I'm new in Category Theory and I just learned about epis and monos. I've been doing some exercises so far, but one of them is confusing me:  Let $i:\mathbb Z \to \mathbb Q$ be the inclusion of integers to rationals (equipped with the usual ring structure):  Show that $i$ an epimorphism Show that it is not surjective   I tried to assume $g,h: \mathbb{Q} \to R$ and that $g \circ i = h \circ i$. But now I have no idea how to get $g=h$. It is clear for me that it has something to do with the fact $i$ is an inclusion, but I just can't construct any argument. I attempted proof by contrapositive as well, assuming $g\neq h$, but the difficulty persists. Thanks in advance!
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I have two ring homomorphisms $f,g\colon \mathbb{Q}\to X$. I know that $f=g$ on the integers. How can I show that $f$ and $g$ agree on the rationals?   My attempt:  let $x,y \in \mathbb{Z}$, $y\neq 0$.  $$\begin{align*} f(x/y)&amp;=f(x)f(1/y) &amp;&amp;\text{because }f\text{ is a ring homomorphism}\\  &amp;=g(x)f(1/y) &amp;&amp;\text{because }f\text{ and }g\text{ agree on the integers} \end{align*}$$  I do not know how to rewrite $f(1/y)$ so that I can replace it with $g(1/y)$. Any hints will be much appreciated. Thanks.
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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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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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$R$ is a domain with characteristic $p$ ($p$ is prime). There is a homomorphism $f : R \to R$, $f(a)=a^p$. $f$ is called the Frobenius endomorphism. And I have known this.   When $R$ which is mentioned above is also a field, it is said that $f$ is an isomorphism. I think I just need to ensure $f$ is bijective. But I don't know how to prove it's surjective.    Actually I have got your ideas. In some of your opinions, if it is finite, then it can be surjective. Then what if it is infinite?  In my question, $R$ is a field as well as a domain. Then I wonder whether the "domain" is helpful for the proof of surjection or not.
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For the surjectivity part, I showed that if $(a,b)\in I$ then $(a,c)\in R_{1}$ and $(c,b)\in R_{1}$ and $(a,c)\in R_{2}$ and $(c,b)\in R_{2}$ for some $c\in S$. Now for arbituary $(x,y)\in R_{1}$ we have $(x,z)\in I$ which implies that $(y,z)\in R_{2}$. But also $(y,z)\in R_{1}$, so $R_{1}$ is surjective. Similar argument for $R_{2}$  And for Injectivity, I've tried to show that if $aR_{1};R_{2}c$ and $bR_{2};R_{1}c$, then $a=b$ but I am not entirely sure what $a=b$ means in this context. Do I have to show that there is an equivalence relation containing (a,b)?  edit: $R_{1};R_{2}$ is a composition of two relations and $I=R_{1};R_{2}$
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Show there is a surjective homomorphism from $\mathbb{Z}\ast\mathbb{Z}$ onto $C_2\ast C_3$, where $\ast$ denotes the coproduct in the category $\mathsf{Grp}$.   Note the exercise in this book (Algebra Chapter 0, Aluffi) is meant, or at least hinted at, to be done using the universal property of the coproduct, ie: coproduct of $A$ and $B$ in $\mathsf{Grp}$ is initial in the category $\mathsf{Grp}^{A,B}$, instead of using the definition of the free product.  By property of the coproduct, for any object $A$ and two homomorphisms $f_1,f_2:\mathbb{Z}\to A$, and two homomorphisms $i_1,i_2:\mathbb{Z}\to\mathbb{Z}\ast\mathbb{Z}$, there exists a unique homomorphism $\sigma:\mathbb{Z}\ast\mathbb{Z}\to A$  such that the diagram commutes, ie: $f_1=\sigma i_1$ and $f_2=\sigma i_2$. Now let $A=C_2\ast C_3$. Now what?
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Is my proof below correct? What specific property of rationals did I exploit in my proof? It looks like the property I exploited is the following: Given any positive rational, I can always write it as sum of arbitrary number of positive rationals, whereas given any positive integer I cannot write it as a sum of arbitrary number of positive integers. Has it got to do with the fact that $\mathbb{Q}$ is a field?       Problem Prove that the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic.   Solution Let there exist an isomorphism between $\mathbb{Z}$ and $\mathbb{Q}$. Now consider the element $1_{\mathbb{Q}}$. We then have $\phi(1_{\mathbb{Q}}) = z \in \mathbb{Z}$. Since $\phi$ has to be a bijection, $z$ cannot be zero, since $\phi(0) = 0$.  Now consider the element $\left(\dfrac1{z+1}\right)_{\mathbb{Q}}$. We now have $$z = \phi(1_{\mathbb{Q}}) = \phi\left(\underbrace{\left(\dfrac1{z+1}\right)_{\mathbb{Q}} + \left(\dfrac1{z+1}\right)_{\mathbb{Q}} + \cdots + \left(\dfrac1{z+1}\right)_{\mathbb{Q}}}_{z+1 \text{ times }} \right) = (z+1) \phi\left(\left(\dfrac1{z+1}\right)_{\mathbb{Q}}\right)$$ However, there is no element in $y \in \mathbb{Z}$ such that $(z+1)y = z$.    First update  Actually I realize that I complicated it unnecessarily. Instead, we can do like this. Since $\phi$ is an isomorphism, we have $\phi(q_{\mathbb{Q}}) = 1_{\mathbb{Z}}$ for some $q \in Q$. However, $$\phi(q) = \phi(q/2+q/2) = 2\phi(q/2)$$ And there is no $y \in Z$, such that $2y=1$. Hence, $\phi(q/2)$ remains unmapped.    Thanks
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The question says it all, but let me recall the definitions. A magma $(X,\cdot)$ is a set with a binary operation $\cdot:X\times X\to X$ (without any further assumptions like associativity). A morphism between magmas is a map $f:X\to Y$ with $f(a\cdot b)=f(a)\cdot f(b)$ for all $a,b\in X$. A morphism $f$ is epi [mono] if one can cancel it from the right [left] ($g\circ f =h\circ f$  $\Rightarrow g=h$ [respectively $f\circ g=f\circ h$ $\Rightarrow g=h$]).  Using free magmas with one generator shows that monomorphisms are injective. The question is whether epimorphisms are always surjective (in many other categories this fails, e.g. in MONOID).
