Given the set of \*-homomorphisms between two $C^\*$-algebras $A$ and $B$, we may define a metric on it by setting $d(f,g):= \sup\_{0<\|a\|\le 1}\|f(a)-g(a)\|$. Could it be true that, for each \*-homomorphism $f\colon A \to B$ there exists such $\epsilon >0$ that if $d(f,g)<\epsilon$, then there exists a homotopy between $f$ and $g$, i.e. a \*-homomorphism $H\colon A\to C([0;1];B)$ such that $ev\_0H=f$ and $ev\_1H=g$? If not, could it be true when $B$ is stable?



Is the space of *-homomorphisms between two $C^*$-algebras locally path connected
Let $X$ be compact. Then $\mathrm{Hom}(C(X),\mathbb{C}) = X$, and in that case the metric you give is the discrete metric, which is [as noted by Vahid Shirbisheh below] locally path connected.


[In light of Vahid's comment, with a silly mistake corrected by Eric:]


Suppose that $B$ is commutative; we may then assume $A$ is commutative as well. By Gelfand--Naimark, there are locally compact $X,Y$ such that $A=C\_0(X), B=C\_0(Y)$. Let $f\colon A\to B$ be an algebra homomorphism. Then for any character $\delta\_y \in B^\*$, $\delta\_y\circ f$ is a homomorphism $A\to\mathbb{C}$ so either zero or a character. So we can write $Y = Y\_0 \sqcup Y\_1$ where $Y\_0$ is closed (and the image of $f$ vanishes identically there), and we have a map $f^\*\colon Y\_1 \to X$ so that $f(a) \restriction\_{Y\_1} = a\circ f^\*$. Again the metric is discrete: let $f,g$ be two homomorphisms. If the respective sets $Y\_0$ differ, the there is a point $y \in Y$ where the image of one vanishes identically, but the image of the other doesn't, so the distance is $1$. Otherwise, the maps $f^\*,g^\*$ are defined on the same set $Y\_1$ but differ, say in that $x = f^\*(y)$ and $x'=g^\*(y)$ are distinct. Then if $a\in A = C\_0(X)$ of norm $1$ has $a(x)=1$ and $a(x')=0$ we have $\Vert f(a)-g(a)\Vert\_\infty \geq f(a)(y)-g(a)(y) = f(x) - f(x') = 1$.


