This short note originated from a short talk of basic category theory in 2022. Category theory was first introduced by Samuel Eilenberg and Saunders Mac Lane in the 20th century. It rapidly became a strong tool in almost all domains of mathematics. Category theory studies the abstract structures of different mathematical objects, and it also has applications on computer science, including AI.1. CategoriesIn this section, we will first lay out the general definition of categories, then we give some concrete and common examples of categories.Definition 1. A category $\mathcal{C}$ consists of the following data:A class of objects, denoted by ${\rm{Ob}}(\mathcal{C})$.To each pair of objects $A, B$, a set ${\rm{Hom}}(A, B)$ of morphisms from $A$ to $B$.To each triple of objects $A, B, C$, a composition law $${\rm{Hom}}(A,B)\times {\rm{Hom}}(B, C)\longrightarrow {\rm{Hom}}(A, C),\ (f,g)\longmapsto f\circ g.$$Moreover, it subjects to the following axioms:(1) Composition is associative, i.e. $(f\circ g)\circ h = f\circ (g\circ h)$ for morphisms $f,g,h$.(2) For each object $A$, there is a unique identity morphism $1_{A}: A\rightarrow A$ such that $1_{A}\circ f=f\circ 1_{A}$ if the composition makes sense.Examples 2. (1) The category $\textbf{Sets}$ of sets, whose objects are sets and morphisms are functions.(2) The category $\textbf{Groups}$ of groups, whose objects are groups and morphisms are homomorphisms.(3) The category $\textbf{Ab}$ of abelian groups, whose objects are abelian groups and morphisms are homomorphisms.By Example 2 (3), one can observe that ${\rm{Ob}}(\textbf{Ab})\subset{\rm{Ob}}(\textbf{Groups})$, and for any $x,y\in{\rm{Ob}}(\textbf{Ab})$, ${\rm{Hom}}_{\textbf{Ab}}(x,y)={\rm{Hom}}_{\textbf{Groups}}(x,y)$. This leads to the definition of subcategories.Definition 3. Let $\mathcal{C}$ be a category. A subcategory of $\mathcal{C}$ is a category $\mathcal{D}$ such that ${\rm{Ob}}(\mathcal{D})\subset{\rm{Ob}}(\mathcal{C})$ and ${\rm{Hom}}_{\textbf{Ab}}(x,y)\subset{\rm{Hom}}_{\textbf{Groups}}(x,y)$ for all $x,y\in{\rm{Ob}}(\mathcal{D})$. The subcategory $\mathcal{D}$ is said to be full if we have ${\rm{Hom}}_{\mathcal{D}}(x,y)\cong{\rm{Hom}}_{\mathcal{C}}(x,y)$ for all $x,y\in{\rm{Ob}}(\mathcal{D})$.An invertible morphism in a category is the so-called isomorphism.Definition 4. Let $\mathcal{C}$ be a category and $f$ is a morphism in $\mathcal{C}$. We say that $f$ is an isomorphism if there is a morphism $g$ in $\mathcal{C}$ such that $f\circ g=1$ and $g\circ f=1$ when the composition makes sense. The morphism $g$ is called the inverse of $f$ and is denoted by $f^{-1}$.2. FunctorsLike objects in the category, we could define transformations between categories. In fact, one could view categories as objects of some bigger category.Definition 5. Let $\mathcal{C}$ and $\mathcal{D}$ be categories. A functor $F:\mathcal{C}\rightarrow\mathcal{D}$ is an assignment that To each $x\in{\rm{Ob}}(\mathcal{C})$, it assigns $Fx\in{\rm{Ob}}(\mathcal{D})$.For all $x,y\in{\rm{Ob}}(\mathcal{C})$ and $f\in{\rm{Hom}}_{\mathcal{C}}(x,y)$, it assigns $F(f)\in{\rm{Hom}}_{\mathcal{D}}(Fx,Fy)$ such that $F(1)=1$ and $F(fg)=F(f)F(g)$ when the composition makes sense.A contravariant functor is a functor $G:\mathcal{C}^{opp}\rightarrow\mathcal{D}$.

