Coproduct and inclusion maps

Question

How can I show that there exist injections $j_1 \to A+B$ and $j_2 \to A+B$ where $A$ and $B$ are categories and $A+B$ their coproduct. The claim is that the union of the images of $j_1$ and $j_2$ is all of $A+B$ and the intersection of the images is empty. This reminds me a lot of some sort of equivalence relation, but I'm missing the details.

Answer 1

Maybe the best is if we turn it upside down:

Generally, if we have a guess about what coproducts will be in a certain category, probably the most straightforward way to verify it is by proving the universal property of coproducts for the guessed object.

In this case, we guess/claim that the coproduct in ${\bf Cat}$ is the disjoint union (just as in ${\bf Set}$, ${\bf Top}$ or ${\bf Graph}$).
So, take (small) categories $A,B$ and form their disjoint union - let's denote it $A+B$ as it will be (isomorphic(!)) to the coproduct.

Then the claim is that $A+B$ with the inclusion functors $j_1:A\to A+B$ and $j_2:B\to A+B$ satisfy the universal property: for any $f:A\to C$, $\,g:B\to C$ there is a unique arrow $A+B\to C$ making both arising triangles commutative.

---

After it's shown, if we want to start out from a given coproduct cone (with $j_1:A\to U$ and $j_2:B\to U$), then we will have $U\cong A+B$, and using that, we can prove that these $j_1$ and $j_2$ are indeed injective.

Answer 2

I assume $A$ and $B$ are both nonempty.

We can prove that $A \to A+B$ is injective by considering the map $A + B \to A$ induced by:

• The identity functor $A \to A$
• Any constant functor $B \to A$

The composite $A \to A+B \to A$ is the identity, and thus $A \to A+B$ is monic.

Similarly, $B \to A+B$ is monic.

---

We can prove the images are disjoint by letting $C$ be the two object category with objects $0$ and $1$ and considering the map $A + B \to C$ given by

• $A \to C$ is the constant functor mapping to $0$
• $B \to C$ is the constant functor mapping to $1$

Thus, nothing in $A+B$ can be in the image of both $A$ and $B$, because then its image in $C$ would have to be both $0$ and $1$.