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...