Product/coproduct properties: If $N_1\simeq N_2$ in some category, then $N_1\times N_3\simeq N_2\times N_3$? Excuse the simple question but I am trying to get my head around category theory. In particular, I am currently looking at product/coproduct properties such as:
If $N_1\simeq N_2$ in some category, then $N_1\times N_3\simeq N_2\times N_3$ in that category.
Just to get comfortable with it, I tried some examples. So for example, let's take the category of finitely generated modules over a ring $R$. Start with $N_1, N_2, N_3\leq M$ three submodules of a module $M$. If $N_1\cap N_3=\{0\}$ then we can define the direct sum $N_1\oplus N_3=N$ (say). If $N_1\cong N_2$ then we should also have $N_2\oplus N_3\cong N_1\oplus N_3=N$. To show this isn't too bad (simply extend the isomorphism $N_1\cong N_2$ in the obvious way).
However, I realised that I seem to be assuming that $N_2\cap N_3=\{0\}$ and I am having a hard time to explicitly show this. I presume this matter is dealt with through the usual abstract general nonsense, but as I am trying to see how things work in specific examples, I would like to show this. My plan was to let $f:N_2\to N_1$ be the isomorphism and induce the map $f_\ast:N_2\cap N_3\to N_1\cap N_3$ by $f_\ast(n)=f(n)$. As $N_1\cap N_3=\{0\}$ then it is easy to see this is a bijective map, but can we be sure this is well defined? After all, if $n\in N_2\cap N_3$ then $n\in N_2$ and $f_\ast(n)=f(n)\in N_1$, but does $f_\ast(n)\in N_3$?
I am sure I am overthinking this, and either way this is ott (hence why category theory is so useful) but I would still like to figure this out for my own sanity!
 A: You're mixing things up. Given two modules $N$, $N'$, their product is a module $N \times N'$ that satisfies some universal property that you certainly know. It turns out that this product is given by the external direct product $N \oplus N'$, which is the set of pairs $\{ (x,x') \mid x \in N, x' \in N' \}$, endowed with a module structure that's easy to define. It does not matter whether or not $N$ and $N'$ are submodules of the same module.
This is not the same thing as the internal direct product of two submodules $N, N' \subset M$. Given two such submodules, if they satisfy $N \cap N' = \{0\}$, then they form a direct sum, unfortunately denoted by the same symbol, $N \oplus N'$. This is the set of elements of $M$ that can be written as a sum of an element of $N$ with an element of $N'$. For this construction it's of course crucial that $N$ and $N'$ are submodules of the same module. It is then the case that the internal direct product, if it exists, is isomorphic to the external direct product – phew!
Now, it is indeed the case that if $N_1$ is isomorphic to $N_2$, then the products (which are external direct products in the category of modules) $N_1 \times N'$ is isomorphic to $N_2 \times N'$. This follows from the universal property and it's a good exercise. It works in every category.
However, it is not the case that if $N_1$, $N_2$, $N'$ are submodules of a module $M$ and $N_1$ is abstractly isomorphic to $N_2$, then the internal direct products $N_1 \oplus N'$ and $N_2 \oplus N'$ are both defined at the same time and isomorphic. For example, consider the $\mathbb{Z}$-module $M = \mathbb{Z}^2$, and $N_1 = \langle (1,0) \rangle$, $N_2 = \langle (0,1) \rangle = N'$. Then the internal direct product $N_1 \oplus N'$ is defined and equal to $M$, whereas $N_2$ and $N'$ do not form an internal direct product (their sum $N_2 + N'$ is just $N_2$ itself).
