This question is being asked on behalf of a colleague of mine.
Let $X$ be a topological space. It is well known that the abelian category of sheaves on $X$ has enough injectives: that is, every sheaf can be monomorphically mapped to an injective sheaf. The proof is similarly well known: one uses the concept of "generators" of an abelian category.
It is also a standard remark in texts on the subject that on a general topological space $X$, the category of sheaves need not have enough projectives: there may exist sheaves which cannot be epimorphically mapped to by a projective sheaf. (Dangerous bend: this means projective in the categorical sense, not a locally free sheaf of modules.) For instance, Wikipedia remarks that projective space with Zariski topology does not have enough projectives, but that on any spectral space, a space homeomorphic to $\operatorname{Spec}R$, there are enough projective sheaves.
Two questions:
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Who knows an actual proof that there are not enough projectives on, say, $\boldsymbol{P}^1$ over the complex numbers with the Zariski topology? What about the analytic topology, i.e. $\boldsymbol{S}^2$?
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Is there a known necessary and sufficient condition on a topological space $X$ for there to be enough projectives?
EDIT: I meant the question to be purely for sheaves of abelian groups. Eric Wofsey points out that the results alluded to above on Wikipedia are not consistent when interpreted in this way, since $\boldsymbol{A}^1$ and $\boldsymbol{P}^1$, over an algebraically closed field $k$, with the Zariski topology are homeomorphic spaces: both have the cofinite topology.
I am pretty sure that when my colleague asked the question, he meant it in the topological category, so I won't try to change that. But the other case is interesting too. What if $(X,\mathcal{O}_X)$ is a locally ringed space. Does the abelian category of sheaves of $\mathcal{O}_X$-modules have enough projectives? I think my earlier warning still applies. Since for a scheme $X$ not every $\mathcal{O}_X$-module is coherent, it is not clear that "projective sheaves" means "locally free sheaves" here, even if we made finiteness and Noetherianity assumptions to get locally free and projective to coincide.
Best Answer
About Jon Woolf's answer, it seems to me that the condition that "$x$ is a closed point" was implicitly used: the extension by zero $Z_A$ is only defined for a locally closed subset $A$ (see e.g. Tennison "Sheaf theory," 3.8.6). So $X-x$ must be locally closed. How about the following trivial modification: instead of $Z_X$, consider the sheaf $i_\ast Z$, where $i$ is the inclusion of a point $x$ into $X$.
Suppose that $x$ does not have the smallest open neighborhood and $x$ has a basis of connected neighborhoods. Then $i_\ast Z$ is not a quotient of a projective sheaf $P$. Suppose otherwise. Then for any connected open neighborhood $U$ of $x$, the homomorphism $P(U) \to i_\ast Z(U)$ is zero. This implies that the homomorphism $P \to i_\ast Z$ is zero since it is equivalent to a homomorphism from the stalk $P_x$ to $Z$. Indeed, pick a neighborhood $V$ which is smaller than $U$. We have a surjection $Z_V \to i_\ast Z$. The homomorphism $P \to i_\ast Z$ must factor through $Z_V$, so $P(U) \to i_\ast Z(U)$ must factor through $Z_V(U)$. But $Z_V(U)=0$. This equality may fail if $U$ is not connected however.
So to summarize Jon Woolf and David Treumann, the category of sheaves of abelian groups on a locally connected topological space $X$ has enough projectives iff $X$ is an Alexandrov space.
Surely this must appear in some standard text. Anybody knows a reference? And what about non-locally connected spaces?
For ringed spaces $(X,\mathcal{O}_X)$ one direction is still clear: $X$ being an Alexandrov space implies you'll have enough projectives. But on reflection the other direction, $X$ being a locally connected space without minimal open neighborhoods implies you don't have enough projectives, appears to be rather tricky. One can think of some weird structure sheaves for which the above argument does not go through, in particular $\mathcal{O}_V(U)\ne 0$. So I still wonder what the answer is for ringed spaces.