Let $X$ be an integral scheme of finite type over a field $k$ of dimension $d$ and $U$ an open dense smooth subscheme.

Let $K\in D_{c}^{b}(X,\bar{\mathbb{Q}}_{l})$ such that $K_{U}=\bar{\mathbb{Q}}_{l}[d]$.

Moreover, we assume that $X=\bigcup\limits_{\alpha\in\Lambda} S_{\alpha}$ where $S_{\alpha}$ is a locally closed subscheme and $\Lambda$ a set, non necessarily finite such that :

$i_{\alpha}^{*}K=IC_{S_{\alpha}}$, do we have that $K=IC_{X}$?

Let $X$ be an integral scheme of finite type over a field $k$ of dimension $d$ and $U$ an open dense smooth subscheme.

Let $K\in D_{c}^{b}(X,\bar{\mathbb{Q}}_{l})$ be such that $K_{U}=\bar{\mathbb{Q}}_{l}[d]$.

Moreover, we assume that $X=\bigcup\limits_{\alpha\in\Lambda} S_{\alpha}$ where $S_{\alpha}$ is a locally closed subscheme and $\Lambda$ is a set, not necessarily finite, such that $i_{\alpha}^{*}K=IC_{S_{\alpha}}$.

Do we have that $K=IC_{X}$?

Let $X$ be an integral scheme of finite type over a field $k$ of dimension $d$ and $U$ an open dense smooth subscheme.

Let $K\in D_{c}^{b}(X,\bar{\mathbb{Q}}_{l})$ such that $K_{U}=\bar{\mathbb{Q}}_{l}[d]$.

Moreover, we assume that $X=\bigcup\limits_{\alpha\in\Lambda} S_{\alpha}$ where $S_{\alpha}$ is a locally closed subscheme and $\Lambda$ a set, non necessarily finite such that :

$i_{\alpha}^{*}K=i_{\alpha}^{*}IC_{X}$, do we have that $K=IC_{X}$?

Let $X$ be an integral scheme of finite type over a field $k$ of dimension $d$ and $U$ an open dense smooth subscheme.

Let $K\in D_{c}^{b}(X,\bar{\mathbb{Q}}_{l})$ such that $K_{U}=\bar{\mathbb{Q}}_{l}[d]$.

Moreover, we assume that $X=\bigcup\limits_{\alpha\in\Lambda} S_{\alpha}$ where $S_{\alpha}$ is a locally closed subscheme and $\Lambda$ a set, non necessarily finite such that :

$i_{\alpha}^{*}K=IC_{S_{\alpha}}$, do we have that $K=IC_{X}$?