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 $f:\gamma \mapsto \int_\gamma dz$ ? If once the map $f$ is proved to be surjective, I finish the proof. I already proved this map is well defined, hom, inj. For arbitrary $nω1＋mω2$ from $Λ$, what is the original image of $f$? Thank you for your help.
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.
30181
I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states     Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$.   Thinking about this for a bit, the only reasonable approach I could think of was to construct some surjective homomorphism $\phi\colon G\to K$ for $K\leq G$, and $\ker\phi=H$, and then just use the isomorphism theorems to get the result.  After a while of trying, I've failed to come up with a good map, since $H$ seems so arbitrary. I'm curious, how can one construct the desired homomorphism? This is just the approach I thought of, if there's a better one, I wouldn't mind seeing that either/instead. Thank you.
73037
Let $(e_1, ..., e_n)$ be the standard $\Bbb Z$-basis of $\Bbb Z^n$. Let $x_1,..., x_n$ be elements of an abelian group $G$. Prove that there exists a homomorphism $f$ $:$ $\Bbb Z^n$ $\rightarrow$ $G$ such that $f$$(e_i)$ $=$ $x_i$ for all $i$.   Can anyone please help me out here?
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?
123818
Let $\kappa, \lambda,\theta$ be infinite cardinals of $M$ where $M$ is a transitive model of ZFC. Let $\mathcal{P}=Fn(\kappa\times \omega,2)$ (finite functions from $\kappa\times \omega$ to 2) and G is a (M,P)-generic set. I want to show $(\lambda^\theta)^{M[G]}=(max(\kappa,\lambda)^\theta)^M$. One direction is easy $(\lambda^\theta)^M \leq (\lambda^\theta)^{M[G]}$ and $(\kappa^\theta)^M\leq (\kappa^\theta)^{M[G]}\leq ((2^\omega)^\theta)^{M[G]}=(2^\theta)^{M[G]}\leq (\lambda^\theta)^{M[G]}$ where $(\kappa \leq 2^\omega)^{M[G]}$ is by the forcing construction. But now I am not quite sure how to get the other direction. Any thoughts?
6343
I'm asked to show that $G=GL_2(\Bbb Z/2\Bbb Z)$ is isomorphic to $S_3$. I have few ideas but I don't manage to put them all together in order to obtain a satisying answer. I first tried using Cayley's theorem ($G$ is isomorphic to a subgroup of $S_6$), and I also noticed that $\operatorname{Card}(G)=\operatorname{Card}(S3)=6$ &amp; that they're both non-abelian group.  Is this enough to say that considering $S_3$ is a subgroup of $S_6$ with the same cardinality than $G$, it has to be isomorphic to it ?  Could anyone give me some elements to get a more rigorous proof or lead me to an other path to show this statement ?  Thanks in advance
101349
We know that two Hilbert spaces that have orthonormal bases of the same cardinality are isomorphic (as an inner product spaces).   My question is: what can we say when we know that their Hamel bases have the same cardinality? It clearly implies they are isomorphic as vector spaces (just send a basis to a basis and extend linearly), but are they isomorphic also as inner product spaces? (i.e. via an isomorphism that also preserves the inner-product).
43478
Given the following problem     Let $U$ and $V$ be finite dimensional vector spaces, such that $\dim(U) = \dim(V)$. Prove that $U$ and $V$ are isomorphic.   I was just wondering if somebody could critique and validate my proof.  Proof  Two vector spaces $U$ and $V$ are isomorphic if and only if there exists a bijection $\phi: U \to V$. That is, an invertible linear map between the two. Let $\dim(U) = \dim(V) = n$ and let $\left\{u_1, \dots, u_n\right\}$ and $\left\{v_1, \dots, v_n\right\}$ be bases for $U$ and $V$, respectively. It follows that any vector $u \in U$ can be written as  $$ u = c_1u_1 + \cdots + c_nu_n $$  We define $\phi: U \to V$ as  $$ \phi(c_1u_1 + \cdots + c_nu_n) = c_1v_1 + \cdots + c_nv_n $$  Observe that $\phi$ is invertible. That is, given a vector  $$ v = d_1v_1 + \cdots + d_nv_n $$  in $V$, we define $\phi^{-1}: V \to U$ as  $$ \phi^{-1}(d_1v_1 + \cdots + d_nv_n) = d_1u_1 + \cdots + d_nu_n $$  Since there exists a bijection between $U$ and $V$, they are isomorphic.
1529
Problem statement: I need to find out if $\mathrm{Aut}(D_4)$ is isomorphic to $D_4$ and explain my answer. I already know that it is isomorphic, so now all I need to do is to prove it.   I assume that first we need to look where it sends $s$ and $t$ and it must send them to elements that satisfy the same relations() And then need to show that those two are conjugation by some element in $D_4$(?)  Any help is appreciated. I was hoping for a duplicate post, but couldn't find it.  Thank you!
82810
Let $\phi: R \to S$ be a surjective homomorphism. Prove that $\frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi} \cong \frac{S}{\phi(J)}$ for an ideal $J$ of $R.$  Obviously, $S \cong R/ \ker \phi$(first isomorphism theorem) and $\phi(J)$ is an ideal of $S$, since $\phi$ is surjective( $\forall s \in \phi(J), s = \phi(j)$ for some $j \in J, \forall s' \in S, s' = \phi(r)$ for some $r \in R$, so $ss' = \phi(j)\phi(r) = \phi(jr) \in \phi(J)$, and $ss' = \phi(r)\phi(j) = \phi(rj) \in \phi(J)$ ).  However, I'm having trouble with going on. If it's important, I mean a unitary ring by a ring.
