Def of Homology from ncatlab I have a question about the def of homology of chain complexes from ncatlab.  So the def provided there is below

I don't know how we have $\operatorname{coker}(V_{n+1} \to \ker(\delta_{n-1})) \cong \ker(\operatorname{coker}(\delta_n) \to \operatorname{im}(\delta_{n-1})) $ from the diagram.
I do see that, if we let $f:V_{n+1} \to \ker(\delta_{n-1}), \phi:\ker(\delta_{n-1}) \to \operatorname{coker}(\delta_n)$ and $\pi:\ker(\delta_{n-1}) \to \operatorname{coker}(f)$ be the cokernel of $f$, then we have a unique arrow $h:\operatorname{coker}(f) \to \operatorname{coker}(\delta_n)$ and I can show that $ (\operatorname{im}(\delta_{n-1}) \xleftarrow[]{} \operatorname{coker}(\delta_{n})) \circ h = 0$ so $h$ factors through a unique arrow from $\operatorname{coker}(f) \to \ker(\operatorname{coker}(\delta_{n}) \to \operatorname{im}(\delta_{n-1}) )$.  If I can find an arrow from $\ker(\operatorname{coker}(\delta_{n}) \to \operatorname{im}(\delta_{n-1}) ) \to \operatorname{coker}(f)$, then I am done but I'm not sure how to find that arrow.
Thank you.
 A: Let's write $a\colon\mathrm{coker}(\delta_n)\rightarrow\mathrm{im}(\delta_{n-1})$. Consider $W_n=\ker(V_n\rightarrow\mathrm{coker}(\delta_n)\rightarrow\mathrm{coker}(\delta_n)/\ker(a))$ with its canonical inclusion $W_n\rightarrow V_n$. Using that $\ker(a)\rightarrow\mathrm{coker}(\delta_n)$ is the kernel of $\mathrm{coker}(\delta_n)\rightarrow\mathrm{coker}(\delta_n)/\ker(a)$ and that the composition $W_n\rightarrow V_n\rightarrow\mathrm{coker}(\delta_n)\rightarrow\mathrm{coker}(\delta_n)/\ker(a)$ is $0$, we obtain that $W_n\rightarrow V_n\rightarrow\mathrm{coker}(\delta_n)$ factors as $W_n\rightarrow\ker(a)\rightarrow\mathrm{coker}(\delta_n)$ with $W_n\rightarrow\ker(a)$ surjective. Using that $\mathrm{im}(\delta_n)\rightarrow V_n$ is the kernel $V_n\rightarrow\mathrm{coker}(\delta_n)$, we obtain a factorization $\mathrm{im}(\delta_n)\rightarrow W_n$ which moreover is the kernel of $W_n\rightarrow\ker(a)$ as a diagram chase shows. It follows that the canonical map $\mathrm{coker}(\mathrm{im}(\delta_n)\rightarrow W_n)\rightarrow\ker(a)$ is an isomorphism.
Next, observe that $W_n\rightarrow V_n$ factors through $\ker(\delta_{n-1})\rightarrow V_n$ via a morphism $W_n\rightarrow\ker(\delta_{n-1})$. Now, consider the composite $W_n\rightarrow\ker(\delta_{n-1})\rightarrow\mathrm{coker}(f)$. Observe that the composite $\mathrm{im}(\delta_n)\rightarrow W_n\rightarrow\ker(\delta_{n-1})$ is the same as the map $\mathrm{im}(\delta_n)\rightarrow\ker(\delta_{n-1})$ in the original diagram (this can be tested with the monomorphism $\ker(\delta_{n-1})\rightarrow V_n$). Thus, the composite $V_{n+1}\rightarrow\mathrm{im}(\delta_n)\rightarrow W_n\rightarrow\ker(\delta_{n-1})\rightarrow\mathrm{coker}(f)$ is the same thing as $V_{n+1}\stackrel{f}{\rightarrow}\ker(\delta_{n-1})\rightarrow\mathrm{coker}(f)$, i.e. $0$. Since $V_{n+1}\rightarrow\mathrm{im}(\delta_n)$ is an epimorphism, this implies $\mathrm{im}(\delta_n)\rightarrow W_n\rightarrow\mathrm{coker}(f)$ is $0$, so that $W_n\rightarrow\mathrm{coker}(f)$ factors through a morphism $\mathrm{coker}(\mathrm{im}(\delta_n)\rightarrow W_n)\rightarrow\mathrm{coker}(f)$. The composite $\ker(a)\rightarrow\mathrm{coker}(\mathrm{im}(\delta_n)\rightarrow W_n)\rightarrow\mathrm{coker}(f)$ is the desired map.
To check that this is inverse to your map $\mathrm{coker}(f)\rightarrow\ker(a)$, note that $\ker(\delta_{n-1})\rightarrow V_n$ also factors through $W_n\rightarrow V_n$ and that the resulting map $\ker(\delta_{n-1})\rightarrow W_n$ is inverse to the map $W_n\rightarrow\ker(\delta_{n-1})$ from before. From this, conclude that the maps $W_n\rightarrow\ker(a)$ and $\ker(\delta_{n-1})\rightarrow\ker(a)$ correspond under these isomorphisms. The rest is some more diagram chases.
