Is $ \mathcal{F} \ : \ U \ \mapsto \ \prod_{x \in U} A_x $ a sheaf?

Question

Here is an elementary question about sheaves to a category $\mathcal{C}$ having all its products.

Let $X$ be a topological space and $\{A_x \ | \ x \in X \}$ be a family of objects inside among the objects of $\mathcal{C}$. Now I wonder if the following assignment is a sheaf: $$ \mathcal{F} \quad : \quad U \ \longmapsto \ \prod_{x \in U} A_x. $$ I am almost sure it is since it can also be described as $$ \mathcal{F}' \quad : \quad U \ \longmapsto \ \{ U \stackrel{f}{\rightarrow} Ob(\mathcal{C}) \ | \ f \text{ a map} \} \ $$ ans that must be a sheaf. Is this right?

Answer 1

Yes, this is the product over $x\in X$ of the skyscraper sheaves $A_x$, and sheaves are closed under limits.

Answer 2

We want to prove that for every open covering $U = \bigcup_{i \in I} U_i$ and every family of morphisms $f_i : T \to \prod_{x \in U_i} A_x$ with $f_i |_{U_i \cap U_j} = f_j |_{U_i \cap U_j}$ $(\star)$ there is a unique morphism $f : T \to \prod_{x \in U} A_x$ with $f |_{U_i} = f_i$. By the universal property of products, we may define $f$ by $p_x f : T \to A_x$ for each $x \in U$. Choose some $i \in I$ with $x \in U_i$. We let $p_x f := p_x f_i$. Because of $(\star)$ the choice of $i$ does not matter. We have $f|_{U_i}=f_i$ by construction, and it is clear that this is how we had to define $f$ in order to fulfill this equation.