Are the natural injections always injections in any concrete category?

Question

I am beginning to learn about categories in Aluffi's Algebra: Chapter 0.

In every example I know, the coproduct's natural injections are always injections. Does this happen in any concrete category? If so, in general categories are they always monomorphisms?

Since I don't know a lot about categories yet, I appreciate keeping the language simple. Thanks!

Answer 1

It is not true in general: Consider the concrete category of commutative rings, where coproducts are given by tensoring over $\mathbb Z$. Now consider $\mathbb Z\to \mathbb Z\otimes \mathbb Z/2\mathbb Z=\mathbb Z/2\mathbb Z$, which is clearly not injective.

Answer 2

(Not a complete answer)

Let it be that $A,B$ are objects in a category and that $C$ serves as coproduct with injections $i:A\to C$ and $j:B\to C$.

Now if the homset $\mathsf{hom}(B,A)$ is not empty hence contains some arrow $f$ then according to the universal property of the coproduct there will exist a (unique) arrow $h:C\to A$ with $h\circ i=\mathsf{id}_A$ and $h\circ j=f$.

The equality $h\circ i=\mathsf{id}_A$ ensures that $i$ is a (split) monomorphism.

Answer 3

No, not necessarily. Consider a lattice-seen-as-a-category (so it has all binary coproducts) $L$ with enough objects that the following example works. Our functor $F$ realizing concreteness takes each object of the lattice to a different singleton, except for one element $A \in L$, which it takes to a two-element set. All arrow $F(f)$ that go into $F(A)$, except the identity, are taken to map to the same element of $F(A)$.

Then $F$ is a faithful functor, but if $g : A \to C$ is an arrow in $L$ other than the identity, then $F(g)$ will not be injective.

Answer 4

If your statement about monomorphisms were true, then by the duality principle, the projections of any product would always be epimorphisms.

This is not true: look at $\mathbf{Set}$, any product with $\emptyset$ is $\emptyset$ so is almost never an epimorphism. To get an example with literal coproducts, just take $\mathbf{Set}^{op}$.

The moral of this example is that it's always interesting to look at the dual statement : sometimes it may be easier to understand in usual categories.