Let $h:X\to X$ denote an endomorphism of a module $X$ over $R$ satisfying $h\circ h = h$. Prove  $$X = Im(h)\oplus Ker(h)$$  I have this theorem:     If the composition $h=g\circ f$ of two homomorphisms $f:X\to Y$ and   $g:Y\to Z$ of modules $X,Y,Z$ over $R$ is an isomorphism, then the   following statements hold:      i) $f$ is a monomorphism      ii) $g$ is an epimorphism      iii) The module $Y$ is decomposable into the direct sum of $Im(f)$ and    $Ker(g)$   Wel, the composition $h\circ h$ is surely a bijection, and since i'ts an endomorphism, it's a homomorphism. So should it follow directly from this theorem?
124189
If $f:A\rightarrow A$ is an $R$-module homomorphism such that $ff=f$, show that   $$A=\ker f\oplus{\rm im}~f$$   Here is a part of what I made as a proof.  Let $a\in A$. $f(a)\in{\rm im}~f$, $f(a)\in A$ since $f$ is from $A$ to $A$, $a-f(a) \in A$ since $A$ is a module.  Consider $f(a-f(a))$. Since $f$ is a module homomorphism, $$f(a-f(a))=f(a)-ff(a)=f(a)-f(a)=0,$$ therefore $a-f(a)\in\ker f$. We have $a-f(a)\in\ker f$ and $f(a)\in{\rm im}~f$. $$(a-f(a))+f(a)=a-f(a)+f(a)=a$$ So $a\in A$ can be written as the sum of an element in $\ker f$ and an element in ${\rm im}~f$ and since $\ker f$ and ${\rm im}~f$ are submodules of $A$, we have $$A=\ker f+{\rm im}~f.$$  I was also able to show that the intersection of $\ker f$ and ${\rm im}~f$ is $\{0\}$.  I showed this proof to one of my senior and he said that what I actually did in my proof was show that $A$ is the Internal Direct Sum of $\ker f$ and ${\rm im}~f$.  Looking back at the problem, what is asked is to prove that $A$ is the External Direct Sum of $\ker f$ and ${\rm im}~f$.  I got the internal and external direct sum wrong but I understand them now. Now how would I go on proving the problem?  Here are some of my questions/ideas:   Since I'm trying to prove that $A$ is an external direct sum may I assume that $A=X\oplus Y$ where $X$ and $Y$ are modules? Then try to prove that that $\ker f=X$ and ${\rm im}~f=Y$? Also, if $A$ is the external direct product of $\ker f$ and ${\rm im}~f$ then that means that an element in $A$ is of the form $(x,y)$ where $x\in\ker f$ and $y\in{\rm im}~f$ right? Does that mean that I have to define $f$ in $X$ and $Y$? Is it possible to salvage my proof above? When does the internal direct sum equal to the external direct sum? I know that it is always the case that the two are isomorphic. If I remember correctly in groups it is possible for the internal direct product and the external direct product to be equal.   I hope you guys could clarify some things to me and point me in the right direction. Thank you!
120297
Suppose $T:V\to W$ and that $V$ is finite-dimensional.  I want to prove that $$\text{Im }T'=(\ker T)^0$$ where $T'$ is the dual/transpose map and $(\ker T)^0$ is the annihilator of the kernel.  I know that $\phi \in V'$ is an annihilator of $\ker T$ if and only if $$\phi(v)=0 \space \forall v\in \ker T$$ if and only if $$\phi(v)=0 \space \forall v \in V \text{ such that } T(v)=0$$  Now I also know that $\phi \in \text{Im }T'\subset V'$ if and only if $$\phi = f \circ T \text{ for some } f \in W'$$ I can see that if $v\in \ker T$, then this implies $\phi(v) = f(T(v)) = 0$, so $\phi \in (\ker T)^0.$  However, I'm having trouble showing the other way, that if $\phi \in (\ker T)^0$, then $\phi \in \text{Im }T'.$  Could anybody help?
89680
Let  $f: C \longrightarrow D$ be a morphism in an abelian category $\mathfrak{A}$ with kernel and cokernel both zero. How can I show that it is an isomorphism? I am not able to find it's inverse.   Let $g: C \longrightarrow D$ be a morphism in the abelian category $\mathfrak{A}$. Let $i: K \hookrightarrow C$ be the kernel of $g$. How can I prove that the induced map $\text{coker}\ i \rightarrow D$ is a monic?   I don't want to use Mitchell's embedding theorem.