My question: It is well known that morphisms between the objects of the category are structure-preserving, but I found that in a textbook it said that morphisms are often structure-preserving. Does this mean that there can be a morphism that is not structure-preserving?Answer 1: A category doesn't have to consist of sets with some additional structure and maps between those preserving the structure.Examples for categories that are not of this kind are:Given any group $G$, we can form a category with one object $*$ and for each $g\in G$ a morphism $\varphi_g\colon *\to *$, where composition of morphisms is defined via the group operation and $\operatorname{id}_* = \varphi_{e}$ for $e\in G$ the identity element.Given a poset $(P,\le)$ we can form a category with set of objects $P$ and exactly one morphism $x\to y$ for each $x,y\in P$ with $x\le y$.The homotopy category of topological spaces, where the objects are topological spaces and a morphism $X\to Y$ is a homotopy class $[f]$ of a continuous map $f\colon X\to Y$.Answer 2: I think the issue begins here:It is well known that morphisms between the objects of the category are structure-preservingThis is not the case. The notion of a category generalises the notion of 'sets-with-structure and structure-preserving functions', such as groups and homomorphisms, or topological spaces and continuous maps.But the extent of the generality is extreme: the objects of a category need not even be sets, and the morphisms of a category need not be functions.For example, every monoid can be considered as a one-object category, where the elements of the monoid are the morphisms from the single object to itself. In this case, the 'object' is just a placeholder—it has no notion of 'structure'—and the morphisms are certainly not functions (in general).Categories that 'look like' sets-with-structure and structure-preserving morphisms are called concrete categories. What this means is that the category $\mathcal{C}$ comes equipped with a faithful functor $U : \mathcal{C} \to \mathbf{Set}$. An object $A$ of $\mathcal{C}$ can be thought of has having 'underlying set' $U(A)$, and a morphism $f : A \to B$ can be thought of as having 'underlying function' $U(f) : U(A) \to U(B)$. However, concrete categories are still more general than sets-with-structure and structure-preserving morphisms. There may not actually be any structure to speak of.This question was asked on MathStackExchange on February 19, 2019. At that time, I was in high school and had to balance the college entrance examination and interest. So I didn't have much time to study mathematics, and my math level at that time was not satisfactory, but I couldn't do anything about it.

Let $A,B$ be commutative rings in characteristic $p$. Let $\phi_{A}:A\rightarrow A,\phi_{B}:B\rightarrow B$ be the Frobenius morphisms, i.e. the $p$-power maps. If we have ${\rm{colim}}_{n\in\mathbb{N}}A\cong {\rm{colim}}_{n\in\mathbb{N}}B$, where the transition maps are Frobenius morphisms, can we show that $A\cong B$ ?Answer: No. Recall that an $\mathbb{F}_p$-algebra $R$ is perfect if the Frobenius map $\varphi : R \ni r \mapsto r^p \in R$ is an isomorphism. The colimit of powers of the Frobenius you describe is the perfection of an $\mathbb{F}_p$-algebra, and it is named that because it's the left adjoint of the inclusion of perfect $\mathbb{F}_p$-algebras into $\mathbb{F}_p$-algebras. This makes perfect $\mathbb{F}_p$-algebras a reflective subcategory of $\mathbb{F}_p$-algebras, which among other things implies that perfection fixes any perfect $\mathbb{F}_p$-algebra.This is all abstract context for the following more specific counterexample: take $A = \mathbb{F}_p[x]$, whose perfection is $\mathbb{F}_p[x^{\frac{1}{p^{\infty}}}]$, the ring obtained by adjoining all $p^n$-power roots of $x$. Then take $B = \mathbb{F}_p[x^{\frac{1}{p^{\infty}}}]$ to be the perfection. More generally, we could take $A$ to be any $\mathbb{F}_p$-algebra which is not perfect and $B$ to be its perfection.These notes by Bhatt have a more general claim right before Remark 1.4 about when two algebras have isomorphic perfections but I'm not familiar enough with universal homeomorphisms to say anything about it.

