Is this proof for Mayer-Vietoris theorem correct or wrong?

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Here is a proof for Mayer-Vietoris theorem for cofibre homology theory from TOPOLOGICAL METHODS FOR C*-ALGEBRAS III:AXIOMATIC HOMOLOGY (p426 theorem 4.1)

Theorem 4.1. Let $h_*$ be a cofibre homology theory and let $P\stackrel{g_i}\to A_i\stackrel{f_i}\to B\ (i=1,2)$ be a pullback. Assume $f_1$ and $f_2$ are surjective, then there is a long exact sequence $...\to h_n(P)\to h_n(A_1)\oplus h_n(A_2)\stackrel{-f_{1*}+f_{2*}}\longrightarrow h_n(B)\to ...$

Proof. Note $P=\{(a_1,a_2)\in A_1\oplus A_2:f_1(a_1)=f_2(a_2)\}$. Thus the mapping cone $Cg$ of $g:P\to A_1\oplus A_2$, that is , $Cg=\{(x,\xi_1,\xi_2)\in P\oplus CA_1\oplus CA_2:g_i(x)=\xi_i(0)\}$, equals to $\{(\xi_1,\xi_2)\in CA_1\oplus CA_2:f_1\xi_1(0)=f_2\xi_2(0)\}$.

Now there is a natrual short exact sequence $\ker\psi\to Cg\stackrel{\psi}\to SB$ where $\ker \psi=C\ker f_1\oplus C\ker f_2$ ($\psi$ is the map that attaches $f_1\xi_1$ and $f_2\xi_2$ into one curve in $SB$). Since $h_n(C\ker f)=0$, this yields an isomorphism $h_n(Cg)\simeq h_n(SB)=h_{n+1}(B)$.

Finally, the cofibration $Cg\to P\stackrel{ g}\to A_1\oplus A_2$ gives riseto the long exact sequence $...\to h_n(P)\to h_n(A_1)\oplus h_n(A_2)\to h_{n-1}(Cg)=h_n(B)\to ...$


$CA,SA$ stands for $C_0([0,1))\otimes A$ and $C_0((0,1))\otimes A$ respectively here.

My confusion is that, $g:P\to A_1\oplus A_2$ doesn't seem a cofibration. Moreover, $g$ is not even surjective, therefore the long exact sequence $...\to h_n(Cg)\to h_n(P)\to h_n(A_1\oplus A_2)\to ...$ does not exist even when $h_n$ is a homology theory.

Is the paper wrong, or have I misunderstood anything?