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Let $X$ be a surface over a field $k$ with a smooth $k$-point $x\in X$, and suppose $\mathcal{Y}$ is a proper DM stack over $k$. (I am really thinking of $\mathcal{Y}=\overline{\mathcal{M}_g}$.) Let $f:X-x\to\mathcal{Y}$ be a morphism over $k$. When does $f$ lift to an honest morphism $Bl_x(X)\to\mathcal{Y}$?

I think the following might be the correct condition: for every smooth curve $C\in X$ containing $x$, the morphism $C-x\to X-x\to\mathcal{Y}$ extends over $X$.

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  • $\begingroup$ One way of resolving the indeterminacy locus of a morphism is by considering the graph; see for instance Section 2.3 of arxiv.org/abs/1703.00488 This might be enough for your purposes (unless you really want to resolve by blowing-up). $\endgroup$ Commented Jul 3, 2018 at 16:03

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The condition is insufficient already if $\mathcal{Y}$ is a scheme. (I assume that the final words extends over $X$ were meant to say extends over $C$.)

Indeed, let $\mathcal{Y}$ be the blow-up of ${\rm Bl}_x(X)$ at a point in the exceptional divisor, and let $f\colon X - x\to \mathcal{Y}$ be the natural open immersion. Clearly, $f$ does not extend to ${\rm Bl}_x(X)$. However, for every smooth curve $C$, the restricted morphism extends. Indeed, every map from a punctured smooth curve into a proper scheme extends over the puncture, by the valuative criterion of properness.

It seems that what you want is that $f$ extends to some blowup at $x$, that is that there exists a proper map $g\colon X'\to X$, which is an isomorphism over $X-x$, such that $f$ extends to $X'$. I do not know about stacks, but now at least the answer is clearly yes if $\mathcal{Y}$ is a scheme!

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