Northcott Multilinear Algebra Universal Property Proof Northcott Multilinear Algebra poses a problem. Consider R-modules $M_1, \ldots, M_p$, $M$ and $N$. Consider multilinear mapping
$$
\psi: M_1 \times \ldots \times M_p \rightarrow N
$$
Northcott calls the universal problem as the problem to find $M$ and multilinear mapping $\phi: M_1\times \ldots \times M_p \rightarrow M$ such that there is exactly one R-module homomorphism $h: M\rightarrow N$ such that $h \circ \phi = \psi$.
Northcott claims that if $(M, \phi)$ and $(M', \phi')$ both solve the universal problem then

In this situation there will exist unique R-homomorphisms $\lambda: M\rightarrow M'$ and $\lambda': M' \rightarrow M$ such that $\lambda \circ \phi = \phi'$ and $\lambda \circ \phi' = \phi$.

If $\lambda$ and $\lambda'$ exist I understand why the equalities at the end of the sentence follow, based on the satisfaction of the universal problem. I can't see however why homomorphisms $\lambda$ and $\lambda'$ should exist.
I did more group theory many years ago and this is my first serious foray into "modules" so I wouldn't be surprised if there is something obvious I'm missing.
my thoughts:
Clearly $M$ and $M'$ are both homomorphic to $N$ through $h$ and $h'$, I'm not sure if this says anything about a relationship between $M$ and $M'$ though.
If $h'$ were injective I could say something like $\lambda(m) = h'^{-1}(h(m))$ but I don't know if there is any guarantee that $h'$ is injective..
Likewise, if $\phi$ were injective I could define $\lambda(m) = \phi'(\phi^{-1}(m))$ but again I don't know why this would be the case...
I've tried replacing $M$ and $N$ with more familiar vector spaces and R-module homomorphisms by multilinear maps for better intuition but no luck.. I do know that if $M$ and $M'$ are vector spaces with the same dimension then there is an isomorphism between them. I guess more generally if $M$ and $M'$ have different dimensions (say $\text{dim}(M') > \text{dim}(M)$) then there is a homomorphism from $M$ into a subspace of $M'$ and another homormophism from $M'$ onto $M$. Maybe this carries over to modules and is in the right direction for what I need...?
 A: The existence of $\lambda$ and $\lambda'$ are furnished by the maps $\phi'$ and $\phi$ via the universal property; that is, we are taking $N = M', M$ respectively. To spell this out further, the map
$$\phi' \colon M_{1} \times \cdots \times M_{p} \to M'$$
is bilinear by assumption, and so the universal property of the pair $(M, \phi)$ says that there is exactly one morphism $\lambda \colon M \to M'$ such that $\lambda \circ \phi = \phi'$. You obtain the morphism $\lambda' \colon M' \to M$ analogously by switching the roles of $M$ and $M'$.
This is a typical situation that arises in universal problems: the data of any two solutions contains information which allows you to relate the two solutions through the universal property. In this case, the crucial piece of information is this bilinear map $\phi \colon M_{1} \times \cdots \times M_{p} \to M$ which allows you to relate any two solutions.
A: This answer has the same content as the answer by Alex Wertheim, just explicated in slightly more detail.
In my post I misunderstood the universal problem. I thought the universal problem supplied a $(M, \phi)$ given a particular map $\psi$. But, more powerfully, a solution to the universal problem (for multilinear maps from $M_1\times\ldots\times M_p$) is a pair $(M, \phi)$ with $M$ an R-module and $\phi:M_1\times\ldots\times M_p\rightarrow M$ multilinear, such that for ANY multilinear $\psi:M_1\times\ldots \times M_p\rightarrow N$ there exists a unique R-module homomorphism $h:M\rightarrow N$ such that $h\circ \phi = \psi$.
In this case we have the following. Suppose we have multilinear $\psi:M_1\times\ldots \times M_p \rightarrow N$ and both $(M, \phi)$ and $(M', \phi')$ solve the universal problem. $\phi'$ is a multilinear map from $M_1\times\ldots\times M_p$ into $M'$ so, because $(M, \phi)$ solves the univeral problem from $M_1 \times\ldots \times M_p$, $\phi'$ can be uniquely decomposed as
$$
\lambda \circ \phi = \phi'
$$
With $\lambda:M\rightarrow M'$ a R-module homomorphism.
Likewise for $\phi$ so that
$$
\lambda'\circ\phi' = \phi
$$
with $\lambda':M'\rightarrow M$ also an R-module homomorphism.
