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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?

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I make a simple modification of your example, just to understand the problem. Take $\Gamma = K_5$ and $\Gamma'$ the rose with 6 petals. Embed $\Gamma'$ in a disc contained in a surface $S$ with large genus. To answer your question positively you need to embed $K_5$ in $S$ with a map which is homotopic to a costant map. But that sounds impossible to me. – Bruno Martelli Jun 18 '13 at 10:51
@Bruno: You are absolutely right, this is impossible (one sees this by lifting maps to the universal cover). I am interested in maps which induce epimorphisms of fundamental groups, so I will add this to the question. – Misha Jun 18 '13 at 12:06
Regarding question 1, your two assumptions together imply that $rank(\Gamma)=2genus(S)$, because the first says $1−r \ge 1−2g$ and surjectivity of $i_∗$ implies $r \ge 2g$. Is that the intent? – Lee Mosher Jun 18 '13 at 16:15
Lee: Yes, this was the motivation. – Misha Jun 18 '13 at 16:46
up vote 9 down vote accepted

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.

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Thank you, Bruno! – Misha Jun 18 '13 at 18:45

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