63503
If $\phi$ is a homomorphism from the group $G$ with identity $e$ to the group $G'$ with identity $e'$, then $\phi$ is injective if and only if $\ker\phi=$?  I think the $Ker\phi = \{x\in G:\phi(x)=1\}$  because you are looking at an identity then it is all reals $0$ does not work a because if $x=0, e'x=0$  but then i realize it is injective how does that affect the kernel?
112259
There's a small detail in a proof of the Chinese Remainder Theorem for modules I don't understand when it comes to showing the normally constructed homomorphism is a surjection.     Suppose $R$ is a commutative unital ring, with $I_1,\dots,I_k$ pairwise comaximal ideals. How then for any $j$ can I find $r\in R$ such that $r\equiv 1\pmod{I_j}$ but $r\equiv 0\pmod{I_i}$ for $i\neq j$?    Thank you!
117962
I'm reading through my notes, and I have a theorem as the title says. As proof I have only a picture.    where $v:m\mapsto(0,...,0,1,0,...,0)$ and 1 is at $m$-th position. I know, that $\oplus_M R$ is a free module generated by set  $M$, but is this all? I do not see how the statement is proved.
4519
Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective.  I need advise or correction if something is incorrect with my proof. Thank you beforehand.  We must show that $g$ is injective, i.e for $x,y\in B, g(x)=g(y)\implies x=y$  Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$  $\implies g(f(a))=g(f(b))$  $\implies g\circ f(a)=g\circ f(b)$  $\implies a=b$ (by injectivity of $g\circ f$)  $\implies f(a)=f(b)$  $\implies x=y$  Would appreciate any correction in proof writing also!
27021
Suppose you have a group isomorphism given by the first isomorphism theorem: $$G/\ker(\phi) \simeq \operatorname{im}(\phi)$$ What can we say about the group $\ker(\phi)\times \operatorname{im}(\phi)$? In particular, when does the following hold: $$G\simeq \ker(\phi)\times \operatorname{im}(\phi)?$$ I ask this question because i want to prove that $GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{&gt;0}$, with $GL_n^+(\mathbb{R})$ the group of matrices with positive determinant. I proved that $SL_n(\mathbb{R})$ is a normal subgroup and that $GL_n^+(\mathbb{R})/  SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{&gt;0}$, using the surjective homomorphism $\det(M)$. I tried something with semidirect products but I got stuck.
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!
64810
The homomorphism $a \mapsto [[a \cdot f]]$ is supposed to be an isomorphism between $A$ and $H_n(D^n, S^{n - 1};A)$ for any homeomorphism $f: \Delta^n \to D^n$. I have proven that it is well defined and injective, but why is it surjective?  The group $H_n(D^n, S^{n - 1};A)$ is generated by elements of the form $a \cdot \sigma + b \cdot \tau + c \cdot \partial(\kappa)$ where $\sigma: \Delta^n \to D^n$, $\tau: \Delta^n \to S^{n - 1}$ and $\kappa: \Delta^{n + 1} \to D^{n + 1}$ and I do not see how an element of this form represents the same equivalence class of $[[m \cdot f]]$ for some $m \in A$. We can write $\sigma$ as $f f^{-1} \sigma$ to get $f$ in there, but I don't see that working out. Any hints?
103896
Let $R$ be a commutative ring with unit and $M$ a right $R$-module. Consider the assignment $\varphi: \operatorname{Hom}_{R} (R,M) \rightarrow M$, $f \mapsto \varphi(f) = f(1)$. Show that $\varphi$ is an isomorphism.  My approach: We have to prove $\varphi$ is a morphism, then $\varphi$ is a injection, surjection, hence $\varphi$ is an isomorphism. $\forall f,g \in \operatorname{Hom}_{R} (R,M)$ $\varphi(fg) = (fg)(1) = f(1)g(1) = \varphi(f) \varphi(g)$ $\Rightarrow \varphi$ is a morphism $\forall f,g \in \operatorname{Hom}_{R} (R,M)$ $\varphi(f) = \varphi(g) \Rightarrow f(1) = g(1) \Rightarrow f = g$ $\Rightarrow \varphi$ is injective $\Rightarrow \varphi$ is a homomorphism  How can I prove $\varphi$ is surjective to conclude $\varphi$ is an isomorphism? Sorry for my poor English.
113811
Let $\phi : C \to D$ be an injective ring homomorphism such that $\phi(1)=1$, where $1$ denotes the constant function $1$ and $C$, $D$ are the rings of continuous, respectively differentiable functions on $\mathbb{R}$. If $f \in C ,g \in D$ are such that $\phi(f)=g$, and $t_0\in \mathbb R$, is it true that $\phi (f-g(t_0))=g-g(t_0)$ ? This is used on page 944, . And why if the image of $\phi $ consists of only constant functions, then $\phi$ is not injective ? Please help. Thanks in advance.
