Regarding the injectivity of units of monads on $\mathbf{Set}.$ Given a monad $T$ on $\mathbf{Set}$, it usually seems to be the case that the unit of $T$ is componentwise injective (meaning that for all objects $X$ of $\mathbf{Set}$, the map $\eta^T_X : X \rightarrow TX$ is an injective function.) The only non-example I can think of is the terminal monad $T$, defined such that for all sets $X$, we have  $TX = 1.$
If this is true, it means means that we can think of each non-terminal monads on $\mathbf{Set}$ as corresponding to a notion of "generalized point", because if we're given a set $X$, then we can build a larger set $TX$ of generalized points, and distinct points of $X$ remain distinct under the inclusion $X \subseteq TX$.
But, this seems to fail in most concrete categories. Consider, for instance, the case where $T$ is the abelianization monad on $\mathbf{Grp}$. Then the morphism $G \rightarrow TG$ is injective iff $G$ is itself Abelian. So the concrete category $\mathbf{Grp}$ doesn't have this miraculous property.

Questions. Is it really true that every monad on $\mathbf{Set}$, except for the terminal monad, has a componentwise injective unit?
If so, what are some other examples of concrete categories with this remarkable property?

 A: Let $f : A \to B$ be a function. We have a commutative diagram
$$ \begin{matrix}
A &\xrightarrow{\eta_A}& T(A)
\\ {\ \ } \downarrow {f} & & {\quad}\downarrow {T(f)} & & 
\\ B &\xrightarrow{\eta_B}& T(B)
\end{matrix} $$
In particular, if $x,y \in A$ satisfy $\eta_A(x) = \eta_A(y)$, then $\eta_B(f(x)) = \eta_B(f(y))$
However, for any set $B$ and any pair of elements $u,v \in B$, there is some function $f : A \to B$ such that $f(x) = u$ and $f(y) = v$.
Consequently, if any $\eta_A$ fails to be injective, then $\eta$ equates all of the elements of all sets.
However, that does not imply $T$ is the terminal monad. A simple counterexample is
$$ T(S) = \begin{cases} \varnothing & S = \varnothing \\ 1 & S \neq \varnothing \end{cases} $$
with the unit and multiplication being the only things they can be. This is the monad corresponding to the variety of universal algebras with no constant symbols and one relation $x=y$.
These are the only two examples where $\eta$ fails to be injective; for any monad, $\eta_{T(S)}$ is monic. So given the argument above, we have to have $T(S) \subseteq 1$ for all $S$.
A: Let $\mathcal{C}$ be a category and $U : \mathcal{C} \to \mathsf{Set}$ be a functor with left-adjoint $F$. If there is an $A\in \mathcal{C}$ with $|UA|\geq 2$, then the unit of the adjunction $F \dashv U$ is componentwise injective (see this answer). 
Now observe, that every monad $T$ on a category $\mathcal{A}$ arises from some adjunction between $\mathcal{A}$  and the category of algebras $\mathcal{A}^T$, such that the unit of the monad is the unit of the adjunction.
Amusingly another answer to the same question says, that the unit of an adjunction $F\dashv U$ is componentwise mono if and only, if $U$ is faithful. So this may give you some ideas for possible generalizations of the property you are interested in.
