Does the functor $V\mapsto V^{**}/V$ (for $\dim V=\infty$) reflect monomorphisms, epimorphisms or isomorphisms?

Let $$\textbf{Vect}_\infty$$ be the category of infinite dimensional vector spaces (over some fixed ground field), and consider the endofunctor $$F$$ of $$\textbf{Vect}_\infty$$ whose effect on objects is given by $$F(V):=V^{**}/V$$, and whose effect on morphisms is the obvious one. (Here $$V^{**}/V$$ is the cokernel of the canonical monomorphism $$V\to V^{**}$$.)

Does $$F$$ reflect monomorphisms?

Does it reflect epimorphisms?

Does it reflect isomorphisms?

(Recall the $$F$$ reflect monomorphisms if the condition that $$F(f)$$ is a monomorphism implies that $$f$$ is also a monomorphism. The reflections of epimorphisms and isomorphisms are defined similarly.)

• If $V$ is an infinite dimensional vector space, if $K$ is the ground field, and if $f$ is the endomorphism $(\lambda,v)\mapsto(0,v)$ of $K\times V$, then $F(f)$ is (isomorphic to) the identity of $F(V)$. This shows that the answer to the three questions is No. – Pierre-Yves Gaillard Oct 30 '18 at 9:31

The endofunctor $$F:\mathbf{Vect}_\infty\rightsquigarrow\mathbf{Vect}_\infty$$ does not reflect monomorphisms. Pick $$V$$ to be the vector space of sequences $$(x_1,x_2,x_3,\ldots)$$ with finitely many non-zero terms. Choose $$f:V\to V$$ to be the left shift operator $$(x_1,x_2,x_3,\ldots)\mapsto (x_2,x_3,x_4\ldots).$$
Clearly, $$f$$ is not monic. We claim that $$F(f)$$ is monic.

The kernel of $$F(f)$$ consists of $$\alpha +V$$ with $$\alpha \in V^{**}$$ satisfying $$f^{**}\alpha \in V$$. Because $$f$$ is an epimorphism, $$f^{**}\alpha=f(v)$$ for some $$v\in V$$. Now, for an arbitrary $$\varphi\in V^*$$, we have $$\alpha(\varphi\circ f)=f^{**}\alpha(\varphi)=f(v)(\varphi)=\varphi\big(f(v)\big)=(\varphi\circ f)(v)=v(\varphi\circ f).$$ Therefore, $$(\alpha-v)(f^*\varphi)=(\alpha-v)(\varphi\circ f)=0$$ for all $$\varphi\in V^*$$.

For convenience, let $$e_j\in V$$ denote the sequence $$(0,0,0,\ldots,0,1,0,0,0,\ldots)$$, where there is only one $$1$$ at the $$j$$th term, and other terms are $$0$$. So, $$V=\bigoplus_{j=1}^\infty Ke_j$$ if $$K$$ is the base field, and $$V^*$$ can be identified with $$\prod_{j=1}^\infty Ke_j$$. Hence, we can identify $$V$$ as a subspace of $$V^*$$ as well.

Observe that $$V^*=Ke_1\oplus \operatorname{im}f^*$$. That is, if $$\lambda=(\alpha-v)(e_1)$$, then $$\alpha=v+\big(\lambda-v_1\big)e_1\in V,$$ where $$v=(v_1,v_2,v_3,\ldots)$$. That is, $$F(f)$$ is monic, but $$f$$ is not.

Note that $$F(f)$$ in the example above is actually an isomorphism. To show that $$F(f)$$ is epic, we observe that for $$\beta\in V^{**}$$, there exists $$\alpha \in V^{**}$$ such that $$\beta=f^{**}\alpha$$. We define $$\alpha(\varphi_1,\varphi_2,\varphi_3,\ldots)=\beta(\varphi_2,\varphi_3,\varphi_4,\ldots)$$ for all $$\varphi=(\varphi_1,\varphi_2,\varphi_3,\ldots)\in V^*$$. That is, $$f^{**}\alpha(\varphi_1,\varphi_2,\varphi_3,\ldots)=\alpha(0,\varphi_1,\varphi_2,\varphi_3,\ldots)=\beta(\varphi_1,\varphi_2,\varphi_3,\ldots)$$ for all $$\varphi\in V^{*}$$, so $$f^{**}\alpha=\beta$$. This proves that $$F(f)$$ is epic. Since it is monic, $$F(f)$$ is an isomorphism, but as we learned, $$f$$ is not an isomorphism. Therefore, $$F$$ does not reflect isomorphisms.

