Is the Gysin morphism equivariant?

Let $X$ be a smooth, projective complex variety and $j \colon D \hookrightarrow X$ a smooth divisor. Then we have a Gysin morphism in singular cohomology

$$j_\ast \colon H^{\bullet}(D) \to H^{\bullet+2}(X)$$

Now assume that $X$ is acted upon by a finite group $G$ and that $D$ is stable under this action. Then we get actions $g^\ast$ on the cohomology of $D$ and $X$.

Is is true that $j_\ast g^\ast=g^\ast j_\ast$ for any $g$ in $G$?

This looks like some projection formula, but I'm unable to prove it.

• The answer is yes if $G$ acts by diffeomorphisms, since then each $g$ is transverse to $j$. I don't have these books to hand to check, but I'm sure you'll find a proof in either E. Dyer's "Cohomology Theories" or W. Fulton's "Intersection Theory" Jul 4 '15 at 15:32
• Thanks! I will have a look at Fulton's book. Can you explain what "g is transverse to j" means?
– ter
Jul 4 '15 at 15:40
• Smooth maps $g: M\to X$ and $j: N\to X$ are transverse if whenever $g(m)=j(n)=x$ then the images of the differential of $g$ at $m$ and the differential of $j$ at $n$ together span the tangent space at $x$ of $X$. Jul 4 '15 at 15:50
• Alternatively, you can argue that both the restriction $j^*$ and Poincare duality isomorphisms are equivariant (by naturality). Jul 4 '15 at 18:35
• Oh thanks, Donu, that looks simpler. Could you please develop a little bit?
– ter
Jul 4 '15 at 18:36

Let me expand my comment slightly into an answer. To simplify matters, suppose that the coefficients are $\mathbb{Q}$ or $\mathbb{C}$. The Gysin map is Poincaré dual to the restriction map $j^*$. In more detail, we have isomorphisms $H^i(X)\cong H^{2n-i}(X)^\vee$ and $H^i(D)\cong H^{2n-2-i}(D)^\vee$, where $n=\dim X$, given by Poincaré pairings $(\alpha,\beta) =\int_X\alpha\cup\beta$ etc. Under these isomorphism $j_*$ corresponds to the dual to restriction $j^*$. This is $G$-equivariant by functoriality. I'll let you check that $(g^*\alpha,g^*\beta)=(\alpha,\beta)$, and this implies $G$-equivariance of the duality isomorphisms, and that's all you need.