My question: Let $\cal{C},\cal{D}$ be categories (resp. stacks). Let $F:\cal{C}\rightarrow\cal{D}$ be an essentially surjective functor, i.e. surjective on isomorphism classes of objects. Then is $F$ an epimorphism in the category of small categories (resp. stacks)?Answer: No. For example, any functor between categories with one object is essentially surjective, but e.g. if $M_1, M_2$ are two nonzero monoids then the inclusion $M_1 \to M_1 \oplus M_2$ of a direct summand, thought of as a functor between one-object categories, is not an epimorphism of categories.Keep in mind, though, that "epimorphism in the category of small categories" is not obviously the "right" concept in any particular application, for multiple reasons. It discards natural transformations, so you're ignoring the fact that you are really working in a $2$-category; and there are also various notions of "epimorphism" you might want in any particular case.

My question: Let $S$ be a base scheme and let $(Sch/S)_{fppf}$ be a big fppf site. Let $U$ be a scheme over $S$. Suppose that there is a surjective morphism $\Phi_{U}:U\rightarrow U$. Then can we show that the induced morphism of sheaves $h_{U}\rightarrow h_{U}$ is locally surjective? It seems that this is false.Note that $h_{U}={\rm{Hom}}(-,U)$ is a representable sheaf. A map of sheaves $F\rightarrow G$ on $(Sch/S)_{fppf}$ is locally surjective if for every scheme $U\in{\rm{Ob}}((Sch/S)_{fppf})$ and every $s\in G(U)$, there exists a covering $\{U_{i}\rightarrow U\}_{i\in I}$ such that for all $i$, $s|_{U_{i}}$ is in the image of $F(U_{i})\rightarrow G(U_{i})$.Answer: Let $S:={\rm Spec}(k)$ be a field and let $U={\rm Spec}(k[t]/t^2)$.The ring $k[t]/t^2$ is a $k$-algebra and there is a map of $k$-algebras $k[t]/t^2\to t$ sending $t$ to $0$ so we get morphisms $U\to{\rm Spec}(k)$ and ${\rm Spec}(k)\to U$.Let $a:U\to U$ be the composition of these two morphisms. This is surjective.Now if the answer to your question is positive, there is a faithfully flat morphism $b:V\to U$ and an element $\rho\in{\rm Hom}(V,U)$ such that $a\circ\rho=b$ (applying local surjectivity to ${\rm Id}_U\in{\rm Hom}(U,U)$). In other words, the morphism $a_V:U\times_U V\to V$ has a section. However, if the morphism $a_V$ has a section then so has the morphism $S\times_U V\to V$ obtained by base-change from the morphism $S={\rm Spec}(k)\to U$. The morphism is $S\times_U V\to V$ is a closed immersion, so it is an isomorphism, since it has a section. Isomorphisms descend along faithfully flat morphisms, so we deduce that ${\rm Spec}(k)\to U$ is an isomorphism, which is false. So this provides a counterexample.

