# Understanding Freyed's Theorem in the Category Theory

Let $$\mathscr C$$ be a category. Suppose for any collection $$\{c_i\}_{i\in \mathscr C_1}$$ ($$\mathscr C_1$$ denotes the collection of morphisms in $$\mathscr C$$), the product $$\prod_{i\in \mathscr C_1} c_i$$ exists in $$\mathscr C$$. Then $$\mathscr{C}$$ is a preorder: $$\forall a,b \in \mathscr C$$, $$Hom_{\mathscr C}(a,b)$$ has at most one element.

Proof: Suppose $$\mathscr{C}$$ is not a preorder, then there exists $$a,b\in \mathscr{C}_0$$($$\mathscr C_0$$ denotes the collection of morphisms in $$\mathscr C$$) and $$f,g: a\to b$$ with $$f\neq g$$.

Consider a constant function $$j$$ $$:\mathscr{C_1}\to \mathscr{C_0}$$ by $$j(h)=b, \forall h\in \mathscr {C_1}$$. Let $$(\prod_{i\in \mathscr C_1}b,\{p_j: \prod_{\mathscr C_1 }b\to b\}_{j\in \mathscr{C_1}})$$ be the corresponding product. By the universal property of products, we have a bijection: $$Hom_{\mathscr C} (a,\prod_{i\in \mathscr C_1}b)\to \prod_{i\in \mathscr C_1 }Hom_{\mathscr C}(a,b).$$

$$|\prod_{i\in \mathscr C_1 }Hom_{\mathscr C}(a,b)|\geq 2^{|\mathscr {C_1}|}$$ and $$|Hom_{\mathscr C} (a,\prod_{i\in \mathscr C_1}b)|\leq |\mathscr{C_1}|$$ (contradiction)

My question:

$$1:$$ how to understand the set $$\{c_i\}_{i\in \mathscr C_1}$$? What is the form of the objects in this set? Since the index is a collection of morphisms, it confused me. (Also for the product $$\prod_{i\in \mathscr C_1} c_i$$)

$$2:$$ How can we conclude $$|\prod_{i\in \mathscr C_1 }Hom_{\mathscr C}(a,b)|\geq 2^{|\mathscr {C_1}|}$$ ? What I know is that given a set $$X$$, there is a bijection between $$\mathscr P(X)$$ and $$2^{|X|}$$, where $$2=\{0,1\}$$.

I am new to the category theory. Much appreciated to any help!

1. It's maybe easier to think about $$\{c_i\}_{i\in\mathscr{C}_1}$$ as a choice, namely for each morphism $$f$$ in $$\mathscr{C}$$ you have chosen a particular object $$c_f$$ in $$\mathscr{C}$$. Note that $$c_f$$ and $$f$$ do not need to be related in any way. The set $$\{c_i\}_{i\in\mathscr{C}_1}$$ is then the set whose elements are those chosen $$c_f$$, where $$f$$ ranges over all morphisms in $$\mathscr{C}$$. Likewise, the product $$c:=\prod_{i\in\mathscr{C}_1}c_i$$ is an object of $$\mathscr{C}$$ that comes with projection maps $$\pi_f\colon c\to c_f$$ for each morphism $$f$$ in $$\mathscr{C}$$, and satisfies the universal property of the product: a map $$d\to c$$ in $$\mathscr{C}$$ is the same datum as a collection of maps $$d\to c_f$$ for each morphism $$f$$ in $$\mathscr{C}$$.
2. The universal property of the product is exactly saying that $$\mathrm{Hom}(a,\prod_{i\in\mathscr{C}_1}b)$$ is in bijection with $$\prod_{i\in\mathscr{C}_i}\mathrm{Hom}(a,b)$$ (namely, we use the above description of $$\prod_{i\in\mathscr{C}_1}c_i$$ in the special case where $$c_f=b$$ for any morphism $$f$$ in $$\mathscr{C}$$). As such, $$|\mathrm{Hom}(a,\prod_{i\in\mathscr{C}_1}b)|=|\prod_{i\in\mathscr{C}_i}\mathrm{Hom}(a,b)|=\prod_{i\in\mathscr{C}_i}|\mathrm{Hom}(a,b)|=|\mathrm{Hom}(a,b)|^{|\mathscr{C}_1|}.$$ Since we assumed $$|\mathrm{Hom}(a,b)|\geq 2$$, this yields $$|\mathrm{Hom}(a,\prod_{i\in\mathscr{C}_1}b)|\geq 2^{|\mathrm{Hom}(a,b)|}$$.