Let $f:X\to Y$ be a proper surjection of complex algebraic varieties. Let $H_i$ denote Borel-Moore homology. Then $$ \mathrm{Gr}^W_{-k} H_k(X) \to \mathrm{Gr}^W_{-k} H_k(Y) $$ is surjective.

Question: Does anyone know a reference for this fact?

I have a proof, but it's not as simple as it could be. And it uses generic smoothness of $f$, so it's only valid in characteristic zero.

The cycle class map from Chow groups to Borel-Moore homology lands in the lowest weight part, so the above is somehow analogous to the fact that proper pushforward is surjective in Chow for a surjective map.

If $f$ instead is an open immersion, then the induced map on lowest weights is also surjective. This is similarly analogous to the fact that flat pullback is surjective in Chow for open immersions. But here the proof for the result in Borel-Moore homology is very simple, which is one reason I think there should be a simple proof of the above result, too.

Addendum. Here's my proof, it's very similar to the one linked to by novice. However, the proof in Lewis's book doesn't need generic smoothness; it uses instead the existence of a subvariety mapping generically finitely onto $Y$, as in the suggestion of ACL below.

Note first that the if $f$ is in addition smooth, then $H_k(X) \to H_k(Y)$ is onto and there's not much to it. For the general case take $U \subset X$ where $f$ is smooth, and let $Z = X \setminus U$. Then there is a map between the long exact sequences $$ \cdots \to H_k(Z) \to H_k(X) \to H_k(U) \to H_{k-1}(Z) \to \cdots $$ and $$ \cdots \to H_k(f(Z)) \to H_k(Y) \to H_k(Y \setminus f(Z)) \to H_{k-1}(f(Z)) \to \cdots $$ as follows. The maps $H_\bullet(X) \to H_\bullet(Y)$ and $H_\bullet(Z) \to H_\bullet(f(Z))$ are the obvious ones. The map $H_\bullet(U) \to H_\bullet(Y \setminus f(Z))$ is the composite $$ H_\bullet(U) \to H_\bullet(X \setminus f^{-1}(f(Z))) \to H_\bullet(Y \setminus f(Z)). $$

Now apply $W_{-k}$ to the long exact sequences. The two maps $H_k(Z) \to H_k(f(Z))$ and $H_k(U) \to H_k(Y \setminus f(Z))$ are surjective on lowest weights: the former by noetherian induction and the latter because it's the composition of pullback for an open immersion and a pushforward for a smooth proper morphism. Since in addition $W_{-k}H_{k-1}(Z) = W_{-k}H_{k-1}(f(Z)) = 0$ the result follows by the four lemma.

  • $\begingroup$ Isn't it possible to consider some closed subset $Z$ in $X$ which maps generically finitely onto $Y$ (take the Zariski closure of a multisection) and use the existence of a trace map? $\endgroup$
    – ACL
    Commented Sep 22, 2013 at 10:04
  • $\begingroup$ @ACL: that sounds reasonable. But where would the lowest weight hypothesis come into such a proof? $\endgroup$ Commented Sep 22, 2013 at 16:31

2 Answers 2


See Lewis's book (page 285) for the case of projective varieties

  • $\begingroup$ That's a nice reference. In fact the only point of the argument that uses projectivity is in order to find a closed subvariety $X' \subset X$ mapping generically finitely onto $Y$. $\endgroup$ Commented Sep 23, 2013 at 8:53
  • $\begingroup$ @Dan Petersen: Proper and surjective is all one needs to get such a subvariety. $\endgroup$
    – naf
    Commented Sep 23, 2013 at 11:26
  • $\begingroup$ @ulrich: Oh, even better! Because any proper variety $X$ has a modification $X' \to X$ which is projective? Or is there a simpler argument? $\endgroup$ Commented Sep 23, 2013 at 13:50
  • 1
    $\begingroup$ @Dan Petersen: Choose a closed point on the fibre of $f$ over each (scheme theoretic) generic point of $Y$ and let $Z$ be the Zariski closure of these points. $\endgroup$
    – naf
    Commented Sep 23, 2013 at 14:04

I guess this is standard and there is a citable reference. I think the following is an argument which only uses the formalism (e.g. also works in the étale case). (This is an edited version of my first incorrect answer.)

