# Direct and inverse image terminology

Let $$f\colon X\to Y$$ be a continuous map. Then $$f$$ induces a geometric morphism $$f^\ast\dashv f_\ast\colon \mathrm{Sh}(X)\leftrightarrows\mathrm{Sh}(Y)$$, whose left adjoint is called inverse image and whose right adjoint is called direct image.

Why is the left adjoint called inverse image and why is the right adjoint called direct image?

Especially the following makes it confusing: The direct image $$f_\ast$$ is defined by $$F\in\mathrm{Sh}(X)\mapsto (U\in\mathcal O(Y)\mapsto F(f^{-1}(U))).$$ Usually one calls $$f^{-1}(U)$$ the inverse image of $$U$$ under $$f$$.

• Very simply, because of the direction the arrows point, compared to the continuous map $f : X \to Y$. Commented Dec 23, 2021 at 14:46

There is a precise, almost literal, sense in which $$f^* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$$ generalises the inverse image as defined in elementary set theory. Observe that open subspaces $$V \subseteq Y$$ correspond to subterminal objects in $$\textbf{Sh} (Y)$$: the sheaf of sections of the inclusion $$V \hookrightarrow Y$$ is a subterminal object, and every subterminal object is isomorphic to one of this form. Since $$f^*$$ preserves finite limits, it preserves subterminal objects, and in fact it sends the sheaf corresponding to $$V$$ to the sheaf corresponding to $$f^{-1} V$$. So we may think of $$f^* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$$ as being an extension of the set-theoretic inverse image operation, and this justifies the name "inverse image functor".

Once you have an "inverse image functor", there is a powerful temptation to call its partner the "direct image functor". My personal opinion is that the name "direct image functor" is unsuitable for the right adjoint of the inverse image functor. For one thing, $$f_* : \textbf{Sh} (X) \to \textbf{Sh} (Y)$$ is not a generalisation of the set-theoretic direct image operation, as I will now explain.

Let $$U$$ be an open subspace of $$X$$. What do you suppose $$f_*$$ applied to the subterminal object corresponding to $$U$$ yields? It is a subterminal object, of course, but it is not (the subterminal object corresponding to) the direct image $$\exists_f U = \{ y \in Y : \exists x \in U . y = f (x) \} = \{ y \in Y : \exists x \in X . y = f (x) \land x \in U \}$$ which is not even guaranteed to be open if we do not assume $$f : X \to Y$$ is an open map. Instead, it corresponds to $$\forall_f U = \{ y \in Y : f^{-1} \{ y \} \subseteq U \} = \{ y \in Y : \forall x \in X . y = f (x) \Rightarrow x \in U \}$$ which is always open: indeed, $$\forall_f U$$ is the union of all open $$V \subseteq Y$$ such that $$f^{-1} V \subseteq U$$.

Unfortunately, the reality is that $$f^* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$$ does not always have a left adjoint that generalises $$\exists_f$$, so while it is $$\exists_f$$ that bears the name "direct image" in elementary set theory, for want of a better name it is $$f_*$$ that gets the name in topos theory.

• I think sometimes $\forall_f U$ is called a "dual image". Unfortunately "coimage" is often used to mean something else. Commented Dec 23, 2021 at 16:59