In Paolo Aluffi's Algebra: Chapter 0, there is an exercise which asks     Prove that there is a surjective homomorphism (an epimorphism) from $\Bbb Z*\Bbb Z$ onto $C_2*C_3$, where $A*B$ is the coproduct of $A$ and $B$ in $\mathsf{Grp}$. One can think of $\Bbb Z*\Bbb Z$ as a group with two generators $x, y$ subject to no relations whatsoever.   I know that this question has been asked previously  and . But neither of them have correct answer that uses the machinery that has already been introduced. If you have not read the book, try not to use anything about coproducts in the answer other than the fact that they are initial objects in the category $C^{A,B}$.  Here's what I thought:  I know that since $\Bbb Z*\Bbb Z$ is an initial object, there is a unique morphism from $\Bbb Z*\Bbb Z$ onto $C_2*C_3$ in the category $C^{A,B}$. The only question is how to prove surjectivity of the morphism. We can construct morphisms from $\Bbb Z \to C_2 \text{ and } C_3$ that are surjective. Somehow we need to compose the two to form the required morphism.
103428
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?
54096
Is the following proof correct, or are certain steps illegal?  Suppose $f:G_1\to G_2$ is a surjective group isomorphism, $N_1\lhd G_1, N_2\lhd G_2$ normal subgroups, with $N_1=f^{-1}(N_2)$. Prove that $(G_1/N_1)\cong (G_2/N_2)$.  Proof:  As $f$ is surjective, we an use the first isomorphism theorem to get $G_1/\ker(f)\cong G_2$, which implies that $$(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)$$  Using the first isomorphism theorem we also know that $$N_1/\ker (f)\cong f(N_1)=f(f^{-1}(N_2))=N_2$$  Combining this we can see that $(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)\cong G_2/N_2$.  Using the third isomorphism theorem we have that: $$ (G_1/\ker f)/(N_1/\ker f)\cong G_1/N_1$$  And from the last two lines we have $G_1/N_1\cong G_2/N_2$ $\tag*{$\Box$}$
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I can't figure out why such an epimorphism cannot exist.  I think the fact that a bijection exists between the $\mathbb{Q}$ and $\mathbb{Z}$ throws me off a little. What is it about having a homomorphic surjection that causes problems?
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?
109350
I am trying to determine if $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $. I noticed that $\Bbb Z_6$ has a generator $1$ and $\Bbb Z_3 \times \Bbb Z_2$ has generator $(1,1)$. Now I set up the bijection $f :\Bbb Z_6 \to   \Bbb Z_3 \times \Bbb Z_2 $ where \begin{eqnarray} f(1)&amp;=&amp;(1,1)\\ f(1+1)&amp;=&amp;(1,1)+(1,1) = (2,0)\\ f(2+1)&amp;=&amp;(2,0)+(1,1) = (0,1)\\ f(3+1)&amp;=&amp;(0,1)+(1,1) = (1,0)\\ f(4+1)&amp;=&amp;(1,0)+(1,1) = (2,1)\\ f(0)&amp;=&amp;(1,1)+(2,1) = (0,0)\\ \end{eqnarray} So basically $f(1^n) = (1,1)^n$ where $1^n$ means we add the $1$ to itself $n$ times and take the result mod 6, so as per usual $1^n$ means applying the group operation on 1 to itself for $n$ times. So this is a bijection but I am stuck on showing that $f(ab)=f(a)f(b)$ for all $a,b \in \Bbb Z_6$. I could check all possibilities but there must be a better way to show this. I was thinking something like $f(1^n1^m) = f(1^{n+m})=(1,1)^{n+m}=(1,1)^n(1,1)^m = f(a)f(b) $ because every element in $\Bbb Z_6$ can be expressed as some power of $1$.  Is this right? And is it enough to show $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $?
