Continuous homomorphisms from GL(n,R) to R

Idea. A homomorphism to an abelian group kills commutators. Since $SL_n(\mathbb{R})$ is generated by commutators, $f$ factors through the determinant. The problem then reduces to continuous homomorphisms $\mathbb{R}^\times\to\mathbb{R}^\times$.

Step 1: $f$ is trivial on commutators. For $A,B$ in $GL_n$, $f(ABA^{-1}B^{-1})=f(A)f(B)f(A)^{-1}f(B)^{-1}=1$, because $\mathbb{R}^\times$ is abelian.

Step 2: $SL_n(\mathbb{R})$ is generated by commutators. $SL_n(\mathbb{R})$ is generated by elementary matrices $E_{ij}(t)=I+tE_{ij}$ ($i\ne j$) (row reduction). For $n\ge3$ and distinct $i,j,k$, $[E_{ik}(a),E_{kj}(b)]=E_{ij}(ab)$, so every $E_{ij}(t)$ is a commutator (take $a=t$, $b=1$). For $n=2$, let $D=\operatorname{diag}(2,\tfrac12)\in GL_2$; then $DE_{12}(s)D^{-1}=E_{12}(4s)$, so $E_{12}(3s)=DE_{12}(s)D^{-1}E_{12}(-s)=[D,E_{12}(s)]$ is a commutator in $GL_2$, and so is every $E_{12}(t)$ (take $s=t/3$); similarly $E_{21}(t)$. In all cases the generators of $SL_n$ lie in the commutator subgroup of $GL_n$. By Step 1, $f\equiv1$ on $SL_n(\mathbb{R})$.

Step 3: factorisation through $\det$. Let $\Delta(d)=\operatorname{diag}(d,1,\dots,1)$. For $A\in GL_n$ with $\det A=d$, $A\Delta(d)^{-1}\in SL_n$, so
$$f(A)=f\bigl(A\Delta(d)^{-1}\bigr)f(\Delta(d))=g(d),\qquad g(d):=f(\Delta(d)).$$
Since $\Delta(d)\Delta(d')=\Delta(dd')$, $g:\mathbb{R}^\times\to\mathbb{R}^\times$ is a continuous homomorphism.

Step 4: continuous homomorphisms $\mathbb{R}^\times\to\mathbb{R}^\times$. For $x>0$, $g(x)=g(\sqrt x)^2>0$. So $h(s)=\ln g(e^s)$ is a continuous additive function $\mathbb{R}\to\mathbb{R}$, hence $h(s)=cs$ for some $c$ (Cauchy's equation with continuity). Thus $g(x)=x^c$ for $x>0$. Finally $g(-1)^2=g(1)=1$ gives $g(-1)=\pm1$, and $g(-x)=g(-1)g(x)$. So $g(d)=|d|^c$ (if $g(-1)=1$) or $g(d)=\operatorname{sgn}(d)|d|^c$ (if $g(-1)=-1$).

Conclusion. $f(A)=g(\det A)$ is of the stated form.

Remark. Continuity is needed only to exclude pathological homomorphisms (as for Cauchy's equation). For $n=1$ the same classification holds for continuous homomorphisms of $\mathbb{R}^\times$ to itself, but the factorisation argument is not needed.

Lesson. Homomorphisms into abelian groups factor through the abelianisation; $GL_n(\mathbb{R})^{ab}\cong\mathbb{R}^\times$ via $\det$.