A weak analytic version of the valuative criterion of properness

Question

EDIT: Let $f\colon X\to Y$ be a morphism of complex analytic spaces (not necessarily smooth or reduced). Assume that

(a) $f$ is injective on points;

(b) $f$ is local imbedding near each point $x\in X$;

(c) for any holomorphic germ $\gamma\colon (\mathbb{C},0)\to Y$ with image contained in $f(X)$ there exists a holomorphic germ $\tilde\gamma\colon (\mathbb{C},0)\to X$ such that $f\circ \bar \gamma=\gamma$.

(d) $f(X)$ is a closed subset of $Y$.

Question. Is it true that $f$ is proper?

Remark. (1) In the algebraic situation the answer is positive by the standard valuative criterion of properness.

(2) A priori it is not clear to me that locally on $Y$ the image $f(X)$ is an analytic subset.

If there is any analytic version of the valuative criterion of properness, I would be curious to know anyway. A reference would be helpful.

EDIT: Conditions (a),(c), (d) imply the following property which looks closer to the standard valuative criterion of properness. For any holomorphic germs $\gamma\colon (\mathbb{C},0)\to Y$ and $\beta\colon (\mathbb{C},0)\backslash\{0\}\to X$ such that $f\circ \beta=\gamma|_{(\mathbb{C},0)\backslash\{0\}}$, there exists a holomorphic germ $\bar\gamma\colon (\mathbb{C},0)\to X$ such that $$\bar\gamma|_{(\mathbb{C},0)\backslash\{0\}}=\beta,\, f\circ \bar\gamma=\gamma.$$

Answer 1

Here is a counterexample: $Y=\mathbb C^2$, $X=(\mathbb C\setminus\{0\})\cup\mathbb C$, and $f$ is given by $z\mapsto (z,e^{1/z})$ on $\mathbb C\setminus\{0\}$ and is given by $w\mapsto(0,w)$ on the second copy of $\mathbb C$. This map is certainly not proper. It clearly satisfies conditions (a),(b),(d). Let's check condition (c).

Let $\gamma:(\mathbb C,0)\to Y=\mathbb C^2$ be a germ whose image is contained in $f(X)$. If the germ sends $0$ to a point whose first coordinate is nonzero, then it clearly lifts, so it suffices to check the case when $\gamma(0)=(0,w)$ for some $w\in\mathbb C$. Consider the first coordinate of $\gamma$: if this first coordinate is not identically zero, then it covers an open neighborhood of zero by the open mapping theorem. On the other hand, the function $e^{1/z}$ takes arbitrarily large values near zero since it has an essential singularity there, and this contradicts the fact that the image of $\gamma$ is contained in the image of $f$. Thus the first coordinate of $\gamma$ is identically zero, and thus it trivially lifts to $X$.