Otdfbt

$newcommandTroperatornameTr$
Is there a continuous map $(p,t) mapsto lambda(p,t)$ which, given a path $p: [0,1] to M(2,mathbb R)$ and a $t in mathbb [0,1]$, gives back an eigenvalue of $p(t)$? Additionally, I'd like it that if $p(a) = -p(b)$ for some $a, b in [0,1]$, then $lambda(p,a) = -lambda(p,b)$.

A naive possibility would be $lambda(p, t) = fracTr(p(t)) pm sqrtTr(p(t))^2 - 4det(p(t)) 2$, using the quadratic formula and the characteristic polynomial. But this has got $pm$ in it, which is not a function. Changing $pm$ into $+$, the image of the path $$p(t) = beginpmatrix0 & 1-2t \ 2t - 1 & 0 endpmatrix$$ is seen to violate the second condition: namely, $lambda(p,0)=lambda(p,1)=i$ when what's needed is $lambda(p,0)=-lambda(p,1)$.

This problem is unsolvable when $M(2, mathbb R)$ is changed to $M(2, mathbb C)$, as the existence of $lambda(-,-)$ would violate the simple-connectivity of $SL(2,mathbb C)$.

I originally gave a wrong candidate for an answer, but in fact no such path can exist (at least, assuming that your topologies are such that $t mapsto lambda(p, t)$ is continuous for all paths $p$). Indeed, define $p : [0, 1] to operatorname M(2, mathbb R)$ by
$$
p(t) = beginpmatrix cos(pi t) & sin(pi t) \ sin(pi t) & -cos(pi t) endpmatrix
$$
for all $t in [0, 1]$. Then $lambda(p, t) in pm1$ for all $t in [0, 1]$, but $lambda(p, 1) = -lambda(p, 0)$, which is impossible if $t mapsto lambda(p, t)$ is continuous.

EDIT: I suspected from your mention of the simple connectedness of $operatornameSL(2, mathbb C)$, and see definitely from your comment, that you are interested in the non-simple connectedness of $operatornameSL(2, mathbb R)$. It may be that a particularly simple way to see this is to consider the metaplectic group, which is a non-split double cover of $operatornameSL(2, mathbb R)$. Notice that your original example can be adapted to give a polynomial path in $operatornameSL(2, mathbb R)$ (not just in $operatorname M(2, mathbb R)$, or even just in $operatornameGL(2, mathbb R)$ as my original path was):
$$
t mapsto p(t) = beginpmatrix
t(1 - t) & 1 - t^2(6 - 4t) \
-1 + t^2(6 - 4t) & 4t(1 - t)(3 - 2t)(1 + 2t)
endpmatrix.
$$

Notice that the cover of $operatornameSL(2, mathbb R)$ implicit in your question, namely
$$
E mathrel:= (g, lambda) : text$g in operatornameSL(2, mathbb R)$ and $lambda$ is an eigenvalue of $g$,
$$
is a branched cover; it is a double cover only on the regular semisimple set (of elements with distinct eigenvalues). Its restriction to the regular semisimple set is isomorphic (as a cover of the rss set) to the restriction there of the branched cover
$$
(g, B) : text$g in operatornameSL(2, mathbb R)$ and $B$ is a Borel subgroup of $operatornameSL(2, mathbb C)$ containing $g$
$$
(namely, associate $(g, lambda)$ and $(g, B)$ if and only if the non-trivial eigenvalue of $operatornameAd(g)$ on $operatornameLie(B)$ is $lambda^2$). However, I think that the restriction is not isomorphic to the restriction of the metaplectic group. Indeed, let $lambda$ and $mu$ be distinct elements of $mathbb R$ satisfying $lambdamu = 1$. Then any lift to $E$ of the path
$$
t mapsto operatornameIntbeginpmatrix
cos(pi t) & sin(pi t) \
-sin(pi t) & sin(pi t)
endpmatrixbeginpmatrix
lambda & 0 \
0 & mu
endpmatrix
$$
projects, via $E to mathbb C$, to $lambda, mu$, hence is constantly $lambda$ or constantly $mu$; but, if I've done my computation properly, then
$$
operatornameInt(w(t))Bigl(beginpmatrix
lambda & 0 \
0 & mu
endpmatrix, sqrtmuBigr),
$$
where $w(t) = Bigl(beginpmatrix
cos(pi t) & sin(pi t) \
-sin(pi t) & cos(pi t)
endpmatrix, epsilon_tBigr)$, with $epsilon_t(i) = e^pi i t/2$, is a path in the restriction to the rss set of the metaplectic cover that connects the two pre-images of $beginpmatrix lambda & 0 \ 0 & mu endpmatrix$. (I'm using the description of points in the metaplectic group from Wikipedia.)