Is this a valid way to show $\chi(SL_n(\mathbb{R}))=0$? Why does $\chi(SL_n(R))=0$? I'm going about it like this. Let $X:=SL_n(R)$. 
Define a map $f:X\to X$ such by $A\mapsto BA$, where $B$ is the identity matrix, except with an extra $1$ in the upper right corner entry. So $\det(BA)=1$, and the map is smooth. I also know $f$ has no fixed points, since $A=BA$ implies $B=I_n$ since all the $A$ are invertible. But $f$ is also homotopic to the identity, say through a family of maps $f_t$ where $f_t$ is left multiplication by the matrix which has $1$ on the diagonal, and $t$ in the upper right entry. 
From Lefschetz fixed point theory, I know the Lefschetz number of the identity is equal to the Euler characteristic, and is homotopy invariant. But the Lefschetz number of a map with no fixed points is $0$, so it would follow that 
$$
\chi(X)=L(id)=L(f)=0.
$$
However, all my searching seems to show that Lefschetz fixed point theory is for compact oriented manifolds. But $SL_n(R)$ is not compact, although I think it is orientable. Does this idea work without the compactness assumption, or is there a better way to compute this?
 A: The polar decomposition defines a deformation retract of $SL(n, \mathbb{R}) \to SO(n, \mathbb{R})$, and since Euler characteristic $\chi$ is homotopy invariant, we may just as well compute $\chi(SO(n, \mathbb{R}))$. But $SO(n, \mathbb{R})$ is compact and orientable (all Lie groups are orientable), so your Lefschetz fixed point argument seems to apply to it. In fact, for any Lie group $G$ and any nonidentity element $g \in G$, the map $h \mapsto gh$ fixes no point, so this argument in fact shows that any compact, connected, nontrivial Lie group has Euler characteristic zero.
For the case of $SO(n, \mathbb{R})$ (again, equivalently $SL(n, \mathbb{R})$, here's an alternate argument that uses doesn't use a fixed point argument, and is more "natively topological": The Euler characteristic is multiplicative for nice fibrations. More precisely, given a fibration $E \to M$ with path-connected base $M$ and fiber $F$ (and a technical orientability condition that holds here), the Euler characteristics of $E$, $M$, and $F$ are related by
$$\chi(E) = \chi(M) \chi(F).$$
For $n > 1$, $SO(n)$ is the total space for the fibration $SO(n) \to \mathbb{S}^{n - 1}$ induced by the standard action of $SO(n)$ on $\mathbb{R}^n$, and its fiber is $SO(n - 1)$, so
$$\chi(SO(n)) = \chi(\mathbb{S}^n) \chi(SO(n - 1)).$$
By induction,
$$\chi(SO(n)) = \chi(\mathbb{S}^n) \cdots \chi(\mathbb{S}^1) \chi(SO(0)),$$
but $\chi(\mathbb{S}^1) = 0$, and so
$$\chi(SO(n)) = 0.$$
