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In general, when we define a function or an operator on equivalence classes, if the expression of the function or the operator depends on the representatives of equivalence classes, then one have to show the result is indeed independent of the choice of representatives. This is so called the well-definedness.
和 representative 無關, i.e. equivalent 的元素經過運算後的結果要相同（如下舉例）。
e.g.
For \(\bar a, \bar b \in \mathbb{Z}_n\), define
\[\bar a + \bar b := \overline{a+b}.\]It can be shown that \(+\) is well-defined. (Let \(\bar c = \bar a\) and \(\bar d = \bar b\), and show \(\overline{c+d} = \overline{a+b}\).)
Show well-definedness 看起來就像 one-to-one，只是順序相反。
The function \(\phi: S \to S'\) is an isomorphism between \((S, *)\) and \((S', *')\) if
若這樣的 \(\phi\) 存在，則 \((S, *) \cong (S', *')\)（或說 \(S \cong S'\)）。
讀：\(\phi\) is an isomorphism.
Prove that the binary structure \((\mathbb{R}, +)\) is isomorphic to \((\mathbb{R}_{>0}, \dot{})\).
Proof
We claim that \(\phi:\mathbb{R} \to \mathbb{R}_{>0}\) defined by \(\phi(x) = e^x\) satisfies
Proof of claim 1.
For all \(x, y \in \mathbb{R}\), if \(\phi(x) = \phi(y)\), then \(e^x = e^y, \ln(e^x) = \ln(e^y)\), and \(x = y\).
Proof of claim 2.
For all \(y \in \mathbb{R}_{> 0}\), let \(x = \ln y\). Then \(\phi(x) = e^{\ln y} = y\).
Proof of claim 3.
For all \(x, y \in \mathbb{R}\), \(\phi(x+y) = e^{x+y} = e^x \dot{} e^y = \phi(x) \dot{} \phi(y)\). ◼
Determine whether binary structures \((\mathbb{Z}, \dot{})\) and \((2\mathbb{Z}, \dot{})\) are isomorphic.
Intuition
在 \(\mathbb{Z}\) 中，有一單位元素 \(e=1\) 使得 \(e \dot{}n = n,\ \forall n \in \mathbb{Z}\)；但 \(2\mathbb{Z}\) 中並沒有相應的成員。所以嘗試證明 non-isomorphic。
Proof
Suppose \(\rho: \mathbb{Z} \to \mathbb{2Z}\) is an isomorphism, and \(\rho(1) = a\).
\[a = \rho(1 \dot{} 1) = \rho(1) \dot{} \rho(1) = a^2, \tag{by homomorphism}\]Hence, \(a = 0 \lor 1\); however \(a = 1\) is impossible since \(1 \not \in 2\mathbb{Z}\). Therefore, \(a = 0\).
Suppose \(\rho(2) = b \not = 0\) (since one-to-one). We can show that
\[b = \rho(2 \dot{} 1) = \rho(2)\rho(1) = ba = 0,\]which is a contradiction.
Therefore, such \(\rho\) doesn’t exists；the two structures aren’t isomorphic. ◼
Determine whether binary structures \((\mathbb{R}, \dot{})\) and \((\mathbb{C}, \dot{})\) are isomorphic.
Intuition
在 \(\mathbb{C}\) 中，\(x \dot{} x = c\) 對於所有 \(c \in \mathbb{C}\) 都有解，但在 \(\mathbb{R}\) 中則不然。所以嘗試證明 non-isomorphic。
Proof
Suppose there exists an isomorphism \(\phi: \mathbb{C} \to \mathbb{R}\), and \(\phi(a) = -1\). We further let \(b^2 = a \in \mathbb{C}\).
We can show that
\[\phi(b \dot{} b) = \phi(b)\dot{}\phi(b) = \phi(b)^2 = -1,\]and hence
\[\phi(b) = i \not \in \mathbb{R},\]which is a contradiction. ◼
is multiplication homomorphism, since
\[det(AB) = det(A)det(B);\]and
\[tr: M_n(\mathbb{R}) \to \mathbb{R}\]is addition homomorphism, since
\[tr(A+B) = tr(A) + tr(B).\]Let \(\rho\) be a homomorphism from a binary structure \((S, *)\) to anther binary structure \((S', *')\). Let \(B\) and \(B'\) be subsets of \(S\) and \(S'\) respectively. Then
If \(B\) is closed under \(*\), then \(\rho(B)\) is closed under \(*'\)
If \(B'\) is closed under \(*'\) then \(\rho^{-1}(B')\) is closed under \(*\).
但 \(\rho^{-1}\) 要先存在。