# Monday, February 01

Recall the universal of projective modules.

\begin{tikzcd}
	&& P \\
	\\
	B && C && 0
	\arrow["g", from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow["{\exists \beta}"', dashed, from=1-3, to=3-1]
	\arrow["\gamma", from=1-3, to=3-3]
\end{tikzcd}

:::{.definition title="Enough Projective"}
If \( \mathcal{A}  \) is an abelian category, then \( \mathcal{A}  \) has **enough projectives** if and only if for all \( a\in \mathcal{A}  \) there exists a projective object \( P \in \mathcal{A}  \) and a surjective morphism \( P \surjects A \).
:::

:::{.example title="?"}
\( \modr \) has enough projectives: for all \( A \in \modr \), one can take \( F(A) \surjects A \).
:::

:::{.example title="?"}
The category of finite abelian groups does *not* have enough projectives.

\todo[inline]{Why?}

:::

:::{.lemma title="?"}
$P$ is projective if and only if $\Hom_{\mathcal{A}}(P, \wait)$ is an exact functor.
:::

:::{.exercise title="?"}
Prove this!
:::

:::{.definition title="(Key)"}
Let $M\in \modr$, then a **projective resolution** of $M$ is an exact complex
\[
\cdots
    \mapsvia{d_2} 
P_1 \mapsvia{d_1}  
P_1 \overset{\eps}{\surjects} 
M \to 
0
.\]

We write $P_\wait \overset{\epsilon}{\surjects} M$.
:::

:::{.lemma title="(Key)"}
Every object $M\in \modr$ has a projective resolution.
This is true in any abelian category with enough projectives.
:::

:::{.proof title="?"}
\envlist

- Since there are enough projectives, choose $P_0 \mapsvia{\epsilon_0} M \to 0$.
- To extend this, set $M_0 \da \ker \epsilon_0$, then find a projective cover $P_1 \mapsvia{\epsilon_1}  M_0$
- Use that $d_1 \da \iota_0 \circ \epsilon_1$ and $\im d_1 = M_0 = \ker \epsilon_0$
- Then $d_2 \da \iota_1 \circ \epsilon_2$ with $\im d_2 = M_1$, and $\ker d_1 = \ker \epsilon_1 = M_1$.
- Continuing in this fashion makes the complex exact at every stage.

\begin{tikzcd}
	&& 0 && 0 \\
	&&& {M_1} \\
	\cdots && {P_2} && {P_1} && {P_0} & M & 0 \\
	& {M_2} &&&& {M_0} \\
	0 &&&& 0 && 0
	\arrow["{\varepsilon_0}", two heads, from=3-7, to=3-8]
	\arrow["{\iota_0}", from=4-6, to=3-7]
	\arrow["{\exists d_1}", dashed, from=3-5, to=3-7]
	\arrow["{\varepsilon_0}"{description}, two heads, from=3-5, to=4-6]
	\arrow[hook, from=5-5, to=4-6]
	\arrow[from=4-6, to=5-7]
	\arrow[hook, from=2-4, to=3-5]
	\arrow[two heads, from=3-3, to=2-4]
	\arrow["{\exists d_2}", dashed, from=3-3, to=3-5]
	\arrow[hook, from=1-3, to=2-4]
	\arrow[hook, from=5-1, to=4-2]
	\arrow[hook, from=4-2, to=3-3]
	\arrow[two heads, from=2-4, to=1-5]
	\arrow[dashed, from=3-1, to=3-3]
\end{tikzcd}


> [Link to Diagram](https://q.uiver.app/?q=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)


:::

## Comparison Theorem

:::{.theorem title="Comparison Theorem"}
Suppose $P_\wait \mapsvia{\epsilon} M$ is a projective resolution of an object in \( \mathcal{A}  \) and \( (M \mapsvia{f} N) \in \Mor( \mathcal{A})  \) and $Q_\wait \mapsvia{\eta} N$ a resolution of $N$.
Then there exists a chain map $P \mapsvia{f}  Q$ lifting $f$ which is unique up to chain homotopy:

\begin{tikzcd}
	\cdots & