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?

$\begingroup$ There is an ambiguity in your question. In your question $\epsilon$ is independent of $f$ and $G$. But the locally path connected means for every homomorphism $f$, there is an $\epsilon>0$ such that if $g$ is a homomorphism and $d(f,g)<\epsilon$, then there is a continuous homotopy between $f$ and $g$. If this second version is what you mean, you should edit your question. $\endgroup$– user23860Apr 6, 2014 at 17:47

$\begingroup$ I think Your version of the question is more correct. Thank You! $\endgroup$– Kolya IvankovApr 6, 2014 at 17:54

1$\begingroup$ I'm not sure your notion of homotopy is the most natural one. I would say that a homotopy of $*$homomorphisms $A \to B$ is a 1parameter family $(\varphi_t)_{t \in [0,1]}$ of $*$homomorphisms such that $t \mapsto \varphi_t(a) : [0,1] \to B$ is normcontinuous for each $a \in A$. This reduces to the usual notion of homotopy for maps $X \to Y$ in the commutative case. Unless I misunderstand you, what you have written would correspond to some "uniform" notion of homotopy in the commutative case. $\endgroup$– MichaelApr 9, 2014 at 23:05

1$\begingroup$ @Michael: are $C^*$algebras compactly generated? If so, then the standard topology on $C([0,1], B)$ should give the usual notion of homotopy $H$ because maps $A \times [0,1] \to B$ are the same thing as maps $A \to C([0,1], B)$. No? $\endgroup$– Andrej BauerApr 10, 2014 at 6:24

$\begingroup$ @AndrejBauer: I'm not sure whether C*algebras are compactlygenerated... but in any case, I retract my comment. Indeed, a $*$homomorphism $\varphi : A \to C([0,1],B)$ is the same thing as a family $(\varphi_t)_{t \in [0,1]}$ of $*$homomorphisms $A \to B$ such that $t \mapsto \varphi_t(a)$ is continuous for each $a \in A$. $\endgroup$– MichaelApr 10, 2014 at 21:40
3 Answers
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 GelfandNaimark, 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 $\Vert f(a)g(a)\Vert_\infty \geq f(a)(y)g(a)(y) = f(x)  f(x') = 1$.

1$\begingroup$ It is not a counterexample for the question, because any discrete space is locally path connected. $\endgroup$– user23860Apr 6, 2014 at 20:24

1$\begingroup$ You're right. In fact, this proves the claim for commutative $A$. I'll update my answer. $\endgroup$ Apr 6, 2014 at 20:47

1$\begingroup$ I believe your argument works when $B$ is commutative, but I don't see why it works when only $A$ is commutative. If the image of $A\to B$ is not central, then you may be able to conjugate it by a continuous family of unitaries to get a nontrivial path in $\mathrm{Hom}(A,B)$. $\endgroup$ Apr 7, 2014 at 14:40

1$\begingroup$ You don't actually need $A$ to be commutative, because when $B$ is commutative all homomorphisms from $A$ to $B$ will factor through the Gelfand transform of $A$. $\endgroup$ Apr 7, 2014 at 20:56

1$\begingroup$ That $X$ there should also be Hausdorff. $\endgroup$ Apr 10, 2014 at 6:21
If you restrict to automorphisms of a $C^*$algebra $A$ instead of endomorphisms $f \colon A \to B$, then I think what you suspect is true due to a paper by Kadison and Ringrose called "Derivations and Automorphisms of Operator Algebras" [KadisonRingrose].
In their main result (Theorem 7 in the paper) they prove that
If $\alpha$ is an automorphism of a $C^*$algebra A and $\lVert \alpha  id \rVert < 2$, then $\alpha$ lies on a normcontinuous oneparameter subgroup of Aut$(A)$.
This provides the continuous path $A \to C([0,1],A)$ you are looking for.
When studying the literature for these kind of questions you have to be aware that there are two very different topologies on Aut$(A)$ or the endomorphisms End$(A,B)$: The topology it inherits from being a subset of the Banach space maps $A \to B$ (sometimes called the uniform norm topology) or the socalled pointwise norm topology generated by the semi norms $p_{a}(f,g) = \lVert f(a)  g(a) \rVert$.
There is a paper by Thomsen called "The homotopy type of the group of automorphisms of a UHFalgebra" [Thomsen], where he calculates the homotopy groups of Aut$(A)$ for a UHFalgebra $A$ in both cases.

1$\begingroup$ Sorry for the paywalled links. Maybe someone can exchange them by free ones. $\endgroup$ Apr 17, 2014 at 8:02
If $A$ is semiprojective (the C*analog of an absolute neighborhood retract, ANR), then your question also has a positive answer. The class of semiprojective C*algebras includes also many noncommutative C*algebras (e.g. all finitedimensional algebras, Cuntz algebras).
More precisely, let $A$ be a semiprojective C*algebra and let $B$ be any C*algebra. Let $D=C([0,1],B)$ and consider the surjective homomorphism $\pi = ev_0\oplus ev_1\colon D\to B\oplus B$. Given any homomorphism $f\colon A\to B$, consider the homomorphism $f\oplus f\colon a\mapsto (f(a),f(a)) \in B\oplus B$. This has an obvious lift to a homomorphism $\alpha\colon A\to D$ such that $f\oplus f=\pi\circ\alpha$.
By Theorem 4.1 in Blackadar, "The homotopy lifting property for semiprojective C*algebras", homomorphisms from semiprojective C*algebras can be lifted if they are close enough to a liftable homomorphism. Let $g\colon A\to B$ be a homomorphism close to $f$. Then $f\oplus g$ is close to $f\oplus f$. Since the latter can be lifted to a homomorphism to $D$, so can $f\oplus g$. But this implies that $f$ is homotopic to $g$.