The endofunctor $$F$$ does not reflect epimorphisms. Pick $$V$$ to be the same as above, but now let $$f$$ be the right shift operator $$(x_1,x_2,x_3,\ldots)\mapsto (0,x_1,x_2,x_3,\ldots).$$ Clearly, $$f$$ is not epic. We claim that $$F(f)$$ is epic.

Let $$\beta \in V^{**}$$. We want to find $$\alpha \in V^{**}$$ such that $$\beta-f^{**}\alpha\in V$$. With the identification $$V^*=\prod_{j=1}^\infty Ke_j$$ as before, we take $$\alpha(\varphi_1,\varphi_2,\varphi_3,\ldots)=\beta(0,\varphi_1,\varphi_2,\varphi_3,\ldots)$$ for all $$\varphi=(\varphi_1,\varphi_2,\varphi_3,\ldots)\in V^*$$ with $$\varphi_j=\varphi(e_j)$$. Therefore, for all $$\varphi\in V^*$$, $$\big(\beta-f^{**}\alpha\big)(\varphi)=\beta(\varphi_1,0,0,0,\ldots)=\varphi_1\beta(e_1)=\beta(e_1)\ e_1(\varphi).$$ That is, $$\beta-f^{**}\alpha =\beta(e_1)\ e_1\in V.$$ Consequently, $$\beta+V=F(f)(\alpha +V)$$, and so $$F(f)$$ is epic without $$f$$ being epic.

However, this is true. If $$f:V\to W$$ is such that $$F(f)$$ is monic, then $$W\cap \big(\operatorname{im}f^{**}\big)\subseteq \operatorname{im} f$$. Conversely, if $$f:V\to W$$ is such that $$W\cap \big(\operatorname{im}f^{**}\big)\subseteq \operatorname{im} f$$ with the extra condition that $$f$$ is monic, then $$F(f)$$ is monic.

To show this, note that the kernel of $$F(f)$$ consists of $$\alpha+V$$ with $$\alpha \in V^{**}$$ satisfying $$f^{**}\alpha \in W$$. Because $$F(f)$$ is monic, $$\alpha=v$$ for some $$v\in V$$. As $$f^{**}\alpha=w$$ for some $$w\in W\cap\big(\operatorname{im}f^{**}\big)$$, $$\varphi\big(f(v)\big)=(\varphi\circ f)(v)=v(\varphi\circ f)=\alpha(\varphi \circ f)=f^{**}\alpha(\varphi)=w(\varphi)=\varphi(w)$$ for all $$\varphi \in W^*$$. That is, $$\varphi\big(w-f(v)\big)=0$$ for every $$\varphi\in W^*$$. This shows that $$w=f(v)$$, and so $$w\in\operatorname{im}f$$.

The converse can be proven as follows. Let $$\alpha\in V^{**}$$ be such that $$\alpha+V\in\ker F(f)$$. Then, $$f^{**}\alpha \in W\cap \big(\operatorname{im} f^{**}\big)\subseteq \operatorname{im}f$$, so $$f^{**}\alpha=f(v)$$ for some $$v\in V$$. Thus, $$(\alpha-v)(\varphi\circ f)=f^{**}\alpha(\varphi)-f(v)(\varphi)=0$$ for all $$\varphi\in W^*$$, so the injectivity of $$f$$ implies surjectivity of $$f^*:W^*\to V^*$$, which means $$\alpha=v$$.

• Thank you very much for this magnificent answer!!! However I don't understand the third point. It seems to me $F(f)$ is injective iff $f^{**}(\alpha)\in W$ implies $\alpha\in V$. I don't see why your $w$ is an arbitrary vector of $W$. – Pierre-Yves Gaillard Oct 29 '18 at 14:47
• Ach! You are very right. This part will be modified. – user593746 Oct 29 '18 at 15:03
• Now I understand! Nice! - Just to make sure I understand correctly the status of my 3 questions after your post, do you agree that the reflection of isomorphisms (or conservatism) remains open? – Pierre-Yves Gaillard Oct 29 '18 at 15:36
• All of them have been answered, if I didn't make stupid mistakes. The same example that fails reflection of monomorphisms works as a counterexample for the reflection of isomorphisms too. – user593746 Oct 29 '18 at 15:37
• I will take a look at them later, but I cannot promise to be able to help. – user593746 Oct 29 '18 at 16:00