Consider the Short Exact sequence
$0\rightarrow C_0((0,1))\rightarrow C([0,1])\rightarrow \mathbb{C}\bigoplus\mathbb{C}\rightarrow 0$
where the map from $C([0,1])\rightarrow \mathbb{C}\bigoplus\mathbb{C}$ sends $f$ to $(f(0),f(1))$. How does one show this sequence does not split in terms of C* algebras. i.e. there does not exist a *-homomorphism inducing a right split. I consider the map $(a,b)$ to the continuous function starting at $a$ and ending at $b$ by Tietzes extension theorem. Why can this not be made to be a *homomorphism?
There is no "the" continuous function that you want. So you are not really defining your section. If the sequence were to split, the claim is that for any $a,b$, you can find $g_{a,b}\in C[0,1]$ such that $g(0)=a$, $g(1)=b$ and, more importantly, the map is a $*$-homomorphism: that is, $$g_{a_1+a_1,b_1+b_2}-g_{a_1,b_1}+g_{a_2,b_2}, \ \ \ g_{a_1a_2,b_1b_2}=g_{a_1,b_1}g_{a_2,b_2}.$$ I wouldn't know how to try to construct such an assignment, nor to check that it is possible.
But there are equivalent conditions to the existence of sections. If you had a right section, that means that the sequence splits (hence the name): that is, you should have $$ C[0,1]\simeq C_0(0,1)\oplus\mathbb C^2. $$ This is impossible, since the algebra on the left has two projections, while the algebra on the right has four. So the sequence does not split as algebras, you don't even need C$^*$-algebras.