This is from Categories & Sheaves by Kashiwara & Schapira.

$$\text{Hom}_{\text{Set}}(X, \lim_{\leftarrow} \beta) \xrightarrow{\sim} \lim_{\leftarrow} \text{Hom}_{\text{Set}}(X, \beta) \tag{2.1.1}$$

Here's the exercise:

Let $\varphi : J \to I$ and $\beta : I^{op} \to \text{Set}$ be functors. Denote by $\varphi^{op} : J^{op} \to I^{op}$ the associated functor. Using (2.1.1), we get a natural morphism:

$$ \lim_{\longleftarrow} \beta \to \lim_{\longleftarrow}\beta \circ\varphi^{op} $$

Firstly, I don't see how the left or right is a functor so how can we speak of the natural map?

Secondly, how do you make the map? If we have a natural map $a \in \lim\limits_{\longleftarrow} \beta$ such that $\beta(f) \circ a_i = a_j$ for any $f : j \to i$ in $I$, I'm not seeing how this should work.

  • $\begingroup$ Surely there must be additional conditions on $\phi$? This would certainly fail if, for example, $J$ was the empty category, since it would imply that every limit is the terminal object, and thus that every set is a singleton. $\endgroup$ – Arnaud D. Jun 12 '18 at 14:36
  • 1
    $\begingroup$ Which exercise is this? Regarding the issue of how both sides are functors, in the book they also refer to the functor $X\mapsto \lim \mathrm{hom}(X,\beta)$ as $\lim \beta$, which in the case the limit exists in the respective category is the Yoneda embedding of the limit. $\endgroup$ – asdq Jun 12 '18 at 14:51
  • $\begingroup$ It's on page 36. $\endgroup$ – StudySmarterNotHarder Jun 12 '18 at 14:53
  • $\begingroup$ On page 36, I read a natural morphism (not isomorphism). If $C$ has all $I$-limits, you can make $\varprojlim\beta$ a functor of $\beta$, in other words $\varprojlim:\operatorname{Fun}(I,C)\to C$. Similarly if $C$ has all $J$ limits, you have a functor $\operatorname{Fun}(I,C)\to C$ such that $\beta\mapsto\varprojlim\beta\circ\varphi^{op}$. The claim is that this is a natural transformation. $\endgroup$ – Roland Jun 12 '18 at 17:12
  • $\begingroup$ @Roland I see that part now, thank you. How do I show that there is a map between the two though? $\endgroup$ – StudySmarterNotHarder Jun 12 '18 at 17:49

Let's dispense with all the duals and the values taken in a particular category, which are obscuring the situation a bit. If $D:J\to C$, $u:I\to J$ are functors such that $C$ admits limits of shapes $I$ and $J$, the claim is that we have a natural map $\mathrm{lim} D\to \mathrm{lim}(D\circ u)$. For any $i\in I$, we must give a map $\mathrm{lim} D\to D(u(i))$.

Now, along with $\mathrm{lim} D$ we are given a cone $\lambda$ with components $\lambda_j:\mathrm{lim} D\to D(j)$. So the natural guess for the desired map is $\lambda_{u(i)}$. To prove that these components determine a map into $\mathrm{lim}(D\circ u)$ as desired, we have only to check that $(D\circ u)(f)\circ \lambda_{u(i)}=\lambda_{u(i')}$ for every $f:i\to i'$ in $I$. But this is simply an instantiation of the assumption that $\lambda$ forms a cone.


This proof uses just the definition of $\lim\limits_{\leftarrow} \beta = \text{Hom}_{I^{\wedge}}(\text{pt}, \beta)$ and the fact that $\beta(f) \circ \theta_y = \theta_x$ for all natural maps $\theta$ in the limit, and $f : x \to y$ in $I$.

Suppose also that since $\varphi(g) \in I^{op}, \forall \ g \in J^{op}$, we have $\varphi(g): \varphi(b) \to \varphi(a)$ for all $g \in J^{op}$ so that by letting $f = \varphi(g)$ in the above we get that $\beta(\varphi(g)) \circ \theta_{\varphi(b)} = \theta_{\varphi(a)}$. Thus for any $\theta$ in the first limit, there is $\theta \circ \varphi \equiv \theta_{\varphi(\cdot)}$ which is a natural map in $\text{Hom}_{J^{\wedge}}(\text{pt}, \beta \circ \varphi)$.


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