# Why is this quotient algebra contractible? (Blackadar 21.4.3)

I ran into this question when reading the proof of Theorem 21.4.3 in Bruce Blackadar, K-theory for Operator Algebras The question is as follows:

Given $A$, $B$ Banach algebras, and a surjective homomorphism $q: A \to B$. Define $T=\{(f, g): g(0)=q(f(0))\} \subset C_0((-1, 0], A) \oplus C_0([0, 1), B)$. Then $T/C_0((0,1), B)$ is contractible.

I know the algebra $T$ is kinda like glueing CA and CB along 0, but I don't know how does mod-ing out $C_0((0,1), B)$ help make thing contractible. Any hint / suggestion would be greatly appreciated!

For those who are less familiar with the homotopy of Banach algebra homomorphism, here is the definition [Blackadar 5.2.2]: Two homomorphisms $\phi, \psi: A \to B$ are homotopic if there is a path of homomorphisms $w_t: A \to B$ for $0 \leq t \leq 1$, continuous in $t$ in the topology of pointwise norm-convergence, with $w_0=\phi$, $w_1=\psi$. This is equivalent to the existence of a homomorphism $w: A \to C([0, 1], B)$ with $\pi_0 \circ w = \phi$ and $\pi_1 \circ w =\psi$.
An algebra $A$ is contractible if the identity map is homotopic to the 0 map.

Consider the map: $T \to C_0((0, 1], A)$ by projection to the first component. This map is clearly a continuous surjective algebra homomorphism. Kernel of this map is $C_0((0, 1), B)$ and hence $T / C_0((0,1), B)$ is isomorphic to the image ($=C_0((0, 1], A)$) by the First Isomorphism Theorem of algebra. It is well-known that cone algebra $C_0((0, 1], A)$ is contractible.