110811
Is my proof correct? I have made use of the fact isomorphism preserves order of elements, which I proved couple of exercises back. I am also interested in other ways of proving it. Is there a more explicit way or is this explicit enough?    Problem Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.  Solution Recall that isomorphism preserves order of elements and hence if there exists an isomorphism from $\phi: \mathbb{C}-\{0\} \mapsto \mathbb{R}-\{0\}$, then $x \in \mathbb{C} - \{0\}$, and $\phi(x) \in \mathbb{R} - \{0\}$, then $\vert x \vert = \vert \phi(x) \vert$. Now note that the element $i \in \mathbb{C} - \{0\}$ has order $4$. However, no element in $\mathbb{R}-\{0\}$ has order $4$. Hence, no isomorphism can exist. Hence, the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.    Thanks
84582
Let $C_*$ be a chain complex such that each $C_i$ is a torsion-free, finite-range abelian group with $C_i=0$ for all $i&lt;0$. Suppose that $C_i=0$ for all $i$ is large enough. The Euler characteristic of $C_*$ chain complex is defined as $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(C_i)$$ Prove that $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$$  I have to prove that $\sum_{i\geq 0}(-1)^iRank(C_i)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$, I think that one way to do this is by showing that $Rank(C_i)=Rank(H_i(C_*))$ but I do not know if this is true in general, in a nutshell, I want to show that the cardinality of the base of any $C_i$ is the same as the cardinality of the basis of the corresponding homology, how can I do this? Thank you
45352
I'm working in Lee's book on topological manifolds and have gotten stumped on the first question in chapter 5, the chapter on cell complexes.  The problem is:   Let $D$ and $D'$ be two closed cells not necessarily of the same dimension.     Show that any continuous map $f:\partial D \to \partial D'$ can be extended to a continuous map $F:D \to D'$ such that $F(Int \; D) \subseteq Int \; D'$. Given points $p \in Int \; D$ and $p' \in Int \; D'$, show that $F$ can be chosen so $F(p) = p'$. Show that if $f$ is a homeomorphism then $F$ can be chosen to be a homeomorphism.   I proved the first part roughly as follows: First suppose $D$ and $D'$ are convex and each contain $0$ in their interior (in their respective ambient spaces).  Then every element other than $0$ in $D$ can be expressed uniquely in the form $\lambda q$ where $q \in \partial D$ and $\lambda\in (0,1]$ (an equivalence relation on the chords connecting $0$ to boundary points partitions $D- \{0\}$).  I then define the map $F(\lambda q) = \lambda f(q)$ which is continuous since if $\lambda_n q_n \to \lambda q$ in $D$ then $F(\lambda_n q_n) \to F(\lambda q)$.  Lastly $\lambda q \in Int \; D$ implies $\lambda &lt;1$ and so $\lambda f(q)$ is an interior point of $D'$ since $f(q)$ is a boundary point and $D'$ is convex.    Now if we suppose $D$ and $D'$ are arbitrary closed cells there are homeomorphisms $g_1: \overline{\mathbb{B}^n} \to D$ and $g_2: \overline{\mathbb{B}^m} \to D'$ (where possibly $m=n$) then we have that $g_2^{-1}\circ f \circ g_1$ is a continuous map between the boundaries of two closed balls and so by the first part can be extended to a continuous map $F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}$.  The mapping $g_2 \circ F \circ g_1^{-1}$ is a continuous map that satisfies the desired claim.    My issue is that parts 2 and 3 don't seem (to me at least) to follow that easily from the proof I constructed.  Maybe my proof is wrong and I'm not seeing it, but more likely I think that I'm missing the spirit in which Lee is intending for us to approach this problem.
88985
I think I'm supposed to prove that both $T$ and $I_V$ are isomorphic, but I'm not sure and even if that's the right way to go, how am I supposed to prove that $I_V$ is isomorphic. Or am I going about this all wrong. Please help.
77054
I am trying to show the isomorphism  $$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle.$$  I've tried to define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ an epimorphism with $\ker(\psi)=\langle (a,...,a) \rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
31461
Suppose that $l_1 = \{ (x_n)_{n \in \mathbb{N}} | \sum|x_n| &lt; \infty \}$ and $l_{\infty} = \{ (x_n)_{n \in \mathbb{N}}| \sup|x_n| &lt; \infty \}$.  Show that $(l_1)^* \cong l_{\infty}$, where $(l_1)^*$ is a dual space of $l_1$.  My attempt: Define a map $L: l_{\infty} \rightarrow (l_1)^*$ given by  $$L(x)(y) = \sum_{n \in \mathbb{N}}{x_ny_n}$$  where $y = (y_n)_{n \in \mathbb{N}} \in l_1$.   My aim is to show that $L$ is an isometric isomorphism.   Clearly $L$ is linear and and injective (choose $y = e_n$ for all $n \in \mathbb{N})$.  How to show that $L$ is a surjection and isometry?
88526
We have a function $\varphi:G\rightarrow H$ is an isomorphism, show its inverse $\varphi^{-1}:H\rightarrow G$ is also an isomorphism  I am fine with showing it to be a homomorphism and surjective, this is my attempt at showing injectivity, I'm unsure if they would be sufficient  Assume: $\forall h\in H, \exists g\in G$ s.t. $\varphi(g)=h$ and $\forall g_1, g_2\in G, \varphi(g_1)=\varphi(g_2)\Rightarrow g_1=g_2$  Injectivity - Need to show $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=g_2$  $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=\varphi(\varphi^{-1}(g_2))$ from our surjectivity assumption on $\varphi$ we know there exists such a $\varphi^{-1}(g_2)\in H$ and from our injectivity assumption on $\varphi$ we know that this solution is unique.   [Sufficient? What more could be added? How could I do this clearer?]