lifting property of surjective $*$-homomorphism Given an exact sequence of $C^*$-algebras $0\to J\xrightarrow{\varphi}A\xrightarrow{\pi}B\to 0$, I want to show that $$0\to C([0,1],J)\to C([0,1],A)\to C([0,1],B)\to 0$$ is exact, with the naturally induced maps. 
We can think about $J$ as an ideal of $A$ and $B$ as $A/J$.
I'm trying to show that $C([0,1], A)\xrightarrow{\Pi}C([0,1], A/J)$ is surjective.
 This argument is actually saying that any continuous path $f: [0,1] \to A/J$ can be lifted to a continuous path $F:[0,1]\to A$, i.e. $\pi \circ F=f$.
I can think about one optional solution: the diagram
$$\begin{array}
AC([0,1])\otimes A & \xrightarrow{f\otimes a\to f\otimes \pi(a)} & C([0,1])\otimes B \\
\vphantom{\\ \ \\}\Bigg\downarrow{f\otimes a \to fa} & & \vphantom{\\ \ \\}\Bigg\downarrow{f\otimes b \to fb} \\
C([0,1],A) & \xrightarrow{\Pi} & C([0,1],B) 
\end{array}
$$is commutative and the vertical arrows are isomorphisms, so it's sufficient to show that $f\otimes a \to f\otimes \pi(a)$ is surjective.
All simple tensors of $C([0,1])\otimes B$ are in the range of this map (because $\pi$ is surjective). 
But the image of $*$ homomorphism between $C^*$ algebras is closed, so we can conclude that it is surjective.
However, I would like to see more "direct" and basic proof to this fact, without using tensor product of $C^*$ algebras. I think that compactness of $[0,1]$ and definition of quotient norm $\|a+J\|=\inf\{\|a+j\|  \ \ : j\in J\}$ are enough tools for solving this question, but I didn't succeed. 
 Thank you in advance.
 A: You don't explicitly need to refer to tensor products to run the same argument. Simply define elementary tensors as functions in this context: Let $B$ be a $C^{\ast}$-algebra. For $f\in C[0,1], b\in B$, define $f\cdot b \in C([0,1],B)$ by 
$$
(f\cdot b)(x) := f(x)b
$$
Let $\tilde{B} := \text{span}\{ f\cdot b : f\in C[0,1], b\in B\}$, then we claim that

$\widetilde{B}$ is dense in $C([0,1],B)$

Proof: Fix $f\in C([0,1],B)$ and $\epsilon >0$, then by uniform continuity, $\exists$ a $\delta > 0$ such that
$$
|s-t| < \delta \Rightarrow \|f(s) - f(t)\| < \epsilon
$$
By compactness, cover $[0,1]$ by finitely many intervals of radius $\delta$
$$
[0,1] = \cup_{i=1}^n (t_i - \delta/2, t_i + \delta/2)
$$
Now choose a partition of unity $\{g_1, g_2,\ldots, g_n\} \subset C[0,1]$ such that each $g_i$ is supported in $(t_i-\delta/2,t_i+\delta/2)$ and
$$
g_1+g_2+\ldots + g_n = 1 \text{ in } C[0,1]
$$
Now with $b_i := f(t_i)$, take $g\in \widetilde{B}$ as
$$
g := \sum_{i=1}^n g_i\cdot b_i
$$
then check that $\|f-g\| < \epsilon$

Now complete your proof using the same argument you mentioned. The natural map
$$
\overline{\pi} : C([0,1],A) \to C([0,1],A/J)
$$
is a $\ast$-homomorphism whose image clearly contains the set $\widetilde{A/J}$ since $\pi : A\to A/J$ is surjective. Since the image of a $\ast$-homomorphism between $C^{\ast}$-algebras must be closed, it follows that $\overline{\pi}$ is surjective.