82048
This is Exercise 2.4.E. from .     Show that a morphism of shaves of sets is an isomorphism if and only if it induces an isomorphism of all stalks.    My failed attempts:  I found some similar questions  and . But I still couldn't figure out the details.   I have shown the "only if" part. I have also shown that if the morphism induces an isomorphism of stalks, it is injective.   The surjective part confused me a lot. Here is a commutative diagram:   To show surjectivity, suppose $g\in \mathscr{G}(U)$. Its image in the stalks at $p$ is $(g,U)_p$. Since $\phi_p$ is an isomorphism for all $p$, we have $(f_p,U_p)\in \mathscr{F}_p$, such that $\phi_p(f_p,U_p)=(g,U)_p$ for all $p\in U$. Now I don't know how to glue the $f_p$ together. Since even we have the pairs $(f_q,U_q)\in \mathscr{F}_q$, we cannot show that their restrictions on $U_p\cap U_q$ are equal. We have to find a small enough open set to glue them.   I also tried the open cover argument from the right hand side:  Since $\prod(g,U)_p$ consists of compatible germs, there exists an open cover $U_i$, and section $g_i$ in $\mathscr{G}(U_i)$, such that $(g,U)_p=(g_i,U_i)$ for all $p\in U_i$. But this doesn't help much. Since $\phi_p$ are different maps, we still cannot conclude anything about the $(f_p,U_p)$.    Any help will be greatly appreciated!
18867
I am working on a school assignment and have been stuck on this question for some time.   Let $f : G \rightarrow G_1$ be a surjective homomorphism (also called epimorphism) from $G$ to another group $G_1$. Prove that $f(Z(G)) \subseteq Z(G_1)$  I know we start by assuming we have an element in $f(Z(G))$ then showing it is also in $Z(G_1)$. Also, $Z(G)$ is the center of $G$, so $Z(G)$ is all the elements of $G$ that commute with all elements in $G$. But how do we know that if we map the elements of $Z(G)$ to $G_1$, then their image will also commute with all the elements in $G_1$? Does it have something to do with using commutativity with the homomorphism? i.e. $f(gz) = f(g)f(z) = f(zg) = f(z)f(g)$.
120959
Let fix an arbitraly lattice of $\mathbb{C}$ and call it Λ. How to prove $H_1(\mathbb{C}/\Lambda, \mathbb{Z})$ is isomorphic to the lattice $\Lambda$ via the map $\gamma \mapsto \int_\gamma dz$ ? If once the map is proved to be well defined, the map is obviously group homomorphism and injection, and surjection may be managed. So, I want to check the map is well defined . Why $ \int_\gamma dz$ is in $Λ$? Thank you for your help.
89299
Let $f \colon A \to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B \otimes_A N$. Show that the homomorphism $g \colon N \to N_B$ which maps $y$ to $1 \otimes y$ is injective and that $g (N)$ is a direct summand of $N_B$.   [Define $p \colon N_B \to N$ by $p (b \otimes y) = by$, and show that $N_B = \operatorname{Im} (g) \oplus \operatorname{Ker} (p)$.]   My idea was to define an isomorphism \begin{align*} \phi \colon N_B &amp; \to \operatorname{Im} (g) \oplus \operatorname{Ker} (p) \\ b \otimes y &amp; \mapsto (p (b \otimes y), b \otimes y - 1 \otimes (g \circ p) (b \otimes y)) \end{align*}  It's easy to check that $\phi$ is injective and surjective. If I take the $B$-module structure of $N_B$ to be $$ b' . (b \otimes y) = (b'b) \otimes y $$ I only manage to show that $\phi$ is a homomorphism of $A$-modules, because $g$ seems to be only a homomorphism of $A$-modules: $$ b.g(y) = b.(1 \otimes y) = b \otimes y \neq 1 \otimes (b.y) = g (b.y) $$ (where $\neq$ means "not equal in general").  However, the way the question is phrased seems to suggest that the isomorphism should be one of $B$-modules.  Should I take the $B$-module structure of $N_B$ to be something else, for example $$ b'.(b \otimes y) = b \otimes (b'.y) $$ or should I show the result only for $A$-modules?  I've been confusing myself about this issue of "as $A$-module" or "as $B$-module" and would appreciate if someone could clarify. (I also wonder if, since homomorphisms are only fancy functions of sets, it somehow doesn't matter.)
38705
Let $R$ a commutative ring and $M$ a $R$-module. Prove there is a bijection between the set of $R[x]$-module structures on $M$ and $\mathrm{End}_{R-\mathsf{Mod}}(M)$.  I really don't even know how to start with this question.