Firstly, if $X \to Y \to Z \stackrel{+1}{\to}$ is a distinguished triangle with $X, Y$ of wt $\ge 0$, then $Z$ is of weight $\ge 0$. (This is easy if one thinks about Frobenius eigenvalues.)

Dually, if $X \to Y \to Z \stackrel{+1}{\to}$ is a dt with $Y, Z$ of wt $\le 0$ then $X$ is of wt $\le 0$.

Now the constant sheaf $k_Y$ on $Y$ is of wt $\le 0$. Hence the dualizing sheaf $\omega_Y$ on $Y$ is of wt $\ge 0$.

Let $f : X \to Y$ be as in your question. Consider the distinguished triangle $K \to f_!f^!\omega_Y \to \omega_Y \stackrel{+1}{\to}$ (where $K$ is defined as the shift of the cone over the adjunction morphism $f_!f^! \to id$). The above remarks show that $K$ is of weights $\ge -1$.

Claim: We are done if we can show $K$ is of weights $\ge 0$.

Proof: Pushing to a point we get a long exact sequence

$\dots \to H^k(Y,K) \to H^k(X,\omega_X) \to H^k(Y,\omega_Y) \to H^{k+1}(Y,K) \to \dots$

now everything in $H^{k+1}(Y,K)$ is of weight $\ge k+1$ (because $*$-pushforward preserves wt $\ge 0$). We conclude that we have a surjection

$gr_W^kH^k(X,\omega_X) \to gr_W^kH^k(X,\omega_Y)$

Finally, because $H_k^{BM}(X) = H^{-k}(X,\omega_X)$ (Borel-Moore homology) the claim follows.

We now give a sketch of how to prove the claim. Because this argument is getting more complicated than I had first intended, I'll give a sketch. I can try to provide more details if it is useful for you.

Consider a weight filtration $W$ on $\omega_Y$ (it is not "the" weight filtration because $\omega_Y$ is not necessarily perverse). I claim that $gr^W_{\le 0}(\omega_Y) = gr^W_0(\omega_Y)$ is isomorphic to $IC(Y)[d_Y](d_Y)$ (where $d_Y = dim Y$). This basic idea is that $\omega_Y$ does not have any sections supported on subvarieties, and any other $IC$ in $gr_{\le 0}(\omega_Y)$ would contribute such a forbidden section. Similarly, if $W$ denotes a weight filtration on $\omega_X$ then $gr^W_{0}(\omega_X)$ is $IC(X)[d_X](d_X)$.

Now consider the adjunction map $f_!f^!\omega_Y \to \omega_Y$. The weight zero part is given by the map $f_!IC(X) = f_! gr^W_0(\omega_X) \to gr^W_0(\omega_Y)$. Now by the decomposition theorem (here we use surjectivity) $IC(Y)$ occurs as a summand of $f_!IC(X)$ in smallest degree.

Now one deduces (I can only see how to do this using generic smoothness at the moment) that there exists $IC(Y)[?](?) \to f_!\omega_X$ such that the induced map

$$IC(Y)[?](?) \to gr^W_0(\omega_Y) = IC(Y)[?](?)$$

is an isomorphism. We conclude that the triangle

$K \to f_!\omega_X \to \omega_Y \stackrel{+1}{\to}$

can be replaced by a triangle

$K \to L \to gr^W_{\ge 1}(\omega_Y) \stackrel{+1}{\to}$

with $L$ of wts $\ge 0$ and $gr^W_{\ge 1}(\omega_Y)$ of weights $\ge 1$. We conclude that $K$ has weights $\ge 0$ as claimed.

  • $\begingroup$ Thanks. But it seems you are not using the surjectivity hypothesis anywhere? The result is obviously false without it (think about a closed immersion). $\endgroup$ Commented Sep 22, 2013 at 9:43
  • $\begingroup$ You are right, the problem is "the above remarks imply that $K$ is of weight $\ge 0$", in fact one only gets $K$ of wt $\ge -1$. $\endgroup$ Commented Sep 22, 2013 at 10:31
  • $\begingroup$ Thanks a lot! In fact I think the proof I have is slightly simpler (it doesn't need perverse sheaves or the decomposition theorem), probably I should've included it in the original question. $\endgroup$ Commented Sep 22, 2013 at 16:35
  • $\begingroup$ Unfortunately I can't include the proof now (I'm posting from my phone) but I'll do it later this evening. $\endgroup$ Commented Sep 22, 2013 at 16:40

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