Finding disjoint loops in a graph Let $G$ be a graph with characteristic $\chi= E-V+1$.  Here $E$ is the number of edges and $V$ is the number of vertices.  (I am not sure that "characteristic" is the right term, but "genus" seems to be taken)
My question is very simple: does there exist a constant $c$ (say, $c=0.01$, independent of the graph) such that one can find at least $c\chi$ edge-disjoint loops in $G$?
I have tried using a spanning tree $T$ of the graph, but the problem is that I don't know how to deal with the fact that the edges in the complement of the spanning tree form cycles that overlap in complicated ways.  I suspect for this reason that the answer to my question is "no", but I can't find any promising examples.
 A: Too long for a comment. Not an answer.
Let $f(n,e)$ be the smallest $k$ such that, in every graph with $n$ nodes and $e$ edges, there must be a subgraph of $k$ nodes on which there is a loop (the loop could be of length $\leq \ell.$)
Given $(V,E)$  be a graph with $|V|=n,|E|=e.$ Assume that, given any sub-graph $(S,E_S)$ with $|S|=k,$ there is no loop, so then there must be at most $|E_S|\leq k-1$ So we have:
$$\sum_{|S|=k} |E_S|\leq (k-1)\binom{n}{k}$$
But also, each edge is exactly $\binom{n-2}{k-2}$ subgraphs of size $k.$
So:
$$\sum_{|S|=k} |E_S|= e\binom{n-2}{k-2}$$
So you have:
$$e\leq\frac{(k-1)\binom{n}{k}}{\binom{n-2}{k-2}}=\frac{n(n-1)}{k}$$
So:
$$f(n,e)\leq 1+\frac{n(n-1)}{e}$$
at least for $e\geq n,$ when $f(n,e)$ is defined.
Now. if $L(n,e)$ is the smallest number of disjoint loops possible for graphs of  $n$ nodes and $e$ edges, then:
$$\begin{align}L(n,e)&=0\text{ if }e<n\\
L(n,e)&\geq 1+L\left(n,e-1-\left\lfloor\frac{n(n-1)}{e}\right\rfloor\right)
\end{align}$$
Not sure if that helps.

Some computer calculations does give this much hope of working. This lower bound for $L$ to have $\min_{e}\frac{L}{\chi}$ go to zero as $n$ gets bigger.
We do get: $$n\min\frac{L}{\chi}\approx 1.62$$
using this lower bound, for $n=100,200,300,400,500,1000,5000.$ It appears to be getting smaller as $n$ gets larger, but very slowly.
The problem is that the estimate only gives one loop for nodes up to $1.6n.$ When:
Solving:
$$ e-1-\frac{n(n-1)}{e}=n$$
we get $e^2 -(n+1)e-n(n-1)=0$ or $$ e=\frac{(n+1)+\sqrt{(n+1)^2+4n(n-1)}}{2}=\frac{(n+1)+\sqrt{5n^2-2n+1}}{2}$$
or:
$$\frac{e}{n}\approx \frac{1+\sqrt{5}}{2}\approx 1.618$$
You might find a counterexample by trying to construct a graph with as many edges as possible such that only one loop is possible.
Or it might work to solve your problem by getting better values for $e$ small.
