According to Hartshorne's book a sheafification is:

I saw a different definition elsewhere

**Definition 2**

Given a presheaf $\mathcal F$ in a topological space $X$ we can associate a $\mathcal F$ a sheaf called $\mathcal F^+$:

$\mathcal F^+(U):=\{t\in D_{\mathcal F}(U);t_x\in Im((\varphi_{\mathcal F})_x),\forall x \in U\}$.

What I'm trying to understand is why saying the sheaf he constructs with (1) and (2) in the Hartshorne's proof is equivalent to the sheaf of the definition 2.

**My attempt to solve this question**

$\varphi_{\mathcal F}$ is defined as $\varphi_{\mathcal F}:\mathcal F\to D_{\mathcal F}$ and when we apply an open subset $U$, we have $\varphi_{\mathcal F}(U):\mathcal F(U)\to D_{\mathcal F}(U)$, where $D_{\mathcal F}(U)=\prod_{x\in U} \mathcal F_x$ and we know that $t_x\in Im((\varphi_{\mathcal F})_x),\forall x \in U\Leftrightarrow t\in Im((\varphi_{\mathcal F}))$

Since the image of $\varphi_{\mathcal F}$ is an element of $\prod_{x\in U} \mathcal F_x$ the part (1) is ok, since these functions $s$ can be regarded as elements of $\prod_{x\in U} \mathcal F_x$ because of (1). (see definition of products as functions).

I'm having troubles to prove that the sheaf of the definition 2 has as a property the part 2 in the Hartshorne's proof and vice-versa.

Thanks