I am trying to prove Lemma 2.1.1 of Kashiwara-Schapira "Categories and Sheaves":

Let $\beta:I^{\text{op}}\to\mathbf{Set}$ be a functor ($I$ a small category) and let $X\in\mathbf{Set}$. There is a natural isomorphism $$\text{Hom}_{\mathbf{Set}}(X,\lim_{\leftarrow} \beta)\cong \lim_{\leftarrow}\text{Hom}_{\mathbf{Set}}(X,\beta)$$ where $\text{Hom}_{\mathbf{Set}}(X,\beta)$ denotes the functor $I^{\text{op}}\to\mathbf{Set}$ , $i\mapsto \text{Hom}_{\mathbf{Set}}(X,\beta(i))$.

This comes just after they have proved that $$\lim_{\leftarrow} \beta\cong \lbrace \lbrace x_i\rbrace_i \in\prod_i \beta(i)\;|\;\beta(s)(x_j)=x_i\;\text{for all}\; s:i\to j\;\text{in}\;I\rbrace$$ and they say that the lemma "is obvious". Why is it obvious?

(By the way, why are they using the notation of a projective limit? They are defining a general limit in the categorical sense, right?)

  • 2
    $\begingroup$ Terminological: this is not a limit in a presheaf category, this is a limit of a presheaf in the category $\mathbf{Set}$. $\endgroup$ – Oskar Feb 17 '14 at 22:05
  • $\begingroup$ "Projective limit" is their terminology for what is normally called a "limit". Their "inductive limit" is what is normally called a "colimit". $\endgroup$ – Uday Reddy Feb 18 '14 at 18:56

Remember the universal property of a limit: the maps from $X$ to $\varprojlim \beta$ are in one-to-one correspondence with the cones over $\beta$ with tip $X$. One builds such cones out of maps from $X$ to $\beta(i)$ which commute with the images of morphisms in $I$ under $\beta$. More precisely, such a cone is specified as an element of $\prod_I \hom(X,\beta(i))$ satisfying the same relations as in the explicit definition you give of a limit. This is just a useful and simple rephrasing of the definition of limit, which can easily get cloaked in technicalities.

The projective limit notation is somewhat standard for an arbitrary limit: projective limits are special cases of limits, and inductive limits of colimits, so it's natural to write $\varprojlim$ for a limit and $\varinjlim$ for a colimit. The alternative is to write $\lim$ and $\text{colim}$. Kashiwara-Schapira is generally considered to be pretty difficult reading, incidentally-if you're new to category theory you might spend some time with a more elementary text alongside.

  • $\begingroup$ Yes, but they haven't defined the limit via a universal property. They have just defined it by setting $$\lim_{\leftarrow} \beta=\text{Hom}_{[I^{\text{op}},\mathbf{Set}]}(1,\beta)$$ $\endgroup$ – triwer23 Feb 17 '14 at 21:24
  • $\begingroup$ OK, then a map from $X$ to $\varprojlim \beta$ is a choice for each $x\in X$ of a natural transformation from $1$ to $\beta,$ i.e. a point in $\beta(i)$ for every $i$ commuting with maps in $I$. This gives a function from $X$ to $\beta(i)$ for every $i$ commuting with maps in $I$, which is what a cone is. $\endgroup$ – Kevin Carlson Feb 17 '14 at 21:29

It's only obvious to the authors because they wrote the book. They should have said "it's obvious to us that, but you need to prove..."

The isomorphism in the category Set is given by the definition of $\lim\limits_{\leftarrow} \beta$. Note that $\text{pt}$ just sends all objects of $I$ into $\{*\}$ a singleton set. So draw out the commutative diagram for an element of $\lim\limits_{\leftarrow} \beta$. I.e. it should look like, for all $f : i \to j$ in $I$:

$$ \require{AMScd} \begin{CD} \text{pt}(j) @>\text{pt}(f)>>\text{pt}(i)\\ @V\theta_jVV @V\theta_iVV \\ \beta(j) @>\beta(f)>> \beta(i) \end{CD} $$

But this diagram can be converted into a triangular looking diagram which I can't render here since $\text{pt}(j) = \text{pt}(i) = \{*\}$ and $\text{pt}(f) = \text{id}_{\{*\}}$ by definition. Then you can see that a choice of such natural map is essentially the same thing as picking out elements of $\beta(k), k \in I$ since a map from a singleton into a set essentially picks out an element.

And similarly, the defining condition of an element $\theta \in \lim\limits_{\leftarrow} \beta$ is $\beta(f) \circ \theta_j = \theta_x$, which is true iff $\beta(f)(\theta_j(*)) = \theta_i(*)$. Let $x_k = \theta_k(*), \forall k \in I$ etc.

A choice of elements of $\beta(k), k\in I$ is clearly an element of $\prod_k \beta(k)$. Hence the set builder definition.

Hope that clears some things up.


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.