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.
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.