# Tag Info

4

It is indeed not surjective, for the reason you state. Consider the case of abelian groups: the chain complex $$\cdots \to 0 \to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \to 0 \to \cdots$$ is quasi-isomorphic to the chain complex $$\cdots \to 0 \to 0 \to \mathbb{Z} / 2 \mathbb{Z} \to 0 \to \cdots$$ via an obvious chain map, but the only chain map in the ...

4

The sequence $$0\rightarrow \mathbb Z\xrightarrow{n} \mathbb Z\rightarrow \mathbb Z_n\rightarrow 0$$ is exact in $\mathbb Z\text{-}\mathsf{Mod}$. Passng to torsion we have $$0\rightarrow 0\rightarrow 0\rightarrow \mathbb Z_n\rightarrow 0$$ which is not exact. Passing to free parts we have $$0\rightarrow\mathbb Z\xrightarrow{n}\mathbb Z\rightarrow ... 4 Let's let C be the category of complex vector spaces, and S be the set consisting entirely of the projection operator p:\Bbb C \to 0. I claim that the localization C[S^{-1}] is a new category where we've simply imposed an equivalence relation on Hom-sets:$$ Hom_{C[S^{-1}]}(V,W) = Hom_C(V,W) / \{\text{rank 1 maps} \sim 0\} $$First, since a ... 3 Not in general. For example, let the original commutative square be$$\begin{array}{ccc} 0&\to&X\\ \downarrow & &\downarrow\\ X[-1]&\to&0 \end{array}$$for any non-zero object X. This can be completed to a diagram$$\begin{array}{cccccccc} 0&\to& X&\to& X&\to &0\\ \downarrow & ...

2

Suppose that $K$ is the fraction field of $A$. If $P$ is a finitely generated projective $A$-module, then $P\otimes_AK$ is a finite dimensional $K$-vector space, so $\dim_K P\otimes_AK$ is a non-negative integer. You should be able to show that there is a (unique!) homomorphism of abelian groups $d:K(A)\to\mathbb Z$ such that for each f.g. projective ...

2

The determinant can be determined for every finitely generated projective module, because these are precisely the locally free modules of finite rank (which doesn't have to be constant, but it is locally constant, and on each constant piece we take the corresponding exterior power). It is additive on short exact sequences (see for example Daniel Murfet's ...

2

You have the exact sequence $$\underset{=0}{\underbrace{\tilde{H}_k(B^n)}} \to \tilde{H}_k(B^n,S^{n-1}) \to \tilde{H}_{k-1}(S^{n-1}) \to \underset{=0}{\underbrace{\tilde{H}_{k-1}(B^n)}},$$ so $$\tilde{H}_k(B^n,S^{n-1}) \simeq \tilde{H}_{k-1}(S^{n-1})= \left\{ \begin{array}{cl} \mathbb{A} & \text{if} \ k=n \\ 0 & \text{otherwise} \end{array} ... 1 This implication is not true in general. For an easy counterexample, let's use real vector spaces (modules over \mathbb R) so that we can visualize what's going on. Consider an exact sequence, as in the question, with L=N=\mathbb R and M=\mathbb R^2, with \alpha(x)=(x,0) and \beta(x,y)=y. This gives you a short exact sequence, because \alpha ... 1 There's a spectral sequence$$H^i(G, \operatorname{Ext}^j_\mathbb Z (A,B)) \Rightarrow \operatorname{Ext}_{\Lambda}^{i+j}(A,B)$$for general A and B which, since \operatorname{Ext}_{\mathbb{Z}}^i=0 for i>1, gives a long exact sequence$$0\to H^1(G,\operatorname{Hom}_{\mathbb{Z}}(A,B))\to \operatorname{Ext}_{\Lambda}^1(A,B) \to ...

1

The author is emphasising the difference between a singular simplex and a simplicial simplex. Among other things, a simplicial simplex has to be an injective map when restricted to its interior. This simply isn't the case for singular simplexes - they can be as non-injective (or singular) as they like - we only require that they are continuous maps. It's ...

1

$\psi$ could also precompose with any homomorphism $V \to V$, which is likely to ruin any chances of it being an $f_*$. For example, let $V = R^2$, $M = N = R$. Let $\psi = s^*$, where $s(\langle x,y \rangle) = \langle y,x \rangle$. Now let $\pi_1$ be the first projection, $\pi_1(\langle x, y \rangle) = x$: I claim $\pi_1 \in \mathrm{Hom}_R(R^2, R)$. Then ...

1

The barycentric subdivision of a simplicial complex $K$ is constructed by adding a vertex to the centroid of every simplex and the cone to its boundary. Iterating this operation $n$ times gives the $n^\text{th}$ barycentric subdivision $\operatorname{Sd}^n(K)$. The simplicial approximation theorem states that if $g:|K|\rightarrow |L|$ is continuous, where ...

1

Hint: Since $F$ is free of finite rank, there exists an $R$-basis $\{f_1,\ldots,f_n\}$ for $F$. Since $\phi:M\to F$ is a surjective map, there exist some $m_1,\ldots,m_n\in M$ such that $\phi(m_i)=f_i$. Let $F'$ be the submodule of $M$ spanned by these $m_i$'s. You need to show that $F'$ is isomorphic to $F$ and that $M=F'\oplus \ker(\phi)$; it will help to ...

1

Even if $g$ is an isomorphism, $h$ need not be monic. Consider $$\begin{array}{ccccccccc} 0 & \longrightarrow & 0 & \longrightarrow & \mathbb{Z} & \longrightarrow & \mathbb{Z} & \longrightarrow & 0\\ & & \downarrow & & \text{id}\downarrow & & h\downarrow \\ 0 & \longrightarrow & \mathbb{Z} ... 1 This answer refers to the original question, in which the ideal {\mathfrak p}_2 was (x_1,x_3,x_4,x_5) instead of (x_1,x_3,x_4,x_6). Working out the quotients makes your map$$\begin{align*}k[x_2,x_7] \oplus k \oplus k[x_2,x_4] \oplus k \oplus k[x_2] \oplus k & \to k \oplus k[x_2] \oplus k\\ (a(x_2,x_7), b, c(x_2,x_4), d, e(x_2), f) & \mapsto ...

1

As you know that for any $N$ you get an exact sequence for the $Hom$ modules, try putting different $N$s. In particular $N=M'$, $N=M''$ and $N=M$ together with their identity homomorphisms. More details: My strategy was to use the following. Taking $N=M'$ to obtain a map $r:M \to M'$ such that $r \circ f =id_{M'}$. Together with the given $g:M \to ... 1 Let$f$be an epimorphism. If the image of$f$is the kernel of$g$then$g$is constant and then the image of$g$being the kernel of$h$is trivial which is enough to have$h$as an monomorphism. Let$h$be a monomorphism, then the image of$g$is constant, hence$f\$ is surjective.

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