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let $K$ be a category with equalizers, show that every extremal epimorphism is epic. for composable morphisms $f: A \rightarrow B $ and $g: B \rightarrow C$ in $K$, show that if $gf$ is an extremal epimorphism, then $g$ is an extremal epimorphism. let $K$ be a category with pullbacks.  For composable morphisms $f$ and $g$ as given in #2, show that if $g$ and $f$ are extremal epimorphisms, then $gf$ is an extremal epimorphism. show that if an extremal epimorphism is monic, then it is an isomorphism.   here is what I know:  In any category $K$, an arrow $f: A \rightarrow B$     is called an isomorphism if there is an arrow $g: B \rightarrow A$ in $K$ such that $ g \circ f = I_{A} $ and $ f \circ g = I_{B}$ is called a monomorphism, if given any $g,h : C \rightarrow A, fg=fh$ implies $g=h$ is called an epimorphism, if given any $i,j: B \rightarrow D, if = jf$ implies $i=j$ is an extremal epimorphism, if for each commutative diagram, $f = mh$ where if $m$ is monic, then $m$ is an isomorphism   I have tried to answer #2:     since $gf$ is an extremal epimorphism, then $gf = mh$ where $m$ is monic and an isomorphism. now let $g = nk$ where $n$ is monic.     I now need to show that $n$ is an isomorphism.  my problem is how do I do this?  I'm trying to visualise things using diagrams, but I'm lost.  Here is what I have so far for #1:  let $f: A \rightarrow B$ be and extremal epimorphism.  consider $ g,h: B \rightarrow C$ such that $gf=hf$ so I need to prove $g=h$  let $e: E \rightarrow B$ be the equaliser of $g, h$  now I don't know where to go from here to prove $g=h$.    I know that $f$ factors through the equaliser.  since $e$ is an equaliser, there is a unique morphism  $\varphi : A \rightarrow E$ therefore I now have $f=e \varphi$   which means that $e$ is now isomorphic since $f$ is an extremal epimorphism.    So can I now conclude $g=h$ ?
21328
The isomorphism $\mathbb{Z}[i]/(a+bi) \cong \Bbb Z/(a^2+b^2)\Bbb Z$ is , when the integers $a$ and $b$ are coprime. But what happens when they are not coprime, say $(a,b)=d&gt;1$?      — For instance if $p$ is prime (which is not coprime with $0$) then $$\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong \begin{cases} \mathbb{F}_{p^2} &amp;\text{if } p \equiv 3 \pmod 4\\ \mathbb{F}_{p} \times \mathbb{F}_{p} &amp;\text{if } p \equiv 1 \pmod 4 \end{cases}$$ (because $-1$ is a square mod $p$ iff $(-1)^{(p-1)/2}=1$).  — More generally, if $n=p_1^{r_1} \cdots p_m^{r_m} \in \Bbb N$, then  each pair of integers $p_j^{r_j}$ are coprime, so that by CRT we get $$\mathbb{Z}[i]/(n) \cong \mathbb{Z}[i]/(p_1^{r_1}) \times \cdots \times \mathbb{Z}[i]/(p_m^{r_m})$$  I was not sure how to find the structure of $\mathbb{Z}[i]/(p^{r}) \cong (\Bbb Z/p^r \Bbb Z)[X] \,/\, (X^2+1)$ when $p$ is prime and $r&gt;1$.  — Even more generally, in order to determine the structure of $\mathbb{Z}[i]/(a+bi)$ with $a+bi=d(x+iy)$ and $(x,y)=1$, we could try to use the CRT, provided that $d$ is coprime with $x+iy$ in $\Bbb Z[i]$. But this is not always true: for $d=13$ and $x+iy=2+3i$, we can't find Gauss integers $u$ and $v$ such that $du + (x+iy)v=1$, because this would mean that $(2+3i)[(2-3i)u+v]=1$, i.e. $2+3i$ is a unit in $\Bbb Z[i]$ which is not because its norm is $13 \neq ±1$.  — I was not able to go further. I recall that my general question is to known what $\mathbb{Z}[i]/(a+bi)$ is isomorphic to, when $a$ and $b$ are integers which are not coprime (for instance $a=p^r,b=0$ or $d=(a,b) = a^2+b^2&gt;1$).  Thank you for your help!
