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.)
 A: 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 $\alpha=v$.
