If $M$ is an endofunctor on a category $\cal K$ and $\eta:Id_{\cal K}\to M$ is a natural transorfmation, what is the difference between $\eta M$ and $M\eta$, and how these two (componentwise) are defined on an object $A$ in $\cal K$ ?
13011
So I'm doing a project in Category Theory. I fully understand natural transformations and functors, but what does it mean to compose them, for example in the monad axioms where you have something like $GF\eta \circ G\epsilon F = 1$ etc. What exactly is $F\eta$? Thanks
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Denote by $\mathbf{Cat}$ the category of small categories, i.e. categories whose class of arrows is a set. I am working with classes but I suppose it works the same if one is working in a given universe following the Grothendieck approach.  My question is whether monomorphisms and epimorphisms in $\mathbf{Cat}$ can easily be classified. An arrow between two small categories is just a functor.  In particular I was wondering if the subcategories of a given small category correspond to the category-theoretic subobjects in $\mathbf{Cat}$. Given a fixed object $X$ in a category I call an isomorphism class of monomorphisms with codomain $X$ a $\textit{subobject}$ of $X$.
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I've read in several places that one motivation for category theory was to be able to give precise meaning to statements like, "finite dimensional vector spaces are canonically isomorphic to their double duals; they are isomorphic to their duals as well, but not canonically."  I've finally sat down to work through this, and -  Okay, yes, it is easy to see that the "canonical isomorphism" from $V$ to $V^{**}$ is a functor that has a natural isomorphism (in the sense of category theory) to the identity functor.  Also, I see that there is no way that the functor $V\mapsto V^*$ could have a natural isomorphism to the identity functor, because it is contravariant whereas the identity functor is covariant. My question amounts to:     Is contravariance the whole problem?   To elaborate:   I was initially disappointed by the realization that the definition of natural isomorphism doesn't apply to a pair of functors one of which is covariant and the other contravariant, because I was hoping that the lack of a canonical isomorphism $V\rightarrow V^*$ would feel more like a theorem as opposed to an artifact of the inapplicability of a definition.  Then I tried to create a definition of a natural transformation from a covariant functor $F:\mathscr{A}\rightarrow\mathscr{B}$ to a contravariant functor $G:\mathscr{A}\rightarrow\mathscr{B}$. It seems to me that this definition should be that all objects $A\in\mathscr{A}$ get a morphism $m_A:F(A)\rightarrow G(A)$ such that for all morphisms $f:A\rightarrow A'$ of $\mathscr{A}$, the following diagram (in $\mathscr{B}$) commutes:  $$\require{AMScd}\begin{CD} F(A) @&gt;m_A&gt;&gt; G(A)\\ @VF(f)VV @AAG(f)A\\ F(A') @&gt;&gt;m_{A'}&gt; G(A') \end{CD}$$  This is much more stringent a demand on the $m_A$ than the typical definition of a natural transformation.  Indeed, it is asking that $m_A=G(f)\circ m_{A'}\circ F(f)$, regardless of how $f$ or $A'$ may vary. Taking $\mathscr{A}=\mathscr{B}=\text{f.d.Vec}_k$, $F$ the identity functor and $G$ the dualizing functor, it is clear that this definition can never be satisfied unless $m_V$ is the zero map for all $V\in\text{f.d.Vec}_k$ (because take $f$ to be the zero map). In particular, it cannot be satisfied if $m_V$ is required to be an isomorphism.     Is this the right way to understand (categorically) why there is no natural isomorphism $V\rightarrow V^*$?   As an aside, are there any interesting cases of some kind of analog (the above definition or another) of natural transformations from covariant to contravariant functors?  Note: I have read a number of math.SE answers regarding why $V^*$ is not naturally isomorphic to $V$. None that I have found are addressed to what I'm asking here, which is about how categories make the question and answer precise. ( was closest.) Hence my question here.
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These seem to be very similar as they both separate two operands. When would one be used over the other? Is there any difference in the math spacing surrounding them?  How would the spacing in these relate to to the spacing produced by \mathop?
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Suppose $A, B, C$ are categories. If $A$ and $B$ are equivalent, is it the case that $C^A$ and $C^B$ are equivalent? Also, is it the case that $A^C$ and $B^C$ are equivalent.  I first conjectured that $C^A$ and $C^B$ are equivalent, but then as I tried to prove the conjecture, I started to realize that it might not be the case. So now I'm trying to find a counterexample. So far I have not been able to find one. Maybe I'm stuck with "too complicated" ideas. I strongly believe that there are very simple counterexamples. Please help. (For more detail, I got to the point where I have a cube that I want to be commutative, but I can only show that 4 of its sides are commutative squares. That's when I realize the other two sides may not necessarily commute.)  And of course, there is the second part to the problem that's not just the dual problem. I would like to know about the special case in which $A$ is a preorder and $B$ is a partial order which is the skeleton of $A$. This may have nothing to do with the validity of the statement in question at all, but it is the setting that originated this question.