53911
I've been trying to wrap my head around the basic concepts of category theory, and I thought I would attempt to illustrate what I understand with the category of sets, probably the easiest example. Particularly, I've been trying to prove that $id_A$ (the identity morphism on $A$, for all $A \in Obj(\mathbf{Set})$) is $1_A \colon A \rightarrow A, x \mapsto x$.  This is a very intuitive and reasonable statement, and it's trivial to prove that $1_A$ is indeed an identity morphism on $A$, and I suppose uniqueness of $id_A$ can be demonstrated analogously to uniqueness of the identity element in a monoid (considering the subcategory which has $A$ as its only object, and endofunctions on $A$ as its only morphisms).  In this manner, it is not hard to prove that the proposition in the title is true, but this demonstration requires to make an assumption or guess as to what could $id_A$ be. Specifically, the scheme of the proof is: assume $id_A$ = $1_A$, see that it works with the definition of an identity morphism, show that the identity morphism is unique, and, in conclusion, $id_A$ can only be $1_A$. What I'm looking for, nonetheless, is a somehow more direct proof, that doesn't assume $id_A$ = $1_A$ at the start. I want to place myself in a state of little or no knowledge about sets and functions, and under this assumption, why would I assume $id_A$ = $1_A$ at first? Why not try with $id_\mathbb{Z}$ = $f \colon \mathbb{Z} \to \mathbb{Z}, x \mapsto x^2 + 1$, for example? It wouldn't work, but I don't have any reason to think that $1_\mathbb{Z}$ is a better guess for $id_\mathbb{Z}$.  I suppose that the proof for which I'm asking would work for categories of sets with additional structure, and probably for posets as well, although I'm not clear as to what modifications it would require to work.  Thanks.
93710
Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$.
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I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$  Can you tell me if my proof is right:  $\mathbb{Z}/m\mathbb{Z}$ and $\mathbb{Z} / n \mathbb{Z}$ are both finite free $\mathbb{Z}$-modules with the basis consisting of one single element $\{ 1 \}$. So $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$ has the basis $\{ 1 \otimes 1 \}$.  Therefore, any element in $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$  is of the form $(ab) 1 \otimes 1$ and any element in $\mathbb{Z}/ \gcd(m,n)\mathbb{Z}$ is of the form $k 1 = k$ where $k \in \{ 0, \dots , \gcd(n,m) \}$.  I would like to construct an isomorphism that maps $ab$ to some $k$. Let this map be $ab (1 \otimes 1) \mapsto ab \bmod \gcd(n,m)$.  This is a homomorphism between modules: it maps $0$ to $0$ because it maps the empty sum to the empty sum. It also fulfills $f(a + b) = f(a) + f(b)$ because there is only one element, $a = 1$.   It is surjective. So all I need to show is that it is injective. But that is clear too because if $ab \equiv 0 \bmod \gcd(m,n)$  then both $a \equiv 0 \bmod n$ and $b \equiv 0 \bmod m$ so the kernel is trivial.  Many thanks for your help!!
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The map $\phi: \mathbb{Z} \to \mathbb{Z}$ defined by $\phi (n) = n + 1$ for $n \in \mathbb{Z}$ is one to one and onto $\mathbb{Z}$. Give the definition of a binary operation $*$ on $\mathbb{Z}$ such that $\phi$ is an isomorphism mapping   a) $\langle\mathbb{Z},+\rangle$ onto $\langle\mathbb{Z}, *\rangle$    I absolutely do not understand this. What is the solution doing? Why are we starting out with $m*n$? And why do we have $\phi(m-1) * \phi(n-1)$?
