A right exact module functor without coproduct preservation is not tensor

$F$ is an additive functor: the operations on ${\prod }_{n}V$ are coordinatewise by [F2], and for parallel maps $u,v$ one has $F(u+v)={\prod }_n(u+v)={\prod }_{n}u+{\prod }_{n}v=F(u)+F(v)$, with $F(\mathrm{i}\mathrm{d})=\mathrm{i}\mathrm{d}$ and $F(u^{\mathrm{\prime }}u)=F(u^{\mathrm{\prime }})F(u)$ coordinatewise.

$F$ is left exact: let $0\to V^{\mathrm{\prime }}\stackrel{u^{\mathrm{\prime }}}{\to }V\stackrel{u}{\to }{V}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ be a short exact sequence of $k$-modules. Coordinatewise, ${\prod }_n{u}^{\mathrm{\prime }}$ is injective, and an element $(v_n)$ of ${\prod }_{n}V$ lies in $\ker {\prod }_{n}u$ exactly when $u(v_n)=0$ for all $n$, i.e. $v_n\in \mathrm{im}{u}^{\mathrm{\prime }}$ for all $n$ by exactness at $V$; hence $\ker {\prod }_{n}u=\mathrm{im}{\prod }_n{u}^{\mathrm{\prime }}$ and $0\to {\prod }_n{V}^{\mathrm{\prime }}\to {\prod }_{n}V\to {\prod }_n{V}^{\mathrm{\prime }\mathrm{\prime }}$ is exact. By [F4] this proves left exactness.

Let $c:{⨁}_{j}F(k)\to F({⨁}_{j}k)$ be induced by the maps $F({\mathrm{ȷ}}_j)$. An element of its source is a family of scalar sequences $(s_j{)}_j$ supported on a finite set $J\subseteq \mathbb{N}$ of summand indices. Its image has $n$-th coordinate ${\sum }_{j\in J}{s}_j(n)e_j$, supported in $J$ for every $n$. Conversely, if $(v_n{)}_n$ has every $v_n$ supported in one finite set $J$, define $s_j(n)$ as the $j$-th coefficient of $v_n$ for $j\in J$ and put $s_j=0$ otherwise; then $(s_j{)}_j$ belongs to the source and maps to $(v_n{)}_n$. Thus the image consists exactly of families with supports contained in one fixed finite set of summand indices.

Under AC, $F$ is right exact: if $u:V\to V^{\mathrm{\prime }\mathrm{\prime }}$ is surjective, each fibre $u^{-1}(w_n)$ for $(w_n)\in {\prod }_n{V}^{\mathrm{\prime }\mathrm{\prime }}$ is nonempty, so by [F1] there is a choice function on the family $(u^{-1}(w_n){)}_{n\ge 0}$, whose values form $(v_n)\in {\prod }_{n}V$ with $u(v_n)=w_n$; hence ${\prod }_{n}u$ is surjective, and with the kernel computation of step 1.2 the sequence ${\prod }_n{V}^{\mathrm{\prime }}\to {\prod }_{n}V\to {\prod }_n{V}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ is exact. By [F4], $F$ is right exact. The Axiom of Choice is used exactly here, to select one preimage in each of the countably many fibres; only this countable instance is used.

The element $w=(e_n{)}_{n\ge 0}\in {\prod }_n({⨁}_{j}k)$, where $e_n$ is the $n$-th standard basis vector, is not in the image of $c$: it would be the image of a source family supported on a finite set $S$ of summand indices, forcing $\{n\}=\mathrm{supp}(e_n)\subseteq S$ for every $n$, so $S\supseteq \mathbb{N}$, contradicting the finiteness of $S$. Hence $c$ is not surjective and $F$ does not preserve the coproduct of the family $(k{)}_{j\in \mathbb{N}}$.

By steps 1.1-1.3 and 2.1 the functor $F$ is additive, left exact and right exact, hence exact by [F4] and in particular right exact.

No tensor functor represents $F$: suppose $\sigma :F\Rightarrow T_N$ is a natural isomorphism. Naturality of $\sigma$ at the coordinate inclusions ${\mathrm{ȷ}}_j$ gives $T_N({\mathrm{ȷ}}_j)\circ {\sigma }_k={\sigma }_{{⨁}_{j}k}\circ F({\mathrm{ȷ}}_j)$ for every $j$, so by the universal property in [F2] the comparison maps satisfy ${\sigma }_{{⨁}_{j}k}\circ c=c^{\mathrm{\prime }}\circ ({⨁}_j{\sigma }_k)$, where $c^{\mathrm{\prime }}$ is the comparison of $T_N$; since ${\sigma }_{{⨁}_{j}k}$ and ${⨁}_j{\sigma }_k$ are isomorphisms, $c$ is an isomorphism if $c^{\mathrm{\prime }}$ is. But $c^{\mathrm{\prime }}$ is an isomorphism because $T_N$ preserves coproducts by [F3], while $c$ is not surjective by step 2.2; this is a contradiction. Hence $F$ is not naturally isomorphic to any tensor functor.

Therefore $F$ is additive and exact, hence right exact, but does not preserve coproducts and is not tensor: the coproduct-preservation hypothesis of the Eilenberg-Watts theorem cannot be dropped even for exact functors. The only choice used is the coordinatewise lifting in step 2.1; the rest of the argument is choice-free.