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Let $\mathcal{C}$ and $\mathcal{D}$ be categories, $\mathcal{F}\colon\mathcal{C}\to\mathcal{D}$ be a functor. There is a notion of  of $\mathcal{F}$, namely a pair $(\ell,\varphi)$, where $\ell\in\text{Obj}(\mathcal{D})$ is an object and $\varphi\colon\Delta_{\ell}\to\mathcal{F}$ is a natural transformation from the constant functor to $\mathcal{F}$, such that for any other pair $(d,\alpha)$ of the same type (which is called a cone) there exists a unique morphism $f\colon d\to\ell$, such that $\varphi\circ\Delta_f=\alpha$.  Obviously the concept of limit is a conjunction of the two properties:   1). for every cone $(d,\alpha)$ there exists a morphism $f\colon d\to\ell$, such that $\varphi\circ\Delta_f=\alpha$;  2). if such a morphism exists, then it is unique.  One may be interested what happens if we reject one of these requirements. For instance, if we reject the uniqueness condition, then we come to the concept of .     Question: What concept do we get if we reject the existence condition?   Of course, we may define it and call it a soft limit. We also may find some examples (empty set is a soft terminal object of $\mathbf{Set}$). My question is whether it was studied anywhere or if this is a special case of another known category-theoreric concept.
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Consider the transpose inverse automorphism on $GL_n(\mathbb F)$ where $n\geq2$ and $|\mathbb F|&gt;2$. (i.e. $\mathbb F$ is a field, possibly infinite, with three or more elements). I want to show this automorphism is not inner. I was told to consider $\det(BAB^{-1}) = \det(A)$ and $\det(\,^TA^{-1})=\det(A)^{-1}$ and derive a contradiction.  However, in some fields (where the result holds) I fail to see the contradiction. What about $\mathbb F_3$? An element of $GL_2(\mathbb F_3)$ either has determinant $1$ or $2$, both of which are their own multiplicative inverses in $\mathbb F_3$. I'm not sure if my knowledge of fields and determinants is flawed or the hint is.
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Does anyone know any example that invalidates the following affirmation:     If a morphism $f:A\to A$ induces the identity $\hat f:\operatorname{Spec} \left( A \right) \to \operatorname{Spec} \left( A \right)$ then $f = \operatorname{id} _A $.
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Question. Let $R$ be a ring, $\mathfrak{p}$ a prime, $M$ a finitely-generated $R$-module, and $N$ any $R$-module. Is the natural map $$\textrm{Hom}_R(M, N)_\mathfrak{p} \to \textrm{Hom}_{R_\mathfrak{p}}(M_\mathfrak{p}, N_\mathfrak{p})$$ an isomorphism (of $R_\mathfrak{p}$-modules)?  I can prove this in the case when $M$ is finitely presented: indeed, let $S$ be any flat $R$-algebra, and let $$R^m \to R^n \to M \to 0$$ be a right-exact sequence; tensoring with $S$ gives another right-exact sequence $$S^m \to S^n \to M \mathbin{\otimes_R} S \to 0$$ and applying hom functors, we get left-exact sequences $$0 \to \textrm{Hom}_R(M, N) \to N^n \to N^m$$ $$0 \to \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S) \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ and $S$ is flat, so tensoring the first sequence yields $$0 \to \textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ but extending the sequences by $0$ to the left, and putting in vertical maps between the last two, we conclude that $\textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \cong \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S)$ by the five lemma (modulo checking commutativity of diagrams). This is essentially the proof Eisenbud gives [Commutative Algebra, Prop. 2.10].  The trouble with extending it to a proof for finitely-generated modules is that we have to replace $R^m$ with a potentially arbitrary submodule $K$ of $R^n$, and that may not be free or even finitely-generated without some additional assumptions on $R$. I can't see an abstract nonsense proof of the claim, but I admit I haven't tried a bare-hands proof. However, is the claim even true?
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Can we find a field $K$ and an endomorphism $f$ of $K$, such that $f$ is not trivial and $f$ is not surjective? In other words, can we find an endomorphism of $K$ which is not an automorphism?
