# What is the explicit isomorphism?

Let $$M_i$$ and $$N$$ be $$A$$-modules.

I want to prove that $$\oplus_{i\in I}(N \otimes M_i ) \simeq N \otimes (\oplus_{i \in I}M_i)$$ explicitly. I know that it follows from some abstract nonsense, but I really want the concrete isomorphism (because I needed in a step of a proof). So, I want to show that $$(n_i\otimes m_i) \mapsto (\sum n_i)\otimes m_i$$ is exactly that isomorphism. However, I can't show that is injective. Can you give me a hand?

• You should be able to extract what the map is from the abstract proof. – Eric Wofsey Dec 31 '19 at 3:14
• The proof that I found says that tensor is left adjoint to internal hom. How can I get the morphism from that statement? – HeMan Dec 31 '19 at 3:19
• You should map the other way. A pure tensor of the right hand side is of the form $n\otimes(\oplus m_i)$, which should map to $\oplus(n\otimes m_i)$, and extend linearly. – Arturo Magidin Dec 31 '19 at 3:23
• Upon heeding Arturo's advice, I also recommend reflecting for a moment how the universal properties can be used to prove injectivity directly. However, I think you'll find that constructing the inverse is often somehow more... natural... the deeper you go into abstract mathematics. – RghtHndSd Dec 31 '19 at 3:35

Every "abstract nonsense" proof actually does give you an explicit isomorphism somewhere, if you unwind what the proof says (sometimes this involves unwinding the proofs of tools like Yoneda's lemma). In this case, you say you got this isomorphism from the fact that the tensor products $$N\otimes -$$ is a left adjoint, and therefore preserves coproducts (i.e., direct sums). But the theorem that "left adjoints preserve colimits" doesn't just say that certain objects are isomorphic; it says certain specific maps are isomorphisms. So you just have to unwrap what those maps are in this case.
Specifically, what it means for $$N\otimes -$$ to preserve coproducts is that the maps $$N\otimes M_i\to N\otimes \left(\bigoplus M_i\right)$$ given by applying the functor $$N\otimes -$$ to the inclusion maps $$j_i:M_i\to \bigoplus M_i$$ form a coproduct diagram. In particular, this means that the map $$\bigoplus (N\otimes M_i)\to N\otimes \left(\bigoplus M_i\right)$$ obtained by combining these maps using the coproduct property of $$\bigoplus (N\otimes M_i)$$ is an isomorphism. What is that map? By definition, it is just the map which, restricted to the $$i$$th coordinate of the domain, is the map $$N\otimes M_i\to N\otimes \left(\bigoplus M_i\right)$$ given by tensoring with the inclusion map $$j_i:M_i\to \bigoplus M_i$$. In other words, it is the map that sends $$n\otimes m$$ to $$n\otimes j_i(m)$$. So, our isomorphism $$\bigoplus (N\otimes M_i)\to N\otimes \left(\bigoplus M_i\right)$$ is the map that sends a tuple $$(n_i\otimes m_i)$$ to $$\sum (n_i\otimes j_i(m_i))$$.
(Note that the expression $$(\sum n_i)\otimes m_i$$ you wrote is incorrect and does not make sense--on the left $$i$$ is a bound variable because of the sum, but on the right it is a free variable. Also, I have distinguished an element $$m_i\in M_i$$ from its image $$j_i(m_i)$$ in $$\bigoplus M_i$$ to avoid any risk of confusion. Also, you should be aware that a general element of $$\bigoplus (N\otimes M_i)$$ does not have the form $$(n_i\otimes m_i)$$ but can instead be a sum of elements of this form, but it is enough to describe the map on these elements (or even on elements with just a single nonzero coordinate which is a tensor $$n\otimes m$$ for $$n\in N$$ and $$m$$ in some $$M_i$$) since those elements generate $$\bigoplus (N\otimes M_i)$$.)