There is a vast literature on embeddings of graphs into surfaces. 
I am interested in embeddings of graphs that 
belong to the given homotopy class. Here is the precise formulation. 


I have two finite graphs $\Gamma, \Gamma'$ and a homotopy equivalence $f: \Gamma\to \Gamma'$. 
I know that there exists an embedding $\iota: \Gamma'\to S$, where $S$ is a fixed closed oriented surface. I am interested in finding an embedding $j: \Gamma \to S$ so that $j$ is homotopic to $\iota\circ f$. This is, of course, impossible without further restrictions on topology of $\Gamma$, since one can take, for instance, $\Gamma=K\_5$, $\Gamma'$ to be the rose with $6$ petals and $S=S^2$. 
Edit: Therefore, let us assume, say, that


$$
(\*) \quad \chi(\Gamma) \ge \chi(S)-1
$$ 


and that $\iota\_\*: \pi\_1(\Gamma')\to \pi\_1(S)$ is surjective. 


Question 1. Does the inequality (\*) together with surjectivity assumption above imply that $\iota\circ f$ is homotopic to an embedding for every $f$ and every embedding $\iota: \Gamma'\to S$?


In fact, the inequality (\*) seems to be way too generous. Assuming that $\Gamma=K\_n$ and taking into account the formula for the genus of $K\_n$, one arrives to:


Question 2. Suppose that, $\chi(S)<0$ and replace (\*) with 
$$
\chi(\Gamma)> 3 \chi(S). 
$$
Does it follow that $\iota\circ f$ is homotopic to an embedding? 



embeddings of graphs into surfaces
The answer in general is no. One obstruction is that a 4-valent vertex has three local resolutions as a pair of trivalent vertices, and only two of them may be realized in a given surface.


**Edit** I have simplified the counterexample below.


More precisely, let $\Gamma'$ be a graph that contains two simple closed curves $\alpha$ and $\beta$ intersecting in a single point $p$, which is a 4-valent vertex of $\Gamma'$. Embed $\Gamma'$ in a surface $S$ so that $\alpha$ and $\beta$ are two simple closed curves intersecting transversely at $p$. Among the three resolutions of $p$, one produces a graph $\Gamma$ homotopic to $\Gamma'$ where $\alpha$ and $\beta$ are disjoint. If $\Gamma$ were mapped injectively into $S$ with the same homotopy type as $\Gamma'$, the curves $\alpha$ and $\beta$ would have disjoint representatives. But two simple closed curves intersecting transversely in a point cannot have homotopic disjoint representatives.


As a simple example, one may take $\Gamma' = \alpha\cup\beta$ and embed it in the torus $T$ in the usual way as a spine.


