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I don't quite understand this example given by Mike Haskel. I want to find an example about

$$\operatorname{Hom}_R\left ( M ,\bigoplus_{i\in I} N_{i}\right )\not \cong\bigoplus_{i\in I} \operatorname{Hom}_R\left ( M ,N_{i}\right ).$$

Mike Haskel's example is that:

It's not true when $I$ is infinite, due exactly to the problem you encounter. Consider the case where $R = \mathbb{R}$, $M$ is an infinite dimensional vector space, and each $N_i$ is $\mathbb{R}$, with $I$ infinite. Convince yourself that $\operatorname{Hom}(M,\bigoplus_i N_i)$ corresponds to infinite matrices whose columns each have finitely many nonzero entries, while $\bigoplus_i \operatorname{Hom}(M,N_i)$ corresponds to infinite matrices with only a finite number of nonzero rows.

It's not quite clear to me why "$\bigoplus_i \operatorname{Hom}(M,N_i)$ corresponds to infinite matrices with only a finite number of nonzero rows".

I don't know how to write a formal rigorous proof that $\operatorname{Hom}_R\left ( M ,\bigoplus_{i\in I} N_{i}\right )\not \cong\bigoplus_{i\in I} \operatorname{Hom}_R\left ( M ,N_{i}\right )$ in this case.

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2 Answers2

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I think you're right that the vector spaces in this example will be isomorphic when $I$ is countably infinite. Probably what Mike Haskel had in mind is that the canonical map from the sum-of-Homs to the Hom-into-sum is not an isomorphism.

The example becomes correct if $I$ is chosen more carefully. What one needs is $|I|=\kappa$ for some cardinal $\kappa>|\mathbb R|$ with the additional property that $\kappa^{\dim M}>\kappa$. For simplicity, let me take $\dim M=\aleph_0$ and all $N_i=\mathbb R$ for this example. The required cardinals $\kappa$ exist; indeed the inequality $\kappa^{\aleph_0}>\kappa$ holds whenever the cofinality of $\kappa$ is $\aleph_0$.

Having chosen $\kappa$ this way, we get for the dimension of the Hom-into-sum, $(|\mathbb R|\cdot\kappa)^{\aleph_0}=\kappa^{\aleph_0}$, while the sum-of-Homs has dimension $\kappa\cdot(|\mathbb R|^{\aleph_0})=\kappa\cdot|\mathbb R|=\kappa$. So the dimensions are different.

Andreas Blass
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  • Why isn't the canonical map from the sum-of-Homs to the Hom-into-sum an isomorphism? – No One Nov 09 '17 at 20:41
  • In Mike Haskel's example, after you fix a basis for $M$ so you can identify homomorphisms with matrices as in the material you quoted, the canonical map sends each matrix (with only finitely many non-zero) rows to itself (considered as a matrix in which each column has only finitely many entries). Since $\dim M$ and $I$ are infinite, there are matrices of the second sort that are not of the first sort, so the canonical map is not surjective. – Andreas Blass Nov 09 '17 at 21:03
  • Is it still injective? If $I$ is countably infinite (we agree that two spaces have the same dimension in this case), an injective map will be an isomorphism, right? – No One Nov 09 '17 at 21:10
  • The canonical map is injective. The last part of your comment seems to be based on the theorem that an injective linear map between vector spaces of the same finite dimension is necessarily surjective and therefore an isomorphism. But in that theorem, "finite dimensional" is an essential part of the hypothesis. The result isn't correct in general for infinite-dimensional spaces. – Andreas Blass Nov 09 '17 at 21:14
  • This is mind-blowing... For the courtesy of later reader, let me just put a related thread on this:https://math.stackexchange.com/questions/419500/linear-map-between-the-same-dimensional-spaces?rq=1 – No One Nov 09 '17 at 21:20
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I think they are probably isomorphic to each other. Let me know where I am wrong.

The dimension of LHS is $|\bigoplus_{i=1}^{\infty}\mathbb R|^{\dim(M)}=|\mathbb R|^{\dim(M)}$, while the dimension of the RHS is $|\mathbb N||\dim(M^*)|=|\mathbb N||\mathbb R ^{\dim(M)}|=|\mathbb R ^{\dim(M)}|$.

Reference:

https://mathoverflow.net/questions/168596/dim-homv-w

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