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I know that for two sheaves having the same stalks in a necessary but not sufficient condition to be isomorphic. However, I also know that if I have two subsheaves $\mathcal{F},\mathcal{F}'$ of a sheaf $\mathcal{G}$, then they are equal if and only if they have the same stalks. Here it comes my question: given two sheaves $\mathcal{F}$,$\mathcal{F}'$ on a topological space $X$ I can always regard them as subsheaves of $\mathcal{F} \oplus \mathcal{F}'$. Hence, why it isn't true that two sheaves with the same stalks are isomorphic?

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    $\begingroup$ Nice little trap here! $\endgroup$ – Georges Elencwajg Jun 20 '18 at 18:14
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Let me clear up some confusion:
Fact 1
For two sheaves to be isomorphic it is necessary that their stalks at all points be isomorphic.
Do not say that these stalks are the same : this is meaningless.
The condition is not sufficient: two line bundles on a topological space (or manifold, variety, complex space, scheme,...) have isomorphic stalks but are not isomorphic in general.

Fact 2
Two subsheaves $\mathcal F, \mathcal F'\subset \mathcal G$ of a given sheaf on $X$ are equal if and only they have the same stalks at any $x\in X$: this means that the subsets $\mathcal F_x, \mathcal F'_x\subset \mathcal G_x$ are equal as subsets of the same set, and not only isomorphic.

Answer to your question
If you have two arbitrary sheaves $\mathcal F, \mathcal F'$ you can embed them into $\mathcal G=\mathcal F\oplus \mathcal F'$.
We do have isomorphisms $\mathcal F\cong\mathcal F\oplus 0\;, \mathcal F'=0\oplus \mathcal F'$.
However according to Fact 2 the two sheaves $\mathcal F\oplus 0\;,0\oplus \mathcal F'$ are definitely not equal.
They have isomorphic stalks, but as explained in Fact 1 this says nothing on their being isomorphic or not.
In conclusion, embedding $\mathcal F, \mathcal F'$ into their direct sum $\mathcal G$ has not advanced the problem in the least: there is no free lunch in mathematics!

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  • $\begingroup$ Thank you very much! Very clear answer. $\endgroup$ – Federico Jun 20 '18 at 18:23
  • $\begingroup$ You are welcome, dear Federico. $\endgroup$ – Georges Elencwajg Jun 20 '18 at 18:39

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