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I would like to prove that the additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity.  Now every $X \in \mathbb{Q}/\mathbb{Z}$ is of the form $\frac{p}{q} + \mathbb{Z}$ for $0 \leq \frac{p}{q} &lt; 1$ for a unique $\frac{p}{q} \in \mathbb{Q}.$ This suggest taking the map $f:\mathbb{Q}/\mathbb{Z} \mapsto C^{\times}$ defined with the rule $$f(\frac{p}{q} + \mathbb{Z}) = e^{\frac{2\pi i p}{q}}$$  where $\frac{p}{q}$ is the mentioned representative.   Somehow I have problems showing that this is a bijective function in a formal way. I suspect I do not know the properties of the complex roots of unity well enough.  Can someone point me out (perhaps with a hint) how to show that $f$ is injective and surjective?
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I have a problem with something in abstract algebra. The problem is the following:  Prove or disprove that the mapping $\theta: \mathbb{C} \to \mathbb{C}$ defined by $\theta(a+bi)=a-bi$ is a ring isomorphism.  I know that it should be an isomorphism. I know how to show that the mapping is one-to-one, and I know how to show that the mapping respects both addition and multiplication as defined in $\mathbb{C}$. But how would you clearly show that $\theta$ is onto?  Thanks in advance for any help!
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I have the following problem:  Is it possible to construct ring homomorphism from $M_2(\mathbb Z)\to \mathbb Z$, or in other words, a homomorphism from ring of all $2\times2$ matrices over the integers into integers?  I tried determinant, trace and mapping that maps matrix to it's element in the position $(1,1)$ but none of that obviously works, which led me to believe there might not be such a homomorphism. Determinant doesn't work because it is obviously not a linear map.  Trace doesn't work because it doesn't respect multiplication.  And the mapping to the position $(1,1)$ also doesn't work because $$\begin{pmatrix}1&amp;1\\1&amp;1\end{pmatrix}\begin{pmatrix}1&amp;1\\1&amp;1\end{pmatrix}=\begin{pmatrix}2&amp;2\\2&amp;2\end{pmatrix}.$$
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1) How can I show that any one to one endomorphism of an Artinian module is an automorphism.   2) I also want to show that any onto endomorphism of a Noetherian module is an automorphism.   I need counterexamples for modules without chain conditions.
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Prove that $\mathbb{Q}$ isn't homeomorphic to $\mathbb{Z}$  I know that a homeomorphism is a bijection function from two sets where both the function and its inverse are continuos.   So I figure that if I show that there is no continuos invective mapping from $\mathbb{Q}\rightarrow\mathbb{Z}$ then I have done the proof. But I do not know how to do this (if his is the right way). How can I start to show this?
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$\widehat{\mathbb{R}/\mathbb{Z}}\cong\mathbb{Z}$, 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!
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If $f: \mathbb Q\to \mathbb Q$ is a homomorphism, prove that $f(x)=0$ for all $x\in\mathbb Q$ or $f(x)=x$ for all $x$ in $\mathbb Q$.  I'm wondering if you can help me with this one?
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I have the following three question about flat modules.   Why is not $\mathbb{Z}$ a flat $\mathbb{Z}$-module. Why is $\mathbb{Q}$ a flat module $\mathbb{Z}$-module. I need an example of a module which is not flat, nor injective, nor projective.    Thanks a lot!   Note:  I need only a hint to attack those problems not the complete answer.
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I am having trouble with the following problem.      Let $R$ be an integral domain, and let $a \in R$ be a non-zero element. Let $D = \{1, a, a^2, ...\}$. I need to show that $R_D \cong R[x]/(ax-1)$.    I just want a hint.  Basically, I've been looking for a surjective homomorphism from $R[x]$ to $R_D$, but everything I've tried has failed. I think the fact that $f(a)$ is a unit, where $f$ is our mapping, is relevant, but I'm not sure. Thanks