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Dan Petersen
Dan Petersen
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My idea was to consider $f \colon X \to Y$ of dimension $d$ such that $a_Y^\ast \mathbf Q[d] \cong IC_Y$ but $a_X^\ast \mathbf Q[d] \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q[d] = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.

(Incorrect example removed)

For a concrete example let $X$ be two copies of $\mathbb A^1$$\mathbb A^2$ glued at a point, let $f\colon X \to Y \cong \mathbb A^1$$f\colon X \to Y \cong \mathbb A^2$ be the quotient by the involution switching the two copies. If it were true that $IC_X \cong a_X^\ast \mathbf Q[2]$ then the same should be also true for $\mathbf P^2$ with two points identified, as perversity is a local property. But the cohomology of $\mathbf P^2$ with two points identified does not have Poincaré duality.

My idea was to consider $f \colon X \to Y$ of dimension $d$ such that $a_Y^\ast \mathbf Q[d] \cong IC_Y$ but $a_X^\ast \mathbf Q[d] \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q[d] = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.

For a concrete example let $X$ be two copies of $\mathbb A^1$ glued at a point, let $f\colon X \to Y \cong \mathbb A^1$ be the quotient by the involution switching the two copies.

My idea was to consider $f \colon X \to Y$ of dimension $d$ such that $a_Y^\ast \mathbf Q[d] \cong IC_Y$ but $a_X^\ast \mathbf Q[d] \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q[d] = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.

(Incorrect example removed)

For a concrete example let $X$ be two copies of $\mathbb A^2$ glued at a point, let $f\colon X \to Y \cong \mathbb A^2$ be the quotient by the involution switching the two copies. If it were true that $IC_X \cong a_X^\ast \mathbf Q[2]$ then the same should be also true for $\mathbf P^2$ with two points identified, as perversity is a local property. But the cohomology of $\mathbf P^2$ with two points identified does not have Poincaré duality.

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Dan Petersen
Dan Petersen
  • 40.2k
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  • 114
  • 201

My idea was to consider $f \colon X \to Y$ of dimension $d$ such that $a_Y^\ast \mathbf Q \cong IC_Y$$a_Y^\ast \mathbf Q[d] \cong IC_Y$ but $a_X^\ast \mathbf Q \not\cong IC_X$$a_X^\ast \mathbf Q[d] \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q = f^\ast IC_Y$$a_X^\ast \mathbf Q[d] = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.

For a concrete example let $X$ be two copies of $\mathbb A^1$ glued at a point, let $f\colon X \to Y \cong \mathbb A^1$ be the quotient by the involution switching the two copies.

My idea was to consider $f \colon X \to Y$ such that $a_Y^\ast \mathbf Q \cong IC_Y$ but $a_X^\ast \mathbf Q \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.

For a concrete example let $X$ be two copies of $\mathbb A^1$ glued at a point, let $f\colon X \to Y \cong \mathbb A^1$ be the quotient by the involution switching the two copies.

My idea was to consider $f \colon X \to Y$ of dimension $d$ such that $a_Y^\ast \mathbf Q[d] \cong IC_Y$ but $a_X^\ast \mathbf Q[d] \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q[d] = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.

For a concrete example let $X$ be two copies of $\mathbb A^1$ glued at a point, let $f\colon X \to Y \cong \mathbb A^1$ be the quotient by the involution switching the two copies.

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Dan Petersen
Dan Petersen
  • 40.2k
  • 2
  • 114
  • 201

My idea was to consider $f \colon X \to Y$ such that $a_Y^\ast \mathbf Q \cong IC_Y$ but $a_X^\ast \mathbf Q \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.

For a concrete example let $X$ be two copies of $\mathbb A^1$ glued at a point, let $f\colon X \to Y \cong \mathbb A^1$ be the quotient by the involution switching the two copies.