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As part of a proof of the Chinese-remainder equivalent of groups, I want to show that:

$$ \phi(a \bmod mn)= (a \bmod m, a \bmod n) $$ is a well-defined map

Note that $\gcd(m,n)=1$ so they are coprime.

We usually need to show that given: $a \equiv b \mod {mn}$, we are able to demonstrate that: $$ \phi(a)=\phi(b)$$

So let's start out writing what both this means. $$ (a \bmod m, a \bmod n)=(b \bmod m, b \bmod n)$$ We get the systems of equations: $$ a \bmod m = b \bmod m$$ $$ a \bmod n = b\bmod n$$

Why is this result true, if it is true I can work my way from bottom to top and arrive at the desired result.

Arturo Magidin
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    $mn\mid a-b,\Rightarrow, m,n\mid a-b,,$ i.e. congruences persist mod divisors of the modulus. – Bill Dubuque Feb 23 '19 at 23:05
  • This answers my question, very simple, but very nice. Thank you very much. –  Feb 23 '19 at 23:45
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    Great. Strangely such persistence is often not highlighted early on in elementary number theory courses so many folks first stumble upon it in more complex context such as CRT - making it harder to comprehend. But we know it implicitly long before that, e.g. coputing the parity of an integer from the parity of its decimal units digit. See here for further examples and discussion. – Bill Dubuque Feb 24 '19 at 00:10

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