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If we have two mappings;

$a \:mod \:NM \to a \:mod \: M $

and

$a \: mod \: NM \to a \:mod \:N $

which are both well defined.

Can we then conclude that the mapping

$a \: mod \: NM \to (a \:mod \: N , a \: mod \: M ) $

is also well defined?

Thanks in advance. It is the only step I have left in order to complete a proof for the Chinese remainder theorem.

hello
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  • yes of course (Well you didn't tell as how the third map is defined, but if it is coordinate-wise then it is well defined). – Yanko Jan 15 '19 at 12:56
  • Yes. That is what I meant. I changed it, thanks :) – hello Jan 15 '19 at 13:02
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    Theoretically the notion of a "well defined map" is strange, because itsc meaning is just the same as "map". In other words, a map that is not well defined is not a map at all. So some people think this usage should be avoided. But in practice, as long as you understand what it really means, then it is very useful! – Derek Holt Jan 15 '19 at 13:19
  • See https://math.stackexchange.com/questions/60781/symmetric-groups-on-sets-with-the-same-number-of-elements-are-isomorphic/60788#60788 – Arturo Magidin Jan 15 '19 at 14:09

1 Answers1

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Yes.

From the perspective of category theory, by definition of direct product, if you have a map $f:X\to Y$ and a map $g:X\to Z$, you get a canonical map $X\to Y\times Z$ (which is unique in how nicely it cooperates with $f$, $g$, and the projections from $Y\times Z$ to $Y$ and $Z$, but that's of little impact here). This goes for groups as well as sets, topological spaces, vector spaces, metric spaces, and many others.

In most common categories (for instance, each category that I mentioned in the above paragraph), this canonical map is indeed known as the "coordinate-wise" map.

Arthur
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  • Would I be right in saying that the perspective you are using here is that of category theory? – Shaun Jan 15 '19 at 13:25
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    @Shaun I'll make that clearer. – Arthur Jan 15 '19 at 13:25
  • Thanks! It makes so much more sense now – hello Jan 15 '19 at 13:29
  • @bladiebla When you, in topology, encounter the product topology, and specifically for a product with infinitely many terms, it will have a quirk that many students find strange when they first encounter it (many sets that most would students intuitively guess to be open aren't actually open). The category-theoretical product is the reason for that. – Arthur Jan 15 '19 at 14:27