Künneth formula in Homology I'm trying to understand Künneth formula from Tom Dieck p.$299$


I don't why the connecting homomoprhism should be $(i \otimes 1)_*$. Is the latter define to make the diagram with $(i \otimes 1)$ commute? If this should be check by its definition, I'm unable to do it since the connecting homomorphism's definition seems quite abstract.
Should I use naturality of the connecting homomorphism creating two LES in order to prove it? I'm lost on it.
Any help reference or direct proof would be appreciated.
 A: This actually follows from a pretty straightforward theorem about chain complexes. For completeness I'll define everything used here. Everything in this answer can be found in Weibel's book on homological algebra.
Let $f:B\to C$ be a map of chain complexes. We define the mapping cone of $f$ to be the chain complex denoted by $\mathrm{cone}(f)$ whose degree $n$ part is $\mathrm{cone}(f) = B_{n-1}\oplus C_n$ and whose boundary map is given by the formula $$d_{\mathrm{cone}(f)}:(b,c)\mapsto (-d_B(b),d_C(c)-f(b)).$$ For any such mapping cone there is a short exact sequence of complexes $$0\rightarrow C\rightarrow \mathrm{cone}(f)\rightarrow B[-1]\rightarrow 0$$ where the first map is $c\mapsto (0,c)$ and the second map is $(b,c)\mapsto -b$ (remember that the boundary map of $B[-1]$ is $-d_B$). The theorem I am referring to is as follows:
Theorem. Let $f:B\to C$ be a map of chain complexes. Then for the associated long exact sequence on homology, the connecting homomorphism $\partial:H_{n+1}(B[-1])=H_n(B)\to H_n(C)$ is just $f_*,$ the map induced on homology by $f.$
Proof. If $b\in B[-1]_{n+1}=B_n$ is a cycle, then $(-b,0)$ lifts $b$ to $\mathrm{cone}(f)$ in the short exact sequence of chain complexes. Applying the boundary here gives us $d_{\mathrm{cone}(f)}(-b,0)=(d_B(b),f(b))=(0,f(b))$ which is just the image of $f(b)$ under the map $C\to\mathrm{cone}(f).$ Hence $\partial [b]=[f(b)]=f_*[b].$
In your question you seem to have some confusion about the defintion of the connecting homomorphism, so let's go over this again. The definition of the connecting homomorphism is just this: take a cycle in $B[-1],$ lift it to $\tilde b$ in $\mathrm{cone}(f)$ (which we can do since the map is surjective), then take its boundary $d_{\mathrm{cone}(f)}\tilde b$ in $\mathrm{cone}(f).$ What we get necessarily ends up in the image of $C\to\mathrm{cone}(f)$ so we can again pull back to $c\in C.$ It turns out that $c$ has to be a cycle, so it gives us a homology class $[c].$ We then define $\partial [b]=[c].$ One can check that this is independent of the representative $b$ and all that good stuff. In the proof we simply traced through this definition (with some convenient choices of lifts) to get that in our case $\partial = f_*.$
Okay this is great and all but how does this apply to your question? Well if one looks hard enough then mapping cones will start to appear all over the place here. In particular, $C$ is made up of free objects, so we get splittings $C_n=B_{n-1}\oplus Z_n.$ This looks like a mapping cone of $i:B[-1]\to Z$, and indeed it is. If we let $\pi:C\to B[-1]$ be the obvious map then we can define $C_n\to \mathrm{cone}(i)_n$ by $c\mapsto ((-1)^nd(c),c-\pi(c)).$ We check that this is a chain map and an isomorphism, hence $C\cong \mathrm{cone}(i).$ We can even check with slightly more work that $C\otimes D\cong \mathrm{cone}(i\otimes 1_D).$ Once we do this, we simply apply the theorem to the short exact sequence in question to get that the connecting homomorphism must be $(i\otimes 1_D)_*.$
