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1 vote
Accepted

number of homomorphism from $K_4\to S_4$

First all the images of elements of $K_4$ need to have orders dividing $2.$ (I'll let you supply the reason.) Now I'll take a stab. There are four $K_4$'s in $S_4.$ Then use that $\rm Aut(K_4)\cong ...
i can try's user avatar
  • 9,388
1 vote
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Is restriction map $\text{Hom}(S_3, \Bbb{Z}/2\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z})\to \text{Hom}(H_2,\Bbb{Z}/2\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z})$ injective?

Yes. There's only one non-trivial homomorphism from $S_n$ into an abelian group, the sign homomorphism. What happens to it? $(12)$ is odd. So the restriction is not the trivial homomorphism.
i can try's user avatar
  • 9,388
1 vote

Is there a onto group homomorphism from $\Bbb Q$ to $\Bbb Z$?

The argument given for the first question is correct. Indeed, homomorphic image of a cyclic group is necessarily cyclic. $\mathbb Q$ is non-cyclic so there can't be an onto group homomorphism $\mathbb ...
Nothing special's user avatar
5 votes

Number of homomorphisms from $\DeclareMathOperator{\Z}{\mathbb Z}\Z\oplus \Z$ to $S_3$

For every group $G$ there is a bijection between $Hom(\mathbb Z^2, G)$ and the set $C_2(G)$ of ordered pairs of commuting elements of $G$. The bijection $\beta$ is given by $$ \beta: \phi\mapsto (\phi(...
Moishe Kohan's user avatar
6 votes
Accepted

Screwing up basic short exact sequence $0\to\mathbb Z\leftrightarrow\mathbb Z\to0$

The identity morphism also satisfies the exactness. Your first two observations are valid for all isomorphisms, in particular also the identity morphism. The error lies in translating the sequence $$0 ...
Flynn Fehre's user avatar
2 votes

Orientation of boundary $\partial_2$ on torus (and other CW complexes)?

To expand on the comments, the standard one uses to compute the boundary of a triangle $T$ does not match some pre-determined choice of what you are calling a "spinor". Instead, the boundary ...
Lee Mosher's user avatar
  • 123k
2 votes
Accepted

Orientation of boundary $\partial_2$ on torus (and other CW complexes)?

In order to define a $\Delta$-complex, you need to orient each simplex, and that orientation comes from an ordering on the vertices. In this case it's a little confusing because the vertices are all ...
John Palmieri's user avatar

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