Wang sequence of a mapping torus

Away from the identified endpoint, the quotient projection has product charts. Across the endpoint, use one interval on the $t=1$ side and one on the $t=0$ side, changing the fiber coordinate by $f$ in one direction and by $f^{-1}$ in the other. Thus the displayed map is the standard mapping-torus fiber bundle. Lifting the positive base loop from $t=0$ to $t=1$ starts at $x$ and ends at the class of $(x,1)$, which equals the class of $(f(x),0)$. Hence [F1] gives $T_q=f_{\ast }:H_q(F)\to H_q(F)$.

Substitute $T_q=f_{\ast }$ into the five-term part of [F2]: $H_q(F)\stackrel{1-f_{\ast }}{\to }{H}_q(F)\stackrel{i_q}{\to }{H}_q(T_f)\stackrel{{\mathrm{\partial }}_q}{\to }{H}_{q-1}(F)\stackrel{1-f_{\ast }}{\to }{H}_{q-1}(F)\mathrm{.}$ Exactness gives $\ker i_q=\mathrm{im}(1-f_{\ast })$, so $i_q$ descends to an injection from the displayed cokernel, whose image is $\ker {\mathrm{\partial }}_q$. It also gives $\mathrm{im}{\mathrm{\partial }}_q=\ker (1-f_{\ast }{\mathrm{\mid }}_{H_{q-1}(F)})$, so ${\mathrm{\partial }}_q$ corestricts to a surjection onto the displayed kernel. These two conclusions are exactly the claimed short exact sequence.

For $q=0$, the right group is zero by the stated $H_{-1}=0$ convention, so the conclusion is the literal isomorphism $H_0(T_f)\cong \mathrm{coker}(1-f_{\ast }{\mathrm{\mid }}_{H_{0}F})$. If $F$ is empty, all three nonzero-index groups are zero and the same sequence is exact. If $f=\mathrm{id}$, both Wang maps vanish and the sequence becomes $0\to H_q(F)\to H_q(F\times S^1)\to H_{q-1}(F)\to 0$, still without selecting a splitting. Orientation reversal replaces $1-f_{\ast }$ by $f_{\ast }-1$; multiplication by $-1$ leaves its kernel, image, and cokernel canonically isomorphic. Thus zero maps, identity monodromy, both exactness endpoints, and both orientation signs are covered. The argument uses no choice principle and proves no converse.