The following question is from An Introduction to $K$-theory for $C^{\ast}$-Algebra and an e-copy can be found here. Below is the question (since I do not know how to create a diagram in MS ...)
By the half exactness of $K_0$ and $K_1$ functor, the six sequence above will be exact at $K_0(E_{\phi})$ and $K_1(E_{\phi})$. By definition of $E_{\phi}$, each $\phi(a)$ will be homotopic to $0$ inside $B$ and hence $K_0(\pi)\big[ K_0(E_{\phi}) \big] = \operatorname{Ker}K_0(\phi)$. Since $K_1(E_{\phi})\cong K_1(\overset{\sim}{E_{\phi}})$, a unitary in $\overset{\sim}{E_{\phi}}$ will be in the form of $(F, u)$ where $u$ is a unitary in $A$ and $F(0) = 1, F(1) = u$. Hence, we can conclude that $K_1(\pi)\big[ K_1(E_{\phi}) \big] = \operatorname{Ker}K_1(\phi)$.
Then I do not know how to show that the sequence is exact at $K_0(B)$ and $K_1(B)$. In the question $\theta_B: K_1(B)\rightarrow K_0(SB)$ is a group isomorphism (which can be found in Chapter 10) and $\beta_B: K_0(B)\rightarrow K_1(SB)$ is also a group isomorphism (which can be found in Chapter 12). The $\delta_0: K_0(A)\rightarrow K_1(SB)$, $\delta_1: K_1(A)\rightarrow K_0(SB)$ are also introduced in Chapter 12 (12.3), but I am more convinced that $\gamma_1 = K_0(\iota)\circ\theta_B$ and $\gamma_0 = K_1(\iota)\circ\beta_B$.
If I am not mistaken on the definition of $\gamma_0$ and $\gamma_1$, I wonder, for instance, if given a projection in $M_n(A)\,(n\in\mathbb{N})$, why the equivalence class of $f_{\phi(p)}f_{s\big[\phi(p)\big]}$ will be mapped to zero under $K_0(\iota)$ (the definition of $f_{\phi(p)}$ can be found in Chapter 11, and $s\big[\phi(p)\big]$ is the scalar part of $\phi(p))$. Any hints will be appreciated.
I found a solution from Exercise 6.M in K-theory and $C^*$-Algebra written by Wegge-Olsen, which might answer the question above but there is still a big ambiguity. First, define $Z_{\phi} = \{a\oplus f\in A\bigoplus C\big([0, 1], B\big)\,\vert\, \phi(a) = f(1)\}$. Then the following sequence is exact:
$$ \large 0\rightarrow E_{\phi}\overset{\iota}{\longrightarrow} Z_{\phi} \overset{\pi_B: f\mapsto f(0)}{\longrightarrow} B\rightarrow 0 $$
where for each $b\in B$, the element $b_t:t\mapsto (1-t)b$ is in $C\big([0, 1], B\big)$ and hence $\pi_B(0\oplus b_t) = b$, which shows that $\pi_B$ is surjective. Then, define:
$$ F:A\rightarrow Z_{\phi},\hspace{0.3cm} a\mapsto a\oplus\mathbb{1}\phi(a) $$ where $\mathbb{1}\phi(a)$ is the function in $C\big([0, 1], B\big)$ that is constantly equal to $\phi(a)$. Also, define:
$$ \pi_A: Z_{\phi}\mapsto A,\hspace{0.3cm} a\oplus f\mapsto a $$
Because $[0, 1]$ is contracitble, we then have $\pi_A\circ F = \operatorname{id}_A$ and $F\circ\pi_A$ homotopic to the identity of $Z_{\phi}$, which implies that $A$ is homotopic equivalent to $Z_{\phi}$. Then, we have the following cyclic exact sequence, where $K_i(Z_{\phi})\cong K_i(A)\,(i=0, 1)$:
where the existence of $\delta_0, \delta_1$ can be found in Chapter 12 in the reference given in the question above. Therefore, we then have the following cyclic exact sequence, which is more like the original one:
In this case, it makes more sense to say $\gamma_0 = \delta_1$ and $\gamma_1 = \delta_0$.