In any category, monic/epic morphisms are equivalently characterized as the morphisms that map to injective maps under the co-/contravariant Hom-functors. Now I'm interested in the morphisms that map to surjective maps, i.e. that permit extensions/lifts of any morphisms along them. I will call them $\overline{\text{monic}}/\overline{\text{epic}}$ in the following.  I'd like to know if these concepts already have an established name, and if there is some theory around them: Maybe there is some criterion to determine if some morphism is $\overline{\text{monic}}/\overline{\text{epic}}$ analogous to the usual (co)kernel pair criterion for monic/epic morphisms. Are there other types of $\overline{\text{monic}}/\overline{\text{epic}}$ morphisms, as there are effective/regular/strict/strong/extremal mono- and epimorphisms?
63362
An $A$-element $x$ of $X$, written $x\in_A X$, is a map $x:A\to X$. If $f$ is a map with domain $X$, and $x\in_A X$ is an element, we write $f(x)$ to denote the composite of $f$ and $x$.  Now say that a map $f:X\to Y$ is cool if $$\forall A\quad \forall x, x'\in_A X\quad:\quad f(x) = f(x')\implies x = x'.$$  In the category of sets, a map is cool if and only if it is injective! In an arbitrary category, a map is cool if and only it is an monomorphism!  Call a map $f : X\to Y$ fresh if for all $A$, and for all $y\in_A Y$, there is an $x\in_A X$ such that $f(x) = y$. It would now be quite nice if, in the category of sets, a map is fresh if and only if it is surjective. 1. Is this true? 2. Also, in an arbitrary category, is a map fresh if and only if it is an ?
61846
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?
82693
I have a problem with enumerating a list of axioms in a custom style.   For instance, if I wanted to write down the axioms that a function will have to satisfy for it to be a homomorphism of rings, I'd prefer to call the axioms (HOM1) and (HOM2). I learnt that this is achievable by using this:  A map $f$ between two rings $R$ and $S$ is called a ring homomomorphism if they respect the algebraic structure in both of them. More precisely,  \begin{itemize} \item[HOM1] $f(r_1+r_2)=f(r_1)+f(r_2)$ \item[HOM2] $f(r_1 \cdot r_2)=f(r_1) \cdot f(r_2)$ \end{itemize}   But the problem is, I have absolutely no way of referring to exactly one of these axioms, say in a proof, without having to "brutally" type in the (HOM1) to refer to the axiom.   This looks slightly like not knowing how to refer to an equation when there are several of them, say, in an align construct.   Any help in achieving this will be appreciated.
48378
It seems to be that if a map is both an epimorphism and a monomorphism, it is not necessarily the case that it is an isomorphism.  However, in the category of sets, if a map is both an epimorphism and a monomorphism it is an isomorphism.  Are these accurate statements?
51026
I was thinking about the following: Say we got two groups $A$ and $B$, and we know that for any group C, there is a bijection  $$Hom(A,C) \to Hom(B,C).$$  Are $A$ and $B$ already isomorphic?  If the family of bijections was natural in $C$, this follows from Yoneda's lemma. But does it hold anyway without naturality, just from the cardinalities? I would suspect no but in simple cases (finitely generated). I'd like to hear any thoughts on how to come up with a proof or counterexample.  Also, does the answer change when we consider the contravariant case with an identification $Hom(C,A) \to Hom(C,B)$?
103469
$\widehat{\mathbb{R}/\mathbb{Z}}\cong\mathbb{Z}$, that is, every character of $\mathbb{R}/\mathbb{Z}$ is of the form $x\mapsto e(mx)$ for some integer $m$.   I was considering the dual of ${\mathbb{R}/\mathbb{Z}}$. What confused me is why we can assign an integer $m$ to the character $\chi.$  Thanks in advance! Any comment is appreciated!
55322
Let's fix some terminology first. A category $\mathcal{C}$ is preabelian if:  1) $Hom_{\mathcal{C}}(A,B)$ is an abelian group for every $A,B$ such that composition is biadditive,  2) $\mathcal{C}$ has a zero object,  3) $\mathcal{C}$ has binary products,  4) $\mathcal{C}$ has kernels and cokernels.  A category $\mathcal{C}$ is abelian if it is preabelian and satisfies:  5) every monomorphism is a kernel and every epimorphism is a cokernel.  Define the coimage of a map to be the cokernel of its kernel, and the image to be the kernel of its cokernel. We have the following commutative diagram:    where $\overline{f}$ is the only existing map (because of universality of kernel and cokernel).  I'm having trouble proving the following:     A preabelian category $\mathcal{C}$ is abelian iff $\overline{f}$ is an isomorphism.   The converse is easily shown, I'm having trouble proving $\Rightarrow$...
25509
Suppose we have two short exact sequences in an abelian category $$0 \to A \mathrel{\overset{f}{\to}} B \mathrel{\overset{g}{\to}} C \to 0 $$ $$0 \to A&#39; \mathrel{\overset{f&#39;}{\to}} B&#39; \mathrel{\overset{g&#39;}{\to}} C&#39; \to 0 $$ and morphisms $a : A \to A&#39;, b : B \to B&#39;, c : C \to C&#39;$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence $$0 \to \ker a \to \ker b \to \ker c \to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c \to 0$$ where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f&#39;$ and $g&#39;$.  It is not hard to show that the morphisms induced by $f, g, f&#39;, g&#39;$ exist, are unique, and that the sequence is exact at $\ker a, \ker b, \operatorname{coker} b, \operatorname{coker} c$. With the use of a somewhat large diagram shown , we can even construct the connecting morphism $d : \ker c \to \operatorname{coker} a$. However, I'm stuck showing exactness at $\ker c$ and $\operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?
