Return to Question

Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) of $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for maps: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps $$ g_1^*, g_2^* : f^{-1}\hat{\Omega^*_{X'}} \to \hat{\Omega^*_X} $$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy explicitly if possible as well...

Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) of $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for maps: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps of abelian sheaves on $Y'$: $$ g_1^*, g_2^* : f^{-1}\hat{\Omega^*_X} \to \hat{\Omega^{*}_{X'}} $$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy explicitly if possible as well...

Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) from $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for maps: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps $$ g_1^*, g_2^* : f^{-1}\hat{\Omega^*_{X'}} \to \hat{\Omega^*_X} $$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy explicitly if possible as well...

Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) of $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for maps: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps $$ g_1^*, g_2^* : f^{-1}\hat{\Omega^*_{X'}} \to \hat{\Omega^*_X} $$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy explicitly if possible as well...

Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) from $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for maps: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps $$ g_1^*, g_2^* : f^{-1}\hat{\Omega^*_{X'}} \to \hat{\Omega^*_X} $$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy down explicitly if possible as well...

Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) from $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for maps: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps $$ g_1^*, g_2^* : f^{-1}\hat{\Omega^*_{X'}} \to \hat{\Omega^*_X} $$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy explicitly if possible as well...