My question: Let $\cal{C}$ and $\cal{D}$ be two groupoids, i.e. the category whose morphisms are isomorphisms. Let $F:\cal{C}\rightarrow\cal{D}$ be a fully faithful functor from $\cal{C}$ to $\cal{D}$. Then is $F$ injective on objects? In other words, is the object function $F:{\rm{Ob}}(\cal{C})\rightarrow{\rm{Ob}}(\cal{D})$ injective?Answer: No. Given any set $X$ we can construct a groupoid called the indiscrete groupoid on $X$, which has a unique (iso)morphism $x \to y$ for $x, y \in X$. Every map of sets $f : X \to Y$ induces a fully faithful functor between the corresponding indiscrete groupoids, whether or not $f$ is injective.What is true is that a fully faithful functor $F : C \to D$ between categories (not necessarily groupoids) induces an injection on isomorphism classes of objects. To see, this let $i : F(c_1) \cong F(c_2)$ be an isomorphism. Because $F$ is full, $i = F(i')$ for some $i' : c_1 \to c_2$, and similarly $i^{-1} = F(j)$ for some $j : c_2 \to c_1$. We have $F(i' \circ j) = F(i') \circ F(j) = \text{id}_{F(c_2)} = F(\text{id}_{c_2})$ so, since $F$ is faithful, $i' \circ j = \text{id}_{c_2}$, and similarly for the other composite. So $j = (i')^{-1}$ and $i' : c_1 \cong c_2$ is an isomorphism. (In the indiscrete groupoid every object has been made isomorphic.)

My question: Let $X,Y$ be schemes. Let $X\rightarrow Y,X\rightarrow X, Y\rightarrow Y$ be morphisms of schemes. Why the morphism $U\times_{X}X\rightarrow U\times_{Y}Y$ is the base change of $X\rightarrow X\times_{Y}Y$ by $U\times_{Y}Y\rightarrow Y$?Here is the diagram I try, where the triangle is commutative. But I found that $(U\times_{Y}Y)\times_{Y}(X\times_{Y}Y)=U\times_{Y}X\times_{Y}Y=U\times_{Y}X$, i.e. I can not obtain the desired $U\times_{X}X$. What mistakes do I make?Here is the context in question, from Xinwen Zhu's paper Affine Grassmannians and the geometric Satake in mixed characteristic (arXiv link):Lemma A.2. For any étale morphism of algebraic spaces $X\to Y$, the relative Frobenius morphism $X\to X\times_{Y,\sigma_Y}Y$ induced by $\sigma_X$ is an isomorphism.Proof. We first assume that $X$ is a scheme. Then $X\to X\times_{Y,\sigma_Y}Y$ a schematic radical étale surjective map, and therefore is an isomorphism by \cite[Theorem 17.9.1]{EGA4-4}. For general $X$, choose an étale cover $U\to X$ by a scheme $U$. Then we have $U\to U\times_{X,\sigma_X}X\to U\times_{Y,\sigma_Y}Y$, with the first map and the composition map being isomorphisms. Therefore, the second map is an isomorphism as well. Note that $U\times_{X,\sigma_X}X\to U\times_{Y,\sigma_Y}Y$ is nothing but the base change of $X\to X\times_{Y,\sigma_Y}Y$ along the étale cover $U\times_{Y,\sigma_Y}Y\to Y$. Therefore, $X\to X\times_{Y,\sigma_Y}Y$ is also an isomorphism.Answer: The diagrams are a bit more complicated than you're using here. First, we have a base change diagram defining $X\times_{Y,\sigma_Y} Y$ as follows:$$\begin{CD} X\times_{Y,\sigma_Y} Y @>>> Y\\ @VVV @VV{\sigma_Y}V \\ X @>>> Y \end{CD}$$The morphism $X\to X\times_{Y,\sigma_Y} Y$ is induced by the identity morphism $X\to X$, the morphism $X\to Y$, and the universal property of the fiber product.The diagram defining $U\times_{Y,\sigma_Y} Y$ is as follows:$$\begin{CD} U\times_{Y,\sigma_Y}Y @>>> X\times_{Y,\sigma_Y} Y @>>> Y\\ @VVV @VVV @VV{\sigma_Y}V \\ U @>>> X @>>> Y \end{CD}$$Now base change the entire first diagram plus the induced morphism $X\to X\times_{Y,\sigma_Y} Y$ by the map $U\times_{Y,\sigma_Y} Y\to Y$ to get the claimed map.The previous article What is base change map ? is indeed inspired by this question.