I'm confunsed by the definition of direct image of sheaf which says

Let $f :X \rightarrow Y $ be a continuous map of topological spaces, and let $\mathcal{F}$ be sheaf on $X$. Then there is a sheaf on $Y$ given by $(f_{*} \mathcal{F})(V)=\mathcal{F}(f^{-1}(V))$, where $V$ is an open subset of $Y$.

But $(f^{-1}(V))=U$, and $U$ in an open subset of X. The problem is we will get a sheaf on $X$, not on $Y$. A sheaf on $Y$ should be defined on some open subset of Y and gives as some nice function(holomorphic,....) on that open subset, but here by that definition we get $\mathcal{F}(U)$ which is a collection of functions on $U$ and they have no relation with functions on $V$!

Can some one explain why this definition is true, or give an example to make it clear?

  • 4
    $\begingroup$ It sounds like you believe that sheaves are necessarily sheaves of functions, i.e. that $\mathcal{F}(U)$ should be a set of "nice" functions with domain $U$. No modern treatments of sheaf theory require this assumption, but some introductory material may not include any examples of sheaves which are not (in any way) sheaves of functions. This is a failure on their part—in order to understand modern sheaf theory, it is necessary to consider sheaves which are not sheaves of functions $\endgroup$
    – jackson
    Commented Apr 1, 2021 at 16:02

2 Answers 2


Let me bring you two pictures that maybe help you to understand the situationenter image description here

Sorry it is written in spanish but the text is not the important part of the image. This is a picture I drawn in order to understand the sheaf of regular function over an afin algebraic variety (algebraic geometry). In these example $V$ and $W$ are varieties that are topological spaces with Zariski topology. And sheaf brings open subsets to the set of regular functions over it that is, in fact, a K-algebra. (K algebraic close field where our varieties are defined) enter image description here

About your question I think this second picture will be usefull to sum up that your sheaf is in $Y$ (It corresponds to $W$ on the picture) because to every open on $W$ you have and image across the sheaf even though you have to pass across $V$ (or $X$ in you case) to obtain this image.

The relation lack you claimed is hidden on your continous function $f$ in your case and $\varphi$ on my case.

-Notation changes I have used;

.$X=V$, $Y=W$

.$\mathscr F$ = $\mathcal O_{V}$ for sheafs notation

.$f$=$\varphi$ for continous function between your topological spaces

Sorry for my poor english.


Just like you said, we need to assign open set $V$ of $Y$ some set or abelian group or whatever. Let do sets here. So what set are we assigning to $V$? We are assigning this set: $$\mathcal{F}(f^{-1}(V))$$ Is this a set? Yes it is. If I give you an open set $V$ of $Y$, can you tell me what the above set is? Yes because you know what $\mathcal{F}$ is. So, this indeed defines a sheaf on $Y$ (You need to check the sheaf definitions, but that's not what your question is about).

As an example, consider a topological space $X$, and a point $p \in X$. Consider $\{p\}$ as a space with only one point. Let $S$ be a set and let $\underline{S}$ be the constant sheaf on $\{p\}$. Let $f: \{p\} \rightarrow X$ be just the inclusion. Then we can do the direct image sheaf. For any open set $V$ in $X$, if $V$ contains $p$ that $f^{-1}(V) = \{p\}$, and if not $f^{-1}(V)$ is empty. So the direct image sheaf would be $$f_{*}\underline{S}(V) = S \text{ or } \{?\}$$ where $\{?\}$ is any singleton set. This is the skyscraper sheaf on $X$ at $p$.

  • $\begingroup$ Can you give me an example $\endgroup$ Commented Apr 1, 2021 at 15:02
  • $\begingroup$ Is thinking of a sheaf on a open set of a topological space X is a collection the functions on that open set? $\endgroup$ Commented Apr 1, 2021 at 15:09
  • $\begingroup$ @Abdallahchaibeddrraa See the edits for an example. Well yes but it's more abstract than that. In general, I don't think you can talk about the values of elements in a section at a particular point on that open set. $\endgroup$
    – DuduBob
    Commented Apr 1, 2021 at 15:14

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