Suppose $M$ is a smooth manifold and $N$ is a singular manifold with one singular point $y_0$ (so that $N-y_0$ is a smooth manifold). I want to prove the following statement: If $f:M\to N$ is a continuous map, then there is a map $g:M\to N$ homotopic to $f$, smooth on $g^{-1}(N-y_0)$, and $f=g$ on $f^{-1}(y_0)$.
I think this can be proved using the well-known Whitney approximation theorem (I stated it below), as follows: Let $X=f^{-1}(y_0)$. The restriction $f|_{M-X}:M-X\to N-y_0$ is a continuous map between manifolds, so by the approximation theorem there is a smooth map $h:M-X\to N-y_0$ homotopic to $f|_{M-X}$. Now define $g:M\to X$ by letting $g=h$ on $M-X$ and $g=f$ on $X$. Then clearly $g$ is smooth on $g^{-1}(N-y_0)=M-X$ and $g=f$ on $X=f^{-1}(y_0)$. But I can't see $g$ is continuous on $M$. Should I need a different approach?
Whitney approximation theorem. Suppose $N$ is a smooth manifold with or without boundary, $M$ is a smooth manifold (without boundary), and $F:N\to M$ is a continuous map. Then $F$ is homotopic to a smooth map. If $F$ is already smooth on a closed subset $A\subset N$, then the homotopy can be taken to be relative to $A$.