69989
states that if $f:X \rightarrow Y$ is a morphism of ringed space, $G$ is a locally free $O_Y$-module, then $f^*G$ is a locally free $O_X$ module.     Suppose that $G$ is a locally free $O_Y$ module, hence it is a free $O_Y|_U=O_U$ module. We have an induced map  $$f^{-1}(U) \rightarrow U$$ given by restriction of morphism of ringed spaces.   Hence, as $\bigoplus O_U \simeq G|_U$, $f^*:Mod(O_U) \rightarrow Mod(O_{f^{-1}(U)})$ is functorial, $f^*(G|_U) = (f^*G)|_{f^{-1}(U)}$, we have isomoprhism,  $$ \bigoplus O_{f^{-1}(U)} \simeq (f^*G)|_{f^{-1}(U)} $$
75835
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?
75100
Are all  $\Pi A_\alpha \stackrel{\pi_i}\longrightarrow A_\alpha$ projection maps epic, given that $\Pi A_\alpha$ be the product of $A_\alpha$s? Of course, assuming that the product exists.
35529
There are various characterizations for an $R$-module to be projective. Two of them can be generalized to any category:  i) $P$ is an object such that given morphisms $\alpha: A \rightarrow B$ and $\varphi: P \rightarrow B$ such that $\alpha$ is an epimorphism, there exists a morphism $\widetilde{\varphi}: P \rightarrow A$ such that $\alpha \circ \widetilde{\varphi} = \varphi$. (Lifting property) In other words, the functor $\text{Hom}_{\mathsf{C}}(P,-): \mathsf{C} \rightarrow \mathsf{Set}$ preserves epimorphisms.  ii) $P$ is an object such that every epimorphism with codomain $P$ splits.  It is easy to see that (i) $\Rightarrow$ (ii) holds in general, by taking $\varphi = \text{id}_P$. Also if the category has the property that for every object $X$ there is an epimorphism $\pi:P \rightarrow X$ where $P$ satisfies (i) (an abelian category with this property is said to have  enough projectives), we can show the reverse implication (ii) $\Rightarrow$ (i).  My first question: Does (ii) $\Rightarrow$ (i) hold in general? I suspect the answer is no. If so, under which conditions does it hold?
65737
If $A$ is Noetherian, the surjectivity of a homomorphism from $A$ to itself is necessarily injective.   Thus, it's clear that if $\phi: A \to A$ is surjective, but not injective, the ring in question would have to be non-Noetherian.  Reformulating the question using the first isomorphism theorem. Given $\phi:A \to A$, we know that $A/ \ker \phi \cong \textrm{im} \phi \cong A$. When is it the case that the kernel is nontrivial?   Related Question: If every surjective homomorphism $A \to A$ is an isomorphism, does it follow that $A$ is Noetherian?
23929
I have the following question:  Is $(\mathbb{R}^2, lexicographic) \cong (\mathbb{R}, discrete)\times (\mathbb{R}, usual)$?  Note: the symbol $\cong$ denotes a homeomorphism between topological spaces  What I think:  I think that in fact there is a homeomorphism between those spaces, because is possible to express the lexicographic order as union of vertical lines with open extremes, I mean this is a basis for the lexicographic order. If I consider the Identical function I have the homeomorphism. Am I right?
40881
Let $X,Y$ be varieties over $\mathbb{C}$, $k(X), K(Y)$ be function fields of $X, Y$. Suppose $\pi: X \to Y$ is a dominant,  $\textit{injective}\ $ morphism, why the degree of the function field extension $[K(X) : K(Y)] =1$?  If $\phi : X \to  Y$ is a finite morphism, then the fibre is finite, and by semicontinuity theorem, let $n$ be the number of the points in the generic fibre, then I feel one should similarly have $[K(X) : K(Y)] =n$. But I don't know how to show that. Any suggestions or reference on this question?  $\textbf{Edit}$: I really want the morphism $\phi$ to be a morphism between locally finite type and finite morphism. To be precise, for any affine open set $U=\rm{Spec}(B) \subset Y$, there is an affine open over of $\pi^{-1}(U)$, such that each $\rm{Spec}(A_i)$ in this cover has the property $A_i$ is a finitely generated $B-$module. I don't know the corresponding definition of this sort of morphism, or is it just a finite morpism?
112578
For $M$ a simple $R$ module, and $N_i$ a family of $R$ modules, we have  $$ \hom (M, \bigoplus_i N_i) \cong \bigoplus_i \hom (M, N_i) $$ as abelian groups. Since the direct sum is the coproduct in the category of $R$ modules, we could rewrite this as  $$ \hom (M, \coprod_i N_i) \cong \bigoplus_i \hom (M, N_i). \tag{1} $$ This raises the question whether (1) is true for any abelian category, when $M$ is a simple object (defined I guess as an object such that any [regular?] monomorphism into it is an isomorphism). Does anyone know whether this is true?
61393
I have a question which should be pretty basic commutative algebra, but I can't find a reference and I'm stuck on proving a result myself, so here it goes:  Let $\varphi: S \to R$ be a morphism of commutative rings and let $M$ be a finitely generated $S$-module. Then, we can "extend the scalars of $M$" by considering $M \otimes_S R$, which is naturally an $R$-module. I would like to get a description of $\text{Supp}(M \otimes_S R)$ (as an $R$-module) involving $\text{Supp}(M)$ and $\varphi$.  Maybe this helps: by Eisenbud's Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a description of $\text{Ann}(M \otimes_S R)$ and it is not difficult to see that $\varphi(\text{Ann}(M))R \subset \text{Ann}(M \otimes_S R)$, but I fail to prove the other inclusion.