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Let $f:X\rightarrow Y$ be an arrow in $\mathsf{Set}$. $f$ induces three functors: $$\exists (f),\forall (f):\mathcal P(X)\rightarrow \mathcal P(Y),\; \mathsf I(f):\mathcal P(Y)\rightarrow \mathcal P(X)$$ where the powerset is a poset category. The action on objects is given by $$\exists (f)(A)=f[A],\; \mathsf I(f)(B)=f^\leftarrow (B),\; \forall (f)(A)=f[A^c]^c.$$ and one can prove $$\exists (f)\dashv \mathsf I(f) \dashv \forall (f).$$ In particular, the inverse image functor is right adjoint to the direct image functor.


Now let $f:X\rightarrow Y$ be an arrow in $\mathsf{Top}$. Then $f$ induces the well known direct and inverse image functors $$\mathsf{Sh}(X)\leftarrow \mathsf{Sh}(Y):f^\ast \dashv f_\ast:\mathsf{Sh}(X)\rightarrow \mathsf{Sh}(Y).$$ So in this instance, the inverse image functor is left adjoint to the direct image functor.


Why does the direction of adjunction change? What is the intuition behind this difference?

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  • $\begingroup$ It's just a question of terminology. Nothing more, nothing less. $\endgroup$ – Zhen Lin Jul 7 '15 at 5:20
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Suppose $(X,{\cal T}_X)$ is a topological space. Then ${\cal T}_X$ is a subcategory of $P(X)$, and if $f: (X,{\cal T}_X) \rightarrow (Y,{\cal T}_Y)$ is a continuous map, the set pullback functor $I(f): P(Y) \rightarrow P(X)$ restricts to ${\cal T}_Y \rightarrow {\cal T}_X$. A sheaf of sets ${\cal F}$ on $X$ being a contravariant functor ${\cal T}_X \rightarrow Sets$, the pushforward $f_*{\cal F}$ is the same as ${\cal F} \circ I(f)$, so $f_*$ is a kind of dual to $I(f)$. The point is that passing to sheaves is like dualizing, which reverses direction.

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