Fundamental Lemma of homological Algebra, what is $H_0f$ supposed to mean?

Question

I am currently attending a lecture in homological algebra, where we discussed the fundamental lemma of homological algebra (i will just cite the relevant part of the theorem):

Let $\mathcal{A}$ be an abelian category. Let $M$ be an object in $\mathcal{A}$ and let $P_*\to M$ be a projective resolution of $M$, that is: $$\cdots \to P_n\to \cdots\to P_1 \to P_0\to M \to 0.$$

Let $N$ be another object in $\mathcal{A}$ and let $Q_*\to N$ be any resolution of $N$.

Let $f\colon M\to N$ be a morphism in $\mathcal{A}$. Then the following holds:

(1) There exists a morphism $\varphi\colon P_*\to Q_*$ of chain complexes, s.t. $H_0\varphi = f\colon M\to N$.

I understand everything, except this:

$H_0\varphi = f\colon M\to N$.

What is that supposed to mean? I know that we can apply the Homology-functor $H_*$ to any chain complex in order to measure the exactness of the chain complex in each degree and any morphism $\varphi_*\colon C_*\to D_*$ between two chain complexes $C_*, D_*$ induces a morphism $H_*\varphi_*$ in homology.

But a resolution of an object in $\mathcal{A}$ is exact in each degree anyway, thus, applying the homology functor returns a long exact sequence containing only trivial elements, doesn't it?

How can i obtain $H_0\varphi = f\colon M\to N$?

Thanks for any help!

Answer 1

The morphism $\phi=\phi_\cdot$ is explicitly: $\require{AMScd}$ \begin{CD} \color{blue}{P_\cdot} @. @. 0 @<<< P_0 @<<< P_1 @<<< P_2 @<<< \dots \\ @V \color{blue}{\phi_\cdot} VV @. @. @VV\phi_0V @VV\phi_1V @VV\phi_2V \\ \color{blue}{Q_\cdot} @. @. 0 @<<< Q_0 @<<< Q_1 @<<< Q_2 @<<< \dots \\ \end{CD} No we apply the functor $H_0$ on the above blue morphism.

We obtain a map, $H_0(\phi_\cdot)$, with domain $H_0(P_\cdot)$ and codomain $H_0(Q_\cdot)$.

Since $P_\cdot $ is a (projective) resolution of $M$, we have $H_0(P_\cdot)\cong M$. (So $H_0$ is a well defined functor only when considered to have values in the category of isomorphism classes in the abelian category $\mathcal A$.) We fix this $M$ as a representative for $H_0(P_\cdot)$.

Similarly, since $Q_\cdot $ is a (projective) resolution of $N$, we have $H_0(Q_\cdot)\cong N$, and we fix this $N$ as a representative for $H_0(Q_\cdot)$.

Now the claim in $(1)$ in the OP is a restriction for $\phi$, which constrains the fact, that (after the choices of $M,N$ as representatives, and i will now write $H_0(P_\cdot)= M$, and $H_0(Q_\cdot)=N$, and lift everything to $\mathcal A$, and possibly after composing "inside" $M$ and/or $N$ with some isomorphism) the induced map $H_0(\phi_\cdot)$ from $H_0(P_\cdot)= M$ to $H_0(Q_\cdot)=N$ is the map $f:M\to N$.

This is equivalent to the fact that we have a commutative "extended diagram":

$\require{AMScd}$ \begin{CD} 0 @<<< M @<<< P_0 @<<< P_1 @<<< P_2 @<<< \dots \\ @. @VfVV @VV\phi_0V @VV\phi_1V @VV\phi_2V \\ 0 @<<< N @<<< Q_0 @<<< Q_1 @<<< Q_2 @<<< \dots \\ \end{CD}

In other words, $f$ "lifts" to $\phi_0$.