Let $\phi: R_{m_1 m_2 \cdots m_n} \to R_{m_1} \times \cdots \times R_{m_n}, a \mapsto (a \mod m_1, \ldots, a \mod m_n)$ be the mapping of the Chinese Remainder Theorem ($R = Z$ or $R = F[x]$ for some field $F$).

I could easily follow the prove on why the above mapping is an isomorphism but I'm struggling with the inverse part. In the lecture notes from our professor, this is presented as follows:

$\psi: R_{m_1} \times \cdots \times R_{m_n} \to R_{m_1 m_2 \cdots m_n}, a = some \, presented \, mapping(a_1, \ldots, a_n)$

Given the proposed mapping $\psi$ it suffices to verify that for all $y \in R_{m_1} \times \cdots \times R_{m_n}: \phi(\psi(y)) = y$ to conclude that $\psi = \phi^{-1}$. In order to do this, assume that the presented mapping $\psi$ is isomorphic and...

I can follow everything (including the rest of the proof, starting from ...) except for the bold part:

Q: Why is it acceptable for a proposed inverse mapping of an isomorphism to be assumed isomorphic? Clearly if $\psi = \phi^{-1}$, then $\psi$ must be an isomorphism. However why can I assume this in the verification part?

Note: My background is in electrical engineering and this is my first algebra course (part of error correcting codes).

  • $\begingroup$ I wasn't able to understand your question. So you say you were able to follow the proof of that $\phi$ is an isomorphism right? Then clearly $\phi$ has an inverse (since it is one-to-one and onto). I think I couldn't understand the question in your mind. $\endgroup$ – Levent Feb 28 '16 at 15:56
  • $\begingroup$ I am presented a mapping $\psi$ and I wish to verify whether this is the inverse mapping (which, as you stated, obviously exists) to $\phi$. My question is: In order to verify whether the presented $\psi$ is actually $\phi^{-1}$, can I assume $\psi$ to be isomorphic? $\endgroup$ – R.G. Feb 28 '16 at 16:01
  • $\begingroup$ Could you explain what means $R_{m_i}$? $\endgroup$ – user26857 Feb 28 '16 at 16:08
  • $\begingroup$ Of course: This is the set $R_{m_1 m_2 \cdots m_n} \mod m_i$ $\endgroup$ – R.G. Feb 28 '16 at 16:13
  • $\begingroup$ Oh I see now. Well, actually I think you should not assume (and also you should not need) such an assumption. $\endgroup$ – Levent Feb 28 '16 at 16:18

THEOREM. Let $\psi:A \to B$ be an isomorphism. Then there exists an inverse isomorphism $\phi:B \to A$.


If $\psi:A \to B$ is an isomorphism, then $\psi$ is bijective. Hence there exists a bijective mapping $\phi:B \to A$ such that $\phi(\psi(a)) = a$ for all $a \in A$ and $\psi(\phi(b)) = b$ for all $b \in B$. We need to show that $\phi:B \to A$ is a homomorphism. It will follow that $\phi$ is an isomorphism.

Let $b_1, b_2 \in B$. Let

$$a_1 = \phi(b_1) \quad \text{ and let } \quad a_2 = \phi(b_2)$$

Since $\psi:A \to B$ is an isomorphism, then $\psi(a_1 + a_2) = \psi(a_1) + \psi(a_2) = \psi(\phi(b_1)) + \psi(\phi(b_2)) = b_1 + b_2$.

Thus $\phi(b_1 + b_2) = \phi(\psi(a_1 + a_2)) = a_1 + a_2 =\phi(b_1) + \phi(b_2) $

Hence $\phi$ is a homomorphism.


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