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Prove that an arbitrary function

$$h:X\to P = \Pi_{i\in M} X_i$$

of a module $X$ over $R$ into the direct product module $P$ is a homomorphism $\iff$ the composition

$$p_i\circ h:X\to X_i$$

with the natural projection $p_i:P\to X_i$ is a homomorphism for every $i\in M$

Let's first suppose that the composition is a homomorphism, that is:

$$(p_i\circ h )(a+b) = (p_i\circ h) (a) + (p_i\circ h) (b)$$ $$(p_i\circ h )(\lambda a) = \lambda(p_i\circ h)(a) $$

we must show that $h$ itself is a homomorphism, that is, $h(a+b) = h(a)+h(b)$ and $h(\lambda a) = \lambda h(a)$. If we don't have $h$ homomorphism, then $p_i(h(a+b))\neq p_i(h(a) + h(b))$ but I don't see this as helping in anything. Or maybe it helps... Shouldn't the projection itself be a homomorphism? So if $(p_i\circ h)(a+b) \neq p_i(h(a)+h(b))$ then $(p_i\circ h)(a+b)$ can't possibly be $p_i(h(a)) + p_i(h(b))$, or can it?

Now, for the reverse, I don't even know how to begin, because the definition of a direct product is too obscure for me, as I already questioned here, but didn't understand a thing of the answer.

Poperton
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2 Answers2

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$f\in P$ iff it is a function $M\to\bigcup_{i\in M}X_i$ that satisfies $f(i)\in X_i$ for every $i\in M$.

For $f,g\in P$ we have pointwise addition, so $f+g$ is prescribed by $i\mapsto f(i)+g(i)$.

Likewise $\lambda f$ prescribed by $i\mapsto\lambda f(i)$

For every $i\in M$ we have the projection function $p_i:P\to X_i$ prescribed by $f\mapsto f(i)$.

So observe that $(p_i\circ h)(a)=p_i(h(a))=h(a)(i)$.

What you write as: $$(p_i\circ h )(a+b) = (p_i\circ h) (a) + (p_i\circ h) (b)$$ $$(p_i\circ h )(\lambda a) = \lambda(p_i\circ h)(a) $$are exactly the same statements as:$$h(a+b)(i) = h(a)(i) + h(b)(i)$$ $$h(\lambda a)(i) = \lambda h(a)(i)$$

This being true for every $i\in M$ means exactly that:$$h(a+b)= h(a) + h(b)$$ $$h(\lambda a)= \lambda h(a)$$

Edit

Equivalent are the following statements where $\lambda$ is an element of the ring, $a,b\in X$ and $i\in M$:

  • $h:X\to P$ is a homomorphism
  • $\forall\lambda\forall a,b\left[h(a+b)=h(a)+h(b)\text{ and }h(\lambda a)=\lambda h(a)\right]$
  • $\forall\lambda\forall a,b\forall i\in M\left[h(a+b)(i)=h(a)(i)+h(b)(i)\text{ and }h(\lambda a)(i)=\lambda h(a)(i)\right]$
  • $\forall\lambda\forall a,b\forall i\left[(p_i\circ h)(a+b)=(p_i\circ h)(a)+(p_i\circ h)(b)\text{ and }(p_i\circ h)(\lambda a)=\lambda (p_i\circ h)(a)\right]$
  • $\forall i\forall\lambda\forall a,b\left[(p_i\circ h)(a+b)=(p_i\circ h)(a)+(p_i\circ h)(b)\text{ and }(p_i\circ h)(\lambda a)=\lambda (p_i\circ h)(a)\right]$
  • $\forall i$ $p_i\circ h:X\to X_i$ is a homomorphism.
drhab
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  • So $p_i\circ h$ turns into $h$, so if the composite is a homorphism, then $h$ must be, right? What about the inverse? Is it automatically rue? – Poperton Apr 21 '17 at 21:39
  • $p_i\circ h$ does not turn into $h$ but $(p_i\circ h)(a)$ turns into $h(a)(i)$. Also have a look at my edit. The first and last statement are equivalent. – drhab Apr 22 '17 at 07:53
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Here is a purely categorical proof. The $\implies$ direction is trivial since homomorphisms compose to homomorphisms and projections are homomorphisms. If you haven't proven this, it's straightforward and you should prove it. Categorically this is just saying you do have a category and the product is a categorical product.

The other way is slighty more interesting. The universal property for the categorical product in any category is: given a collection of arrows $h_i : X\to X_i$ for $i\in M$ there exists a unique arrow $h : X \to \prod_{i\in M}X_i$ such that $h_i = p_i \circ h$. So in your case, there necessarily exists a unique homomorphism $h' : X \to \prod_{i\in M}X_i$ such that $h_i = p_i \circ h'$. Now since there are more functions than homomorphisms, it could be the case that $h$ is a function that isn't a homomorphism that just happens to coincide with $h'$ when post-composed with $p_i$. However, the category of modules has an underlying set functor $U : {}_R\mathbf{Mod}\to\mathbf{Set}$ which is a right adjoint. In general, right adjoints preserve limits and particularly products which are a special case of limits. This means that $Uh' : UX \to \prod_{i\in M}UX_i$, which, by the exact same logic as before, must be the unique function such that $Uh_i = Up_i \circ Uh'$, but our characterization of $h_i$ is $Uh_i = Up_i \circ h$ so $Uh' = h$ i.e. $h$ was the underlying function of a homomorphism, namely $h'$.

This argument works for any category and any right adjoint on it, though depending on how we word it we may need the right adjoint to be faithful as well for the $h_i$ to be well-defined. In particular, for most "algebraic" categories (e.g. rings, lattices, modules, groups, semirings, monoids) and some non-algebraic categories, e.g. $\mathbf{Top}$, the argument will apply using the underlying set functor for those categories.

Derek Elkins left SE
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