Since it is very much possible to have functors with $F_0 \dashv F_1 \dashv F_2$ or even $F_0 \dashv F_1 \dashv F_2 \dashv F_3$, it begs the question:     Can there be an example of infinitely many, non-repeating mutually adjoint   functors?   Obviously, "non-repeating" is important, since $\operatorname{id} \dashv \operatorname{id} \dashv \dots$ would otherwise of course be a boring example (and every other pair of pseudo-invers functors would be too).
37787
Suppose we have (some categories and some functors such that) $F_1$ is left adjoint to $G_1$, $G_1$ left adjoint to $F_2$, $F_2$ left adjoint to $G_2$. Will $F_1$ then be equal to $F_2$ (and $G_1$ to $G_2$)?  Or can there exist arbitrarily long 'chains' of adjoint functors that do not trivialize?
82384
The answer seems to be yes, judging from the following exercise I found in the book Mathematical Logic by H.D. Ebbinghaus, J. Flum, and W. Thomas:     Let $S$ be a finite symbol set and let $\mathfrak{U}$ be a finite $S$-structure. Show that there is an $S$-sentence $\varphi _{\mathfrak{U}}$ the models of which are precisely the $S$-structures isomorphic to $\mathfrak{U}$.   I think I have an idea of how to solve this exercise, but I seem to be unable to materialize it. Thanks.
58678
I'm currently self-studying "Categories and Sheaves"by Schapira and Kashiwara, and I've been stuck on problem 1.19 all day today, so I was hoping that someone could help me out.  Let $\mathcal{C}$, $\mathcal{C&#39;}$ be categories and $L_v:\mathcal{C} \rightarrow \mathcal{C&#39;}$, $R_v:\mathcal{C&#39;} \rightarrow \mathcal{C}$ be functors such that $(L_v,R_v)$ is a pair of adjoint functors (v=1,2). Let $\epsilon_v: id_\mathcal{C} \rightarrow R_v L_v$ and  $\eta_v: L_v \circ R_v  \rightarrow id_{\mathcal{C&#39;}}$ be the adjunction morphisms. Prove that the two maps $\lambda$,$\mu$ : $$Hom_{Fct(\mathcal{C},\mathcal{C&#39;})}(L_1,L_2) \rightleftarrows^{\lambda}_{\mu} Hom_{Fct(\mathcal{C&#39;},\mathcal{C})}(R_2,R_1)$$ given by : $$\lambda(\varphi):R_2 \rightarrow^{\epsilon_1 \circ R_2} R_1 \circ L_1 \circ R_2 \rightarrow^{R_1 \circ \varphi \circ R_2} R_1 \circ L_2 \circ R_2 \rightarrow^{R_1 \circ \eta_2} R_1$$ for $\varphi \in Hom_{Fct(\mathcal{C},\mathcal{C&#39;})}(L_1,L_2)$ $$\mu(\psi) : L_1 \rightarrow^{L_1 \circ \epsilon_2} L_1 \circ R_2 \circ L_2 \rightarrow^{L_1 \circ \psi \circ L_2} L_1 \circ R_1 \circ L_2 \rightarrow^{\eta_1 \circ L_2} L_2$$ for $\psi \in Hom_{Fct(\mathcal{C}&#39;,\mathcal{C})}(R_2,R_1)$ are inverse to eachother.  So far, I've been trying to play around with the zig-zag identity, but it doesn't seem to lead me anywhere. Any help will be greatly appreciated!
75149
I hope this doesn't turn out to be a silly question.  There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms.  But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.    My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?    For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following.  Roughly, the idea is to find an ordered family of topologies $\tau_i$ ( $i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself.  Let $S = \mathbb{Z} \coprod \mathbb Z$.  The topology $\tau_i$ is as follows: if $i&lt;0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete.  The space $(S, \tau_0)$ is then indiscrete.  For $i&gt;0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement.  Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism.
40868
Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ that takes each value in $\mathbb{R}$ three times? If not, how could I prove that such a function does not exist?
75927
I am reading Homological Algebra by J.J. Rotman and am unable to do this problem.  Given an adjoint pair $(F,G)$ where $F : \mathcal{C} \to \mathcal{D} $ and $G : \mathcal{D} \to \mathcal{C} $ are two covariant functors, we can obtain a natural transformation $ \eta : \mathbb{1_{\mathcal{C}}} \to GF $.  I understand the definition of $\eta$ but am unable to prove that the transformation is natural. Couls someone please help me prove that for $f \in Hom (C,C')$, $GF(f)\eta_C=\eta_{C'}f$ ? Thanks !
15583
Often we have adjoint pairs $(A, B)$ (meaning $A$ is left adjoint to $B$). Sometimes we have adjoint triplets $(A,B,C)$ (meaning $A$ is left adjoint to $B$ and $B$ is left adjoint to $C$. No adjoint relation between $A$ and $C$ obviously, since they have the same source and target).  So the first question is: can we have quadruplets $(A,B,C,D)$ ? (meaning that $C$ is also left adjoint to $D$).  Second question: It would be then possible that we have $(A,D)$ besides $(A,B)$. But I think not, because I think that it is not possible to have a functor $A$ with two different right adjoints, $B$ and $D$. Is this correct?