71829
(1) Prove or disprove: For any sets $X,Y,Z$ and any maps $f:X \to Y$ and $g:Y \to Z$, if $f$ is injective and $g$ is surjective, then $g \circ f$ is surjective.  So i proved previously that $f$ is injective if $g \circ f$ is injective, but then i also proved that if $g \circ f$ is surjective then $g$ is surjective. then this question appears, so how do i prove it that it is true or false? I'm saying it's true, because $g \circ f$ is bijective, so how do i go around proving it? Or my idea is already wrong? I need some help thanks.  (2) Show that (it is true in general that) for any sets $A,B$, one has $P(A) \cup P(B) \subseteq P(A \cup B)$.  I know that power set has exactly all the subsets of $A$ for $P(A)$. but I dont know how to start.
6082
I was reading an example about describing homomorphisms and I'm having a bit of trouble with this one.  Show there is no group homomorphism $f : \mathbb{Z}_{10} \to \mathbb{Z}_{25}$ such that $f(1) = 3$. The first step of the proof is  $|1| = 10$ in $\mathbb{Z}_{10}$, so $|f(1)| \in\{ 1, 2, 5, 10\}$.  Does this mean that $|f(1)|$ can be $1$, $2$, $5$ or $10$? I see that $|1| = 10$ since $1$ is a generator of $\mathbb{Z}_{10}$, and that $|f(1)|$ seems to be a factor of $10$. If $f(1)$ is the image of $1$ under $f$, how would I find the order of $f(1)$?  Thanks for your help.
113703
I am doing an exercise about algebraic geometry, where the exercise  tell us to provide example of the image of polynomial map $f:\mathbb{C}^{m}\rightarrow \mathbb{C}^{n}$ that is not an algebraic set. But I am thinking that the image of every polynomial is $\mathbb{C}$ thus every polynomial map is surjective. What have I misunderstand?
15293
In John Fraleigh's book, A First Course In Abstract Algebra Exercises 15.7 and 15.11, one shows that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (1,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \  \mbox{ and  } \ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$  One does this with the first isomorphism theorem. With the same idea I proved for example that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \ \mbox{ and }\ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,4)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$ So I conjectured that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_{k}$ where $k=\mathrm{gdc}(m,n)$. For the previous four cases, the homomorphism $\phi: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} \times\mathbb{Z}_k$ given by $$ \phi(x,y)=\left(\frac{nx-my}{k}, \ x \ \ (\mathrm{mod} \ k) \right) $$ is surjective with kernel =$\langle (m,n)\rangle$. However, this is not the case when $(m,n)=(4,6)$. So, I cannot use what I did for the four cases to prove the general case.     What I want to know is if my conjecture is true. If so, how can I give a general homomorphism? If it is not true, how can I find $k$, such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$?   Thanks in advance for any help/hint/comment!
23648
I am trying to prove a series of propositions:  Given any homomorphism p from $\mathbb{R}$[X] to $\mathbb{R}$[X], show that it is equal to $\phi_g$ for a unique g in $\mathbb{R}$[X], with $\phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(\sum_{i=0}^n a_iX^i) =\sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = \sum_{i=0}^n p(a_i)p(X)^i = \sum_{i=0}^n a_ig(X)^i = \phi_g(f)$.  Show that if h,g $\in \mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $\in$ R$^x$ and b$\in \mathbb{R}$.
16562
I think I may have caught a mistake in my professor's lecture notes, and just want to be sure I'm not mistaken.   We want to construct $\mathbb{Q}$ from $\mathbb{Z}$, and we consider the mapping $$f: \mathbb{Z} \to \mathbb{Q}, \; n \mapsto \frac{n}{1}.$$ He writes that because $f$ is surjective, $\mathbb{Z} \subset \mathbb{Q}$.   However, it is surely not surjective because there are surely rational numbers whose  (unique) simplest form is $\frac{a}{b}$ for $b \neq 1$. The function is injective, surely, because the simplest representative in $\mathbb{Q}$ is unique, but that alone would not imply that $\mathbb{Z} \subset \mathbb{Q}$.   I think he meant to say that because $f$ is well-defined over the entirety of $\mathbb{Z}$, meaning we actually can identify each integer with a single rational number via the above inclusion, $\mathbb{Z} \subset \mathbb{Q}$.   Am I correct or glossing over something?