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Consider the short exact sequence of $C^\ast$-algebras

$$ 0 \rightarrow C_0(0,1)\overset{\imath }{\rightarrow} C[0,1] \overset{\psi}{\rightarrow} \mathbb C \oplus \mathbb C \to 0, $$ where $\imath: C_0(0,1) \to C[0,1]$ is the inclusion and $\psi(f) = (f(0),f(1))$ for each $f \in C[0,1]$.

I'm trying to proof that this exact sequence does not split.

Supposing that it does split, exists $\lambda: \mathbb C \oplus \mathbb C \to C[0,1]$ an $\ast$-homomorphism such that $\psi \circ \lambda = id|_{\mathbb C \oplus \mathbb C}$. Then, given $a,b \in \mathbb C$, we have

$$ (a,b) = \psi \circ \lambda (a,b) = ([\lambda (a,b)](0), [\lambda (a,b)](1)). $$ Hence, $\lambda (a,b) : [0,1] \to \mathbb C$ is a continuous path from $a$ to $b$.

How can I get a contradiction?

user 242964
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    $\lambda(a,b)(t)=(1-t)a+tb$ satisfies $\psi\circ\lambda=id|_{\mathbb{C}\oplus\mathbb{C}}$. – YAlexandrov Mar 15 '18 at 12:07
  • And $\gamma:C[0,1]\to C_0[0,1]$ defined by $\gamma(f)(t)=f(t)-(1-t)f(0)-tf(1)$ satisfies $\gamma\circ \iota(f)=f$ for $f\in C_0[0,1]$. – YAlexandrov Mar 15 '18 at 12:11

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