of $P$ as follows: The objects are all pairs $(C,p)$ where $C$ is an object of $\mathcal{C}$ and $p$ is an element $p\in P(C)$. Its morphisms $(C',p')\to (C,p)$ are those morphisms $u:C'\to C$ of $\mathcal{C}$ for which $pu=p'$. I do not understand the morphisms of this category. $p$ is a set and $u$ a morphism in $\mathcal{C}$ so what does $pu$ mean? Do they mean $P(u)(p)=p'$ maybe?

Thanks for any help.

-

$u:C'\to C$ is a morphism, and $P$ is a presheaf, thus $P(u):P(C)\to P(C')$ is just a function between two sets, which you can apply to $p\in P(C)$ and get an element in $P(C')$, that is simply the element $P(u)(p)$, and the requirement is, as you write, just $P(u)(p)=p'$.

Notice that there is no guess work involved here and really there is no other way to interpret it. Also, notice that they introduce this shorthand notation a few pages earlier when they introduce presheaf categories.

-
Thanks. I guess I skipped the material until this section because it seemed like things I knew. But apparently there were some things I missed. – Mark Isleng Nov 2 '13 at 8:40

Yes, that's what they mean.

Writing $up$ or $u(p)$ would seem to make more intuitive sense, as this feels more like $u$ is "acting" on $p$ (and since we only have one gadget, $P$, for turning $u$ into a function on sets, your definition is really the only way), but since we are using contravariant functors, these definitions would cause composition to look backwards.

If if $u' : C'' \to C'$ is another morphism, then writing this action as $u(p)$ would give:

$$u'(u(p)) = (uu')(p)$$

If we use the definition from the text, however, we get the much more reasonable-looking:

$$p(uu') = (pu)u'$$

-
Ok thanks! Your explanation on the notation was much appreciated, makes much more sense now. – Mark Isleng Nov 2 '13 at 8:38