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I am reading about additive categories from https://ncatlab.org/nlab/show/additive+category and trying to prove that in any Ab-enriched category, if the coproduct and product of A,B exist, then there is an isomorphism between them. This is proposition 2.1 in the linked page.

In particular, I am trying to understand equation (**) in 2.1. We are showing that $(id\times 0)\pi_1 + (0\times id)\pi_2 = id_{X_1\times X_2}$ by composing both sides of this equation with every possible morphism from $R\rightarrow X_1\times X_2$ and checking that they both give the same result. Why is this enough to check that these morphisms are the same? The linked page says that this is the "Yoneda Lemma on Generalized Elements," but I cannot find any information on what that means.

My understanding of category theory & the Yoneda lemma is very limited, so any details that can be spelled out here are very much appreciated.

2 Answers 2

1

Invoking the Yoneda lemma for this is more of a habit than a necessity. Given two arrows $f,g : X \to Y$, the following are equivalent:

  • $f = g$
  • $f \circ 1_X = g \circ 1_X$
  • $f \circ h = g \circ h$ for all $h$ with codomain $X$
  • $f_*, g_* : \hom(-, X) \to \hom(-, Y)$ are the same natural transformation

For the example at hand, it suffices to use the second bullet; i.e. take $R = X_1 \times X_2$, and the generalized element named by the identity map, written as $(\alpha, \beta)$ where $\alpha$ and $\beta$ are the two projections $R \to X_1$ and $R \to X_2$.

Of course, such a proof looks basically the same as a proof of the third bullet or of the fourth bullet, so with experience one tends to conflate these ideas rather than carefully distinguishing between them, which is why the Yoneda lemma is invoked as the rationale.

2

Generalizing, consider two arrows of a category $\mathcal{C}$, $g, h : A \to B$. We have the hom-functors $\mathcal{C}(-,A)$ and $\mathcal{C}(-,B) : \mathcal{C}^{op}\to\mathbf{Set}$. The Yoneda embedding turns each arrow into a natural transformation $g\circ - : \mathcal{C}(-,A) \to \mathcal{C}(-,B)$ similarly for $h$. As a simple corollary of the Yoneda lemma, the mapping $g \mapsto g \circ -$ is a natural bijection.

$\sigma = \tau$ holds for $\sigma, \tau : \mathcal{C}(-,A)\to\mathcal{C}(-,B)$ if and only if for all $R \in \text{Ob}(\mathcal{C})$ and for all $f \in \mathcal{C}(R,A)$, $\sigma_R(f) = \tau_R(f)$. For the natural transformations above, this is $g\circ f = h\circ f$ for all $f \in \mathcal{C}(R,A)$ for all $R \in \text{Ob}(\mathcal{C})$. The (natural) bijection given by the Yoneda lemma then implies that this means $g = h$. Of course, this is obvious because we can choose $R = A$ and $f = id$ which is nothing more than half of the proof of the Yoneda lemma.

The story is basically the same in the $\mathbf{Ab}$-enriched case.