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Let $\mathcal{C}$ be a small category. Functoriality for presheaves says that for any functor $u\colon\mathcal{C}\to\mathcal{D}$ the precomposition functor $u^*\colon PSh(\mathcal{D})\to PSh(\mathcal {C})$ has two adjoints $u_!, u_*$ on the left and on the right correspondingly, see SGA 4, Exposé I, Proposition 5.1 or ncatlab or stacks 1, stacks 2. It is a particular case of the Kan extension. (Here we consider only presheaves of sets.) These functors can be defined as follows: $$u_!F(Y)=\varinjlim_{Y\to u(X)} F(X)$$ $$u_*F(Y)=\varprojlim_{u(X)\to Y} F(X)$$ for any $Y\in \mathcal{D},\; F\in PSh(\mathcal{C})$, see Theorem 2.3.3 on Kan extensions in Kashiwara-Schapira "Categories and Sheaves" (here the limits are reversed since we work with contravariant functors = presheaves).

I try to understand this construction in the case of topological spaces. Let $f\colon Y\to X$ be a continuous map of topological spaces. Let $Op(X)$, $Op(Y)$ be the categories of open subsets where morphisms are inclusions. Then there is a functor $u\colon Op(X)\to Op(Y)$ which sends $U\subset X$ to $f^{-1}U\subset Y$. Then we get a functor $u^*\colon PSh(Y)\to PSh(X)$ which is the usual pushforward $f_*$. The functor $$u_!F(V)=\varinjlim_{V\subset f^{-1}(U)} F(U)=\varinjlim_{f(V)\subset U} F(U)=f^{-1}F(V)$$ is the usual pullback for presheaves. However, I cannot unfold the definition of the second adjoint $u_*$. By the general definition above, one should have $$u_*F(V)=\varprojlim_{f^{-1}(U)\subset V} F(U).$$ But then it is not clear how the restriction maps look like: for an inclusion $V\subset V^\prime$ of open subsets in $Y$ we should have a map $$u_*F(V^\prime)=\varprojlim_{f^{-1}(U)\subset V^\prime} F(U) \to \varprojlim_{f^{-1}(U)\subset V} F(U)=u_*F(V).$$ However, each $f^{-1}(U)$ contained in $V$ is contained in $V^\prime$ and not vice versa. So the map seems to be in the opposite direction $$\varprojlim_{f^{-1}(U)\subset V} F(U) \to \varprojlim_{f^{-1}(U)\subset V^\prime} F(U).$$ Thus, what we have constructed is not a presheaf of sets but rather a covarinat functor ${Op(Y)}\to Sets$. I would be grateful if you could tell me where is my mistake.

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2 Answers 2

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In general, if $J \subseteq I$, and you have an inverse system of algebraic objects $X_i$, then you have a canonical map $$\varprojlim_{i\in I} X_i \to \varprojlim_{i\in J} X_i.$$ One way to see this is as a restriction of the "selection of a subset of coordinates" operation $\prod_{i\in I} X_i \to \prod_{i\in J} X_j$ (and verify that if you have the consistency condition for an element of $\prod_{i\in I} X_i$ to be in $\varprojlim_{i\in I} X_i$, then its image satisfies the consistency condition to be in $\varprojlim_{i\in J} X_i$). Another way is to construct it using the universal property of $\varprojlim_{i\in J} X_i$.

In your case, you have that $\{ U \in \operatorname{Op}(X) \mid f^{-1}(U) \subseteq V \} \subseteq \{ U \in \operatorname{Op}(X) \mid f^{-1}(U) \subseteq V' \}$.

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  • $\begingroup$ Thank you very much! This answers my question. I got confused with the direction of arrows. $\endgroup$
    – JfR_01
    Apr 21 at 18:35
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Hint: The limit of a diagram $D$ has a canonical map to the limit of a subdiagram $D'$, induced from projection maps out of the limit of $D$, using the universal property of the limit of $D'$.

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