标签：推出 Case group 同态 times cdots pushout Lambda
以下定理在我学习低维拓扑时遇到(推出的构造来自Van Kampen定理), 但并没找到详细证明. 我在这里给出一个不那么繁琐的证明过程.
Theorem 1
If \(A\to B, A\to C\) are both group monomorphisms, then the pushout \(B\to B\sqcup_{A} C, C\to B\sqcup_{A} C\) are also monomorphisms.
(假装这里有一个交换图表)
Proof
We attempt to construct a group satisfying the properties of pushout.
Step 1 Define the group as a set.
(by Choice Axiom) \(C = \bigsqcup_{c_\alpha\in \Lambda_C} c_\alpha f(A), B = \bigsqcup_{b_\alpha\in \Lambda_B} b_\alpha g(A)\) are divided into some cosets. WLOG \(1_C\in \Lambda_C, 1_B\in \Lambda_C\), Let \(\Lambda = \Lambda_B\sqcup \Lambda_C\), and define a boolean map \(\mathbf{\delta}:\Lambda\times\Lambda\to\{0,1\}\), such that \(\delta(g,h)=1\) if and only if \(g\) and \(h\) lie in \(\Lambda_B, \Lambda_C\), respectively.
Let \(G\) consists of all words of the form \(g_1\cdots g_na\) such that
(C1) \(n\ge 0, g_i\in \Lambda-\{1_B,1_C\}, a\in A\),
(C2) \(\delta(g_i,g_{i+1})=1\).
Step 2 Define the multiplication.
Consider
\[g_1\cdots g_{n_1}a \times h_1\cdots h_{n_2}b, \]where \(g_i, h_i\in \Lambda-\{1_B,1_C\}, a,b\in A\). We'll define the product inductively(on \(n_2\)).
Case 2.1
For \(x\in A\), define
Case 2.2
For \(x\in \Lambda - \{1_B, 1_C\}\),
Case 2.2.1
If \(n_1=0\) or \(n_1>0, \delta(g_{n_1}, x)=1\), then WLOG \(x\in \Lambda_B\), calculate \(f(a)x = x'f(a')\) (obviously \(x'\neq 1_B\)) and define
Case 2.2.2
If \(n_1>0, \delta(g_{n_1}, h_1)=0\), WOLG \(g_{n_1}, h_1 \in \Lambda_B\), calculate \(g_{n_1}f(a)x = x'f(a')\). Let \(X = \begin{cases}
a', &x'=1_B;\\
x'a', &x'\neq 1_B.
\end{cases}\) and define
Now define the following word as the product by induction
\begin{equation}
g_1\cdots g_{n_1}a \times h_1\cdots h_{n_2}b =
\begin{cases}
(g_1\cdots g_{n_1}a \times h_1) \times h_2\cdots h_{n_2}b, &n_2>0;\
g_1\cdots g_{n_1}a \times b, &n_2=0.
\end{cases}
\end{equation}
Step3 Verify the associative law.
Consider
\[(f_1\cdots f_{n_1}a \times g_1\cdots g_{n_2}b) \times h_1\cdots h_{n_3}c, \]\[f_1\cdots f_{n_1}a \times (g_1\cdots g_{n_2}b \times h_1\cdots h_{n_3}c). \]We'll also show that they are equal by induction on \((n_3, n_2)\).
Notice that for \(g_i\in \Lambda_B\sqcup A\), \(g_1\times\cdots\times g_m\) does not depend on the order of the multiplication since the product in \(B\) satisfies the associative law. For example, we have
that is to say, for any \(F\in G, a\in A, g\in \Lambda\),
\[(F\times a)\times g = F\times (a\times g). \]Denote \(f_1\cdots f_{n_1}a\) as \(F\).
Case 3.1
If \(n_2=n_3=0\), then
Case 3.2
If \(n_2>0, n_3=0\), then by induction hypothesis on \((0, n_2-1)\),
Case 3.3
If \(n_3>0, n_2=0\), by induction hypothesis on \((n_3-1, 1)\), we have
Case 3.4
If \(n_3>0, n_2>0\), then by induction hypothesis on \((n_3, n_2-1)\), \((n_3-1, n_2')\) and \((n_3-1, n_2'+1)\),
Now \(G\) is actually a group since every element in \(\Lambda\cup A\) has inverse element in \(G\).
Step 4 Verify the properties of pushout.
Since the elements in \(G\) is made up of elements in \(A, B\) and \(C\), there're natural injections from \(A,B,C\) to \(G\). There's also a natural homomorphism from \(G\to B\ast C/N, x\mapsto\begin{cases} f(x)+N = g(x)+N, &x\in A;\\ x+N, &x\in \Lambda;\\ \end{cases}\). Thus, \(G\) inherits the property of pushout.
Now the injectivity of \(B\to B\sqcup_{A} C, C\to B\sqcup_{A} C\) is obvious.
\[\newcommand{\laa}{\langle\hspace{-0.2em}\langle} \newcommand{\raa}{\rangle\hspace{-0.2em}\rangle} \newcommand{\genr}[1]{\langle#1\rangle} \]事实上, 这个结果不能推广到只有一边是单射的情形.
Example 2
In this example, we attempt to show that in the group category, if \(D\) is the pushout of \(f:A\to B, g:A\to C\), then \(g\) is injective cannot guarantee that \(B\to D\) is injective.
Let \(A=\Z/mn\Z\), \(B=\Z/n\Z, g:x\mapsto mx\) and denote \(c=f(1)\). Then the pushout \(D\) is just isomorphic to \(C/\laa c^m \raa\) where \(\laa c^m \raa\) is the minimum normal subgroup of \(C\) containing \(c^n\). Notice that if we let \(C\) be a simple group, then the pushout must be a trivial group and the homomorphism \(h\) must not be injective.
In order to guarantee the injectivity of \(f\), we should find a simple group containing a cyclic subgroup of a composite order. In fact, \(A_{8}\) has such a subgroup \(\langle\sigma^2\rangle\cong \Z/4\Z\), where \(\sigma\) is an order \(8\) element in \(S_{8}\).
Combining all the discussion above, we give such an example. \(A=\Z/4\Z=\genr{\alpha}, B=\Z/2\Z=\genr{\beta}, C=A_{8}, \sigma = (1\hspace{0.5em}2\cdots 8)\) with \(f(\alpha)=\beta, g(\alpha)=\sigma^2\), then \(f\) is injective while \(D=1\), which implies that \(h\) is not a monomorplism.
标签：推出,Case,group,同态,times,cdots,pushout,Lambda 来源： https://www.cnblogs.com/qrcer/p/15731070.html
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