# Right-adjoint to the inverse image functor

Let $X$ be a set. We can turn $\mathcal P(X)$ (the power set of $X$) into a category by taking inclusion maps as morphisms. Now consider a function $f : X \to Y$, which induces the functor $f^{-1} : \mathcal P(Y) \to \mathcal P(X)$. Now we have the identities $$f^{-1} \left( \bigcup_{\alpha} V_{\alpha} \right) = \bigcup_{\alpha} f^{-1}(V_{\alpha}), \quad f^{-1} \left( \bigcap_{\alpha} V_{\alpha} \right) = \bigcap_{\alpha} f^{-1}(V_{\alpha}).$$ A limit/colimit in the category $\mathcal P(X)$ is a intersection/union of subsets of $X$, so the first equation says that $f^{-1}$ commutes with all limits/colimits. If our categories are nice enough (which we assumed, since everything here is small, c.f. this Wikipedia page), the Special Adjoint Functor Theorem tells us that $f^{-1}$ should have a left-adjoint and a right-adjoint functor. This : $$f(U) \subseteq V \quad \Longleftrightarrow \quad U \subseteq f^{-1}(V)$$ tells us that $$\mathrm{Hom}_Y(f(U),V)) \simeq \mathrm{Hom}_X(U, f^{-1}(V))$$ so I assumed that $f : \mathcal P(X) \to \mathcal P(Y)$ mapping $U \mapsto f(U)$ was the left-adjoint functor I was looking for, so that $f^{-1}$ has a left-adjoint and preserves all limits, we are happy.

Now the thing is I can't find the right adjoint ; it's not $f$, and I've tried several other things, doesn't work.

Question : First of all, did I understand all the above correctly, or did I make a mistake somewhere? I am not quite acquainted with the concepts of limits/colimits and the theorem, I am still learning this stuff.

Second of all, assuming the answer to the first question is yes, what is the right-adjoint of $f^{-1}$?

• I thought the set-theory tag was relevant because we actually thought about set-theoretic operations when we tried to figure out the right-adjoint. – Patrick Da Silva Sep 1 '14 at 0:50
• I discovered the right-adjoint $g$ by considering two simple examples. Let $f_1$ be the function from the set of two points to the singleton set. Let $f_2$ be the function $\{a,b,c\}\to\{u,v\}\colon a,b\mapsto u,c\mapsto v$. Perhaps these considerations will help you too. – Karl Kroningfeld Sep 1 '14 at 1:07
• Hint: consider the function $f_*:\mathcal P(X)\rightarrow \mathcal P(Y)$ which sends a subset $A$ of $X$ to the largest subset $B$ of $Y$ such that $f^{-1}(B)\subseteq A$. – user123641 Sep 1 '14 at 1:23
• @Bryan : Is it a surprise that $f_*(A)$ is defined by a colimit in $\mathcal P(Y)$? – Patrick Da Silva Sep 1 '14 at 2:56
• Denoting $\mathbf 2$ the category $0 \to 1$, you have an equivalence of categories between $\mathcal P(X)$ and $\mathbf 2^X$ (regarding $X$ as a discrete category). Then the theory of Kan extensions apply, giving you the left and right adjoint of $f^{-1} \colon \mathbf 2^Y \to \mathbf 2^X$. – Pece Sep 1 '14 at 7:30

Powersets, as posets / categories, are self-dual via taking complements. This implies that the right adjoint is the complement of the image of the complement.

A conceptual way to think about the left and right adjoints of taking inverse image is that they are given by fiberwise existential vs. universal quantification. That is, if $f : X \to Y$ is a map of sets and $f^{\ast} : 2^Y \to 2^X$ the inverse image map, its left adjoint takes a subset $S \subseteq X$ to the set

$$T = \{ y \in Y : \exists x \in f^{-1}(y) : x \in S \}$$

while its right adjoint takes a subset $S \subseteq X$ to the set

$$T = \{ y \in Y : \forall x \in f^{-1}(y) : x \in S \}.$$

• there are two notations used here. The first one f* and then later f^-1. Are these meant to be the same? – Henry Story Apr 3 '18 at 16:35

Following your thoughts, we look for a functor $F:\mathcal P(X)\to\mathcal P(Y)$ which satisfies $$\hom_{\mathcal P(X)}(f^{-1}(V),\,U) \simeq \hom_{\mathcal P(Y)}(V,\,F(U))\,.$$ Since both categories in question are partial orders by inclusion, this means exactly that $$f^{-1}(V)\subseteq U\ \iff\ V\subseteq F(U)$$ for all $U\subseteq X,\ V\subseteq Y$.
This suggests $F(U):=\bigcup\{V\,:\,f^{-1}(V)\subseteq U\}\ =\ \{y\in Y\,:\,f^{-1}(y)\subseteq U\}$. $\$(See also the comments.)

Note that, when viewing $f:X\to Y$ as an $Y$-indexed collection of its fibers $\{f^{-1}(y)\}_{y\in Y}$, then $F(U)$ just picks the indices of those fibers which are fully contained in $U$.

• No offense, but I did perfectly fine with Bryan's comment. His hint was more than enough! But thanks for the full description. – Patrick Da Silva Sep 3 '14 at 0:38
• @Patrick Way better than leaving the question unanswered. If you know the answer, you should consider writing one yourself – roman Sep 3 '14 at 10:49