41851
(I'll explain what I know first and then I'll ask the questions). Given a finite dimensional vector space $V$, it is often remarked that there is no "natural" isomorphism from $V$ to $V^*$ (I guess this means a basis independent isomorphism?). I understand that one typically constructs an isomorphism $V \to V^*$ by fixing a basis for $V$, call this $B = \{v_1, v_2, \ldots, v_n\}$, and then mapping $v_i \mapsto \delta_i$ where $\delta_i$ is the linear functional given by $\delta_i(v_i) = \delta_{ij}$. Here are my questions:  (1) Why is there no natural isomorphism $V \to V^*$?  (2) If $V$ is a finite dimensional inner product space, we can map each $v \mapsto \langle v, \cdot \rangle$. Is this not "natural"? (I know we can consider sesquilinear forms, but let's keep the discussion to this inner product for simplicity).
105170
In one of the books on algebraic topology (I don't remember exactly which one) there was an exercise to build an example of two topological spaces having two continuous bijections between them which are not homotopy equivalent. To be honest, this exercise confuses me a little because, as I understand, each pair of homeomorphic spaces is homotopy equivalent by construction. On the other hand, the existence of bijective continuous mapping between spaces automatically provides their homeomorphism (correct me if I'm wrong). Thus, in this logic, if we have two continuous bijections between topological spaces this will inevitably lead to the homotopy equivalence between them. I guess, however, that there is a simple counterexample related to the discrete topologies which breaks such a reasoning (see, for example, this: ), but I have certain difficulties in discovering it. Is there any suggestion?
50780
Let $R$ be a commutative ring and $M$ and $N$ be two finitely generated $R$-modules. I wanna construct an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated.   It's well-known that if $R$ be a Noetherian ring and $M$ and $N$ be two finitely generated $R$-modules then $\operatorname{Hom}_R(M,N)$ is finitely generated. So in this example $R$ must be a Non-Noetherian ring.
36035
I'm working through the following,     Exercise 2.1.iv. A functor $F$ defines a subfunctor of $G$ if there is a natural transformation $\alpha: F \Rightarrow G$ whose components are monomorphisms. In the case of $G: C^{op} → \operatorname{Set}$, a subfunctor is given by a collection of subsets $Fc \subset Gc$ so that each $Gf : Gc \longrightarrow Gc'$ restricts to a function $Ff: Fc → Fc'$. Characterize those subsets that assemble into a subfunctor of the representable functor $C(−, c)$.   I have two questions. The first one is, when we say that a subfunctor for $G$ is given by subsets of each $Gc$, we are identifying each $Fc$ with its image via $\alpha_c$, right? So we would have uniqueness up to isomorphic objects.   Secondly, for the concrete exercise regarding $C(-,c)$, could you provide any hints on how to characterize such sets?   So far I've only noted that given $\mu: d' \rightarrow d$, we need the precomposition,  $$ \mu^{*}: f \in C(d,c) \mapsto f\mu \in C(d',c) $$   to restrict to $Fd$ and corestrict to $Fd'$. Thus, $\mu^*(Fd) \subseteq Fd'$, and so  $$ \bigcup_{\forall \mu: \ d' \to d \\ \forall f\in Fd} f\mu \subset Fd' $$  but I haven't got from there to any fruitful conclusions.
26753
Can you show me a reasonably simple (using only elementary group-theoretic tools) example of infinite group with just 2 conjugacy classes ?
16591
I don't know if this question is appropriated for this site. Anyway, I'm searching for an isomorphism of order $f:K \longrightarrow \epsilon_o $, such that $(K, \leq)$ is a subset(proper or not) of $(\mathbb{Q}, \leq)$ and $\epsilon_o = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, ... \}$. Actually,  I'm not finding neither an isomorphism between $K$ and $\omega^{\omega}$.  Thanks in advanced.
105994
Are there any theorems characterizing the continuity of the inverses of bijections between certain classes of particular spaces. E.g. it seems to me the inverse of any continuous bijection between connected subsets of $\mathbb{R}$ is continouos. It seems that most counterexamples require highly specific spaces. I'm thinking there might be something along the lines of any continuous bijection between homeomorphic spaces has a continuous inverse. Any of this make sense?
66238
My question  raised another one. How many differents vector space structures over a field $\mathbb{F}$ we may have on an abelian group?  I know that there are abelian groups that we can not endow it with a structure of vector space over any field, for example $\mathbb{Z}_{6}$. But if an abelian group has a structure of a vector space over a field $\mathbb{F}$, is there an upper bound for ways we can define a different structure? Such as the number of automorphisms of the field $\mathbb{F}$. I am conjecturing according to the answer given before.
116901
Let  $f:X\times Y\to \mathbb{R}$. I could not come up with an example that satisfies $$\sup_{y\in Y}\inf_{x\in X}f(x,y)&lt;\inf_{x\in X}\sup_{y\in Y}f(x,y).$$ Any help would be appreciated. Thanks!
57724
Is it possible to stack three relations or symbols ontop of one another? Specifically I would like to stack three symbols, R, \rightleftarrows and L in order to define a pair of adjoint functors $R:C\to D$ and $C\gets D:L$.  Also, is there a \rightleftarrow with long arrows? EDIT I actually meant to ask wether there is a version of \rightleftarrows with long arrows.  I'm not a regular user, so I don't know how to format my question adequately, could someone convert what's inside the dollar so that it appears as latex please?