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The category Set has the property that any morphism f can be decomposed into an epimorphism $e$ and a monomorphism $m$ with $f$ = $m\circ e$. (The intermediary set object is the image of $f$.)  Is there a name for this construction?  Is there a name for this property of categories?  In Set, the intermediary object is unique up to isomorphism. Is this the case for all categories with the property?
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Let $M$ be an A-module and $f \in \text{End}(M)$ an idempotent endomorphism. Show that $M = \ker(f) \oplus \ker (f-\text{id}_M)$. I don't know how to approach this exercise, using only properties of modules and direct sums, without projections. Can you help me? Thanks in advance.
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Given some category ${\mathcal C}$, then the opposite category will consist of the same objects with the morphisms "turned around."  Given $f:A\rightarrow B$, for $A,B$ objects of ${\mathcal C}$, then in general do we have a canonical $f^{op}:B\rightarrow A$ which is induced by $f$?    Slight Motivation:  This is something that's just been bothering me recently, and I feel like I am thinking about this concept incorrectly.  To make this question slightly less vague, if we had, say, the category of rings and we had the mapping $f:{\mathbb Z}\rightarrow {\mathbb Z}/2{\mathbb Z}$ by $f$ sends even numbers to 0 and odd numbers to 1, then what is the corresponding map in the opposite category, $f^{op}:{\mathbb Z}/2{\mathbb Z}\rightarrow {\mathbb Z}$ induced by $f$?  Also, I always see this $f^{op}$ that I've mentioned just written as $f$; is this shorthand, or is there something deeper there?
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I am wondering if there is a standard definition for a category with objects as first order logical (FOL) expressions e.g. $\neg x \vee y$. It seems to me that these logical expressions would be part of a concrete category as they can be mapped (forgetfully) to sets, and as many concrete categories (groups etc.) have homomorphisms as morphisms that perhaps homomorphisms are the morphisms in this FOL category.  I am having difficulties visualizing what these homomorphisms would look like - perhaps just some operation $ (\circ,\bullet) $ such that   $f((x \wedge y) \vee z) \rightarrow (f(x) \bullet f(y)) \circ f(z)$ where $(\vee,\wedge) \rightarrow (\circ,\bullet)$  ?  Thanks,  Brian
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Let $A$ be a commutative ring and $M$ an $A$-module. Let $I$ be any ideal of $A$.  We have an epimorphism $M \otimes_A I \rightarrow IM$. It seems to me that this is not in general an isomorphism.   Q1: Any counterexample?   If $M$ is flat, then $M \otimes_A I \cong IM$. However, flatness seems to be a too strong condition for this equality to be true for any ideal $I$. I am interested in finding a characterization of $M$ such that $M \otimes_A I \cong IM$ for any ideal $I$ of $A$.   Q2: Any suggestions?
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I have long been confused about this notion. I know that for a statement within a single category, forming the dual statement is just reversing every arrows. But what about a statement concerning several categories and functors between them?   Mac Lane suggests reversing all arrows in all categories and leaving the functors invariant in his book "Categories for the working mathematician" (page 32), but it seems that Mac Lane himself doesn't always follow this discipline (when he states "the dual of Yoneda Theorem", he doesn't reverse the arrows in $\mathbf{Set}$ !) Can anyone give me a formal definition of "dualize a statement in category theory"?
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This is from  .      Let $\mathcal{A}$ be a locally small category, and let $A,A' \in \mathcal{A}$ with $\operatorname{Hom}(-,A) \cong \operatorname{Hom}(-,A')$. Prove directly that $A \cong A'$.     My thoughts:      Let $\eta$ denote the natural isomoprhism between the two functors.    We have    $$H(A,A) \xrightarrow{ \eta_A} H(A,A')$$   So $\eta_A(id_A): A \rightarrow A'$ is a candidate. Now I want to construct inverse.    $$ H(A',A') \xrightarrow{\eta_{A'}} H(A',A)$$   Then the map $\eta^{-1}_{A'} (id_{A'}):A' \rightarrow A$.    So I'd like to show that the composition is identity. By the naturality condition of $\eta$, from $H_A(A) \rightarrow H_{A'}(A')$ (and similarly on other direction)    $$ \eta_A(id_A) \circ \eta^{-1}_{A'}(id_{A'}) = \eta_{A'}(\eta_{A'}^{-1}(id_{A'}))= id_{A'} $$   We deduce these two maps are inverses. Hence $A' \cong A$.    Is this correct? Or is there a neat way to see this?