(Disclaimer: I suspect the answer is no, but I fail to see why)
Here is a nice picture by David Epstein (taken from Wikipedia) illustrating Hadwiger's conjecture:
The point is that if in a given graph $G$ some-one gives you four connected subgraphs $G_1$, $G_2$, $G_3$, $G_4$ (here illustrated with grey boxes) such that for each of the six combinations $(i, j)$ there is at least one edge in $G$ connecting a vertex in $G_i$ to one in $G_j$, then the conjecture claims that there is no 3-coloring of the big graph $G$.
Suppose that Hadwiger's conjecture is true. Wouldn't this put
3-coloring in coNP? At least to me it seems that once someone gives you $G_1, G_2, G_3, G_4$ their connectedness and existence of 6 edges connecting the four subgraphs can be checked in time quadratic in the number of vertices of the big graph $G$.
But if indeed 3-coloring is in coNP, wouldn't that put all of NP in coNP thus making these complexity classes equal?
This seems wrong, my impression was that both Hadwiger's conjecture and $NP \neq coNP$ are widely believed, so most likely something is wrong with my reasoning above.