I am currently reading about the Eilenberg-Watts theorem. I got the original work from Charles Watts "Intrinsic Characterizations of Some Additive Functors" and theorem 1 is what I am interested in. Alternatively the proof is stated here and the important part here in Lemma 1.

The proof in Watts says the following (a few, very small changes are made be me to fit it in my paper):

Consider a functor $F: \mathrm{mod-}A \rightarrow \mathrm{mod-}A$, a module $M \in \mathrm{mod-}A$, $m \in M$ and the homomorphism $\phi_m: A \rightarrow M, a \mapsto a\cdot m$.

Since $\phi_m \in \mathrm{Hom}(A, M)$ we can apply $F$ on $\phi_m$ and get $F(\phi_m): F(A) \rightarrow F(M)$.

We define $$\Psi_0^M: M \times F(A) \rightarrow F(A), (M, \tilde{A})\mapsto F\phi_M(\tilde{a}),$$ where $a \in A$ and $\tilde{a} \in F(A)$.

Since $A$ is a bimodule over itself consider $\Psi_0^A: A \times F(A)\rightarrow F(A)$ and by that $F(A)$ gets a left module structure. This is compatible with the right module structure such that $F(A)$ is a bimodule.

By the universal property of the tensor product we get $\Psi^M: M\otimes F(A) \rightarrow F(M)$.

Next we show that $(\Psi^M)_{M \in \mathrm{mod-}A}$ is a natural transformation [...]

Since $F$ and $- \otimes F(A)$ commute with direct sums, it follows that $\Psi^M$ is an isomorphism whenever $M$ is a free module. Now take an arbitrary $N \in \mathrm{mod-}A$. Look at a chosen exact sequence $$ 0 \rightarrow R \rightarrow M \rightarrow N \rightarrow 0 $$ where $M$ is free. Since $F$ and $- \otimes F(A)$ are right-exact, we get the following diagram with exact rows: enter image description here

With a diagram chase we can conclude that $\Psi^N$ is an epimorphism. Since $N$ was chosen arbitrary it follows that $\Psi^R$ is an epimorphism. Again with a diagram chase it follows that $\Psi^N$ is a monomorphism and hence an isomorphism.

My questions to that are the following:

  • Why is $\Psi^M$ an isomorphism if $M$ is free?
  • What is the module $R$?
  • Why does such an exact sequence $0 \rightarrow R \rightarrow M \rightarrow N \rightarrow 0$ exist?
  • Why is $\Psi^R$ an epimorphism since $N$ was chosen arbitrary?

If you want to check the other source (Specksmath) I gave, my questions would be:

  • What are $F_0$ and $F_1$?
  • Why can we say that $\sigma$ is an isomorphism for both of them?
  • Why does such an exact sequence $F_1 \rightarrow F_0 \rightarrow M \rightarrow 0$ exist?

Maybe someone can help me understand this :D


The exact sequences exist because for any module $N$, we can always take

  • $M=F(I)$ to be the free module on some set $I$ of generators of $N$
  • $F(I)\to N$ to be the map induced by the inclusion $I\hookrightarrow N$ and the universal property of the free module $F(I)$

and this is a surjective morphism, because $I$ generates $N$. Then if you take $R$ to be the kernel of $M\to N$, with its inclusion into $M$, you have by construction a short exact sequence $$0\rightarrow R \rightarrow M \rightarrow N \rightarrow 0.$$ This is what happens in the article. You can also apply the same trick to obtain a surjective map $F(J)\to R$, and then composing this with the inclusion $R\hookrightarrow F(I)$ gives you an exact sequence $$F(J) \rightarrow F(I) \rightarrow N \rightarrow 0,$$ and that's what happens in the blog post (with slightly different notation).

Now in the article, $\Psi^A$ is the canonical isomorphism $A\otimes_A F(A)\to F(A)$, since the structure of left $A$-module of $F(A)$ is exactly the bilinear map that induces $\Psi_A$. Moreover if $M=F(I)=\bigoplus_{i\in I}A$, then since $\_\otimes_AF(A)$ and $F$ commute with direct sums we have canonical isomorphisms $$\bigoplus_{i\in I}(A\otimes_A F(A))\to M\otimes_A F(A)\quad \text{and}\quad \bigoplus_{i\in I}F(A)\to F(M),$$ which make the diagram $$\require{AMScd}\begin{CD}\bigoplus_{i\in I}(A\otimes_A F(A)) @>>> M\otimes_A F(A)\\ @V{\oplus_i \Psi^{A}}VV @VV{\Psi^M}V \\ \bigoplus_{i\in I}F(A) @>>> F(M)\end{CD}$$ commute. In particular, $\Psi^M$ is then an isomorphism.

This implies that the composite $M\otimes_A F(A)\to F(M)\to F(N)$ is surjective, and then so is $\Psi^N$, thanks to the commutative diagram you've written. This is true for any module $N$; in particular, it's also true for $R$.

  • $\begingroup$ Thanks a lot... A few questions to your notation: In my question, $M$ is the free module and $N$ an arbitrary one. In your answer $M$ ist the arbitrary one in the beginning. Later you say that $M = \oplus A$ is $M$ still the arbitrary module or the free module? $\endgroup$ – P. Schulze Mar 4 at 17:04
  • 1
    $\begingroup$ @P.Schulze I started with notation that matched the blog post and then inadvertently switched to the article, so $M$ was an arbitrary module in the first part and then a free module. I've edited to keep the notation closer to the article. $\endgroup$ – Arnaud D. Mar 4 at 17:21

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