Is the inclusion into a coproduct an equalizer?

Question

Let $\mathcal{C}$ be a category with finite limits and binary coproducts. Let $1$ denote the terminal object, and let $\langle\rangle_A : A \rightarrow 1$ denote the unique map from $A$ to $1$. If we have two objects $A,B$, then we can form their coproduct $A+B$, and we have a map $\langle\rangle_A + \langle\rangle_B : A+B \rightarrow 1+1$. However, we can also map all of $A+B$ to the terminal object, and include that into $1+1$ on the left side, giving us the map $i_1 \circ \langle\rangle_{A+B} : A+B \rightarrow 1+1$. My question is: is $i_1 : A \rightarrow A+B$ the equalizer of these two arrows?

Intuitively, this makes sense to me. The "$A$ part" of $A+B$ is precisely what's getting sent to the left component of $1+1$ by $\langle\rangle_A+\langle\rangle_B$, so $A$ should be the equalizer. However, I don't see how it's possible to recover a map $X \rightarrow A$ from a map $X \rightarrow A+B$. This leads me to believe this is not true in general. However, I'm not familiar enough with category theory to come up with a counterexample.

Answer 1

This need not be true.

For example, in an abelian category, $1 + 1 \simeq 1 \times 1 \simeq 1,$ so your diagram will look like $A + B \rightrightarrows 1,$ for which the equalizer is necessarily $A + B.$

Answer 2

This is actually true in extensive categories. We'll use the definition given by Carboni, Lack, and Walters, in their paper Introduction to extensive and distributive categories. The property of extensivity that we need is found in proposition 2.2: if a category $\mathcal{C}$ (assumed to have binary coproducts) is extensive, then for any commutative diagram as below, where the bottom row is a coproduct,

the two squares are pullbacks iff the top row is a coproduct.

To see how this can be applied to the problem at hand, we consider the following diagram.

We know both rows are coproducts, and it's easy to check that the diagram commutes. Therefore, by the above extensivity property, the left square is a pullback.

It remains to show that the left square being a pullback implies the equalizer condition we want. Concretely, the pullback property states that $\forall X$, $\forall f : X \rightarrow 1, \; g : X \rightarrow A+B$ such that $i_1 \circ f = (\langle\rangle_A + \langle\rangle_B) \circ g$, there exists a unique $h : X \rightarrow A$ such that $f = \langle\rangle_A \circ h$ and $g = i_1 \circ h$.

But, because $1$ is a terminal object, this $f$ must be $\langle\rangle_X$. The equality $f = \langle\rangle_A \circ h$ automatically holds, and $i_1 \circ f = i_1 \circ \langle\rangle_X = i_1 \circ \langle\rangle_{A+B} \circ g$. Therefore, the condition can be rewritten as: $\forall X$, $\forall g : X \rightarrow A+B$ such that $(i_1 \circ \langle\rangle_{A+B}) \circ g = (\langle\rangle_A + \langle\rangle_B) \circ g$, there exists a unique $h : X \rightarrow A$ such that $g = i_1 \circ h$. This is precisely the equalizer condition.