Is the space of *-homomorphisms between two $C^*$-algebras locally path connected 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?
 A: 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 $\Vert f(a)-g(a)\Vert_\infty \geq f(a)(y)-g(a)(y) = f(x) - f(x') = 1$.
A: 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 norm-continuous one-parameter 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 so-called 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 UHF-algebra" [Thomsen], where he calculates the homotopy groups of Aut$(A)$ for a UHF-algebra $A$ in both cases.
A: 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 non-commutative C*-algebras (e.g. all finite-dimensional 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 $g$.
