I am reading Hartshorne's proof of Lemma 3.3 on Chapter III, that asserts that for every injective module $I$, the natural map $I\rightarrow I_f$ is surjective.
The key of his proof is that for every element $y/f^n \in I_f$, he defines a map $\varphi:(f^{n+r})\rightarrow I$, sending $f^{n+r}$ to $f^ry$, where $r$ is the "upper bound" of the chain of annihilators of the $f^i$ in $A$. He claims that "this is possible" because the annihilator of $f^{n+r}$ annihilates $f^ry$.
I do not understand this argument. More precisely, I do not see where do you have to use the fact concerning the annihilators to define the map $\varphi:(f^{n+r}) \rightarrow I$.
The idea is to make it well-defined. We want to show there exists a module homomorphism $\phi : (f^{n+r}) \rightarrow I$ that sends $f^{n+r}$ to $f^r y$.
We might define it in a way like : let $a \in (f^{n+r})$, so we can write it in the form $a = f^{n+r}*b$ for some $b \in A$. Then, $\phi$ sends $a$ to $b f^r y$.
The problem is that if $a = f^{n+r}*b = f^{n+r}*c$ for $b, c \in A$, we have that $\phi(a) = b f^r y = c f^r y$. This is only valid if $b f^r y = c f^r y$
If $f^{n+r}*b = f^{n+r}*c$, then $(b-c)f^{n+r} = 0$ so $b-c \in Ann(f^{n+r}) = Ann(f^n)$. Thus, $(b -c)f^r y = 0$ so $b f^r y = c f^r y$. Therefore, $\phi$ is well-defined.