Let φ: 𝒫 →𝒫^' be a presheaf map. Prove that the following...

Let $\varphi: \mathcal{P} \rightarrow \mathcal{P}^{\prime}$ be a presheaf map. Prove that the following statements are equivalent: (i) $\varphi$ is an isomorphism; (ii) $\varphi \mid \mathcal{P}(U): \mathcal{P}(U) \rightarrow \mathcal{P}^{\prime}(U)$ is an isomorphism for every open setU; (iii) $\varphi \mid \mathcal{P}(U): \mathcal{P}(U) \rightarrow \mathcal{P}^{\prime}(U)$ is a bijection for every open set $U$.

Let $\varphi: \mathcal{P} \rightarrow \mathcal{P}^{\prime}$ be a presheaf map. Prove that the following statements are equivalent:
(i) $\varphi$ is an isomorphism;
(ii) $\varphi \mid \mathcal{P}(U): \mathcal{P}(U) \rightarrow \mathcal{P}^{\prime}(U)$ is an isomorphism for every open setU;
(iii) $\varphi \mid \mathcal{P}(U): \mathcal{P}(U) \rightarrow \mathcal{P}^{\prime}(U)$ is a bijection for every open set $U$.

Key Concepts

Presheaf

A presheaf is a contravariant functor from the category of open sets (of a topological space) to another category (like sets, groups, etc.), which assigns to each open set an object and to each inclusion of open sets a corresponding restriction map. This structure captures the idea of local data that can be restricted from larger to smaller open neighborhoods.

Presheaf Map

A presheaf map is a natural transformation between two presheaves. It consists of a family of maps (one for each open set) that are compatible with the restriction maps of the presheaves. In other words, if a section is restricted to a smaller open set, applying the map either before or after the restriction yields the same result, ensuring coherent transformation of data across the space.

Isomorphism

In the context of presheaves, an isomorphism is a presheaf map that admits an inverse presheaf map. This is equivalent to the condition that on every open set, the map induces a bijection between the sections of the two presheaves. Therefore, checking if the individual maps on open sets are bijective is sufficient to conclude the presheaf map is an isomorphism.

Transcript

00:01 Okay, so here's a diagram of the functions.

00:03 In part a, we want to conclude that f is unto.

00:06 So let's take an arbitrary element x of c.

00:11 We know that there exists y in a, such that f composed with g evaluated at y equals x.

00:19 This is because of the assumption that f composed with j is under.

00:26 Now f composed with y of y can be written as f of y of y, and notice that this g of y is an element of v.

00:35 So we have found an element of v such that its image is under f is x and this means that f is on to x x observatory.

00:46 Now for part v we have to conclude that use one to one.

00:53 So let's take two elements x and y elements of a and let's suppose that g of x is equal to y and we want to prove that x is equal to y.

01:05 So let's consider, well this if this two, if this equation is valid then we can take f on both sides, we have this, and well we can write f of g of x as f composed with g of x and the same on the other side.

01:26 And because of the assumption that f composed with g is one to one, we conclude from here that x is equal to y and well we started assuming that g of x was equal to y and we concluded that x was equal to y that means that j is one to one and now for part c we have an if and relief so the first part is to prove this if f composed with g is a bijection and f is one to one then we have to prove that g is unto.

02:05 So let's consider an element, an arbitrary element of b.

02:13 Remember that the domain of g was a and it goes to b and f goes from b to c.

02:22 Yeah, so let's consider an arbitrary element of b and then noted y...

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