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?