I am self-studying topology with only background in physics. I follow this lecture notes. On page 4, proposition 1.4 says:
Proposition: Let $M^{\prime}$, $M$, and $M^{\prime\prime}$ be differential $R$-modules, with differential morphisms $d^{\prime}:M^{\prime}\rightarrow M^{\prime}$, $d:M\rightarrow M$, and $d^{\prime\prime}: M^{\prime\prime}\rightarrow M^{\prime\prime}$, such that $$d^{\prime}\circ d^{\prime}=d\circ d=d^{\prime\prime}\circ d^{\prime\prime}=0.$$ Let the following diagram $$0\xrightarrow{}M^{\prime}\overset{i}{\hookrightarrow}M\overset{p}{\twoheadrightarrow}M^{\prime\prime}\xrightarrow{}0$$ be a short exact sequence, and $H(M^{\prime},d^{\prime})$, $H(M,d)$, and $H(M^{\prime\prime},d^{\prime\prime})$ the cohomology groups, respectively. Then, there exists a morphism $\delta:H(M^{\prime\prime},d^{\prime\prime})\rightarrow H(M^{\prime},d^{\prime})$ such that the following diagram $$\require{AMScd} \require{cancel} \def\diagdownarrow#1{\smash{ \raise.6em\rlap{\scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}} }} \begin{CD} && H(M^{\prime},d^{\prime})\\ & \diagdownarrow{\delta} @VV H(i) V \\ H(M^{\prime\prime},d^{\prime\prime}) @<< H(p)< H(M,d) \end{CD}$$ is exact.
My question is about its proof:
To begin with, we can pick up an element $[m^{\prime\prime}]\in H(M^{\prime\prime},d^{\prime\prime})$, where $m^{\prime\prime}$ is a cocycle in $M^{\prime\prime}$. Since $p$ is surjective, there exists an element $m\in M$ such that $p(m)=m^{\prime\prime}$. Then, it follows that $$p\circ d(m)=d^{\prime\prime}\circ p(m)=0.$$
This means $d(m)\in\ker(p)=\mathrm{im}(i)$, i.e there exists $m^{\prime}\in M^{\prime}$ such that $d(m)=i(m^{\prime})$. Then, one finds $$i\circ d^{\prime}(m^{\prime})=d\circ i(m^{\prime})=d\circ d(m)=0.$$
But since $i$ is injective, the above identity means $d^{\prime}(m^{\prime})=0$. i.e $m^{\prime}$ is a cocycle in $M^{\prime}$.
Now, to show that $m^{\prime}$ defines a morphism $H(M^{\prime\prime},d^{\prime\prime})\rightarrow H(M^{\prime},d^{\prime})$, I must first show that for each $[m^{\prime\prime}]\in H(M^{\prime\prime},d^{\prime\prime})$, there exists at most one $[m^{\prime}]\in H(M^{\prime},d^{\prime})$. But this isn't so obvious to me because the preimage of $p(m)$ isn't unique.
So my question is how to prove that the morphism $$\delta:H(M^{\prime\prime},d^{\prime\prime})\rightarrow H(M^{\prime},d^{\prime})$$
exists?
Given $[m''] \in H(M'', d'')$, the definition of $\delta([m''])$ involves making two choices: we chose a representative element in $M''$ of the cohomology class $[m'']$, and we chose a pullback $m$ of $m''$. To see that it is well defined we have to make sure that our class $[m']$ is independent of these choices.
For the latter say we have $m_1, m_2$ with $p(m_1) = p(m_2) = m''$, and $m_1', m_2' \in M'$ with $i(m_1') = d(m_1), i(m_2') = d(m_2)$. Then $p(m_1 - m_2) = p(m_1) - p(m_2) = 0$ so $m_1 - m_2 = i(m')$. Now $i(d'(m')) = d(i(m')) = d(m_1 - m_2) = d(m_1) - d(m_2) = i(m_1') - i(m_2') = i(m_1' - m_2')$, so $d'(m') = m_1' - m_2'$ as $i$ is injective. Then $[m_1' - m_2'] = 0$ so $[m_1'] = [m_2']$ and hence two pullbacks of $m''$ define the same cohomology class in $H(M', d')$.
Now we need to show that $\delta([m''])$ is independent of the choice of representation for our cohomology class. For this take $m_1'', m_2''$ with $[m_1''] = [m_2'']$. We take $m_1, m_2$ with $p(m_1) = m_1''$ and $p(m_2) = m_2''$ and $m_1', m_2'$ with $i(m_1') = d(m_1), i(m_2') = d(m_2)$. Since $[m_1''] = [m_2'']$ we have $m_1'' - m_2'' = d''(m'')$. Take some $m$ with $p(m) = m''$, so that $p(d(m)) = d''(p(m)) = d''(m'') = p(m_1 - m_2)$. Then $p(m_1 - m_2 - d(m)) = 0$ and there exists $m' \in M'$ with $i(m') = m_1 - m_2 - d(m)$. Thus $i(d'(m')) = d(i(m')) = d(m_1 - m_2 - d(m)) = d(m_1) - d(m_2) = i(m_1') - i(m_2')$, so by injectivity $m_1' - m_2' = d'(m')$ and $\delta([m_1'']) = [m_1'] = [m_2'] = \delta([m_2''])$.
We conclude $\delta$ is well defined. It's probably worth mentioning that there's a purely categorical version of this argument coming from applying the snake lemma to a commutative diagram given by stacking one copy of your SES on top of the other to get two rows and then connecting them vertically with the differential maps, if that means anything to you.