While proving the equivalence of Sheaf cohomology (defined by using injective resolutions), Tennison (in his book 'Sheaf Theory' on page 146) says :

If we let $S$ be a presheaf such that the sequence $ 0 \to R' \to R \to S \to 0 $ is exact in Presh/X, we see that $S$ has as sheafification the zero sheaf and there is an exact sequence < long exact sequence corresponding to the above short exxact sequence>.

Here $R' \to R$ is the sheafification map of the presheaf $R'$. I can understand that $S$ can be shosen to be the Presheaf cokernel of the sheafification map. My question is : But how do we know that this map is exact at $R'$ ? Edit : As pointed out in the comment, this need not be exact at $R'$. So the question is : How do we tweak the proof in the book so that it becomes valid ?

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    $\begingroup$ Unless $R^\prime$ is separated, meaning that two sections over an open which agree on some covering of that open must be the same, the map from $R^\prime$ to its sheafification will not be injective. $\endgroup$ – Keenan Kidwell Mar 2 '16 at 13:09
  • $\begingroup$ @KeenanKidwell Thanks. That is right, I have edited the question to ask : how do we tweak the proof in the book then ? $\endgroup$ – user90041 Mar 4 '16 at 6:32
  • $\begingroup$ You could consider instead the short exact sequences $0\to \text{ker}(s)\to R^{\prime}\to \text{im}(s)\to 0$ and $0\to \text{im}(s)\to R\to\text{coker}(s)\to 0$ where $s: R^{\prime}\to R$ is the sheafification map, using that the sheafifications of $\text{ker}(s)$ and $\text{coker}(s)$ vanish. $\endgroup$ – Hanno Mar 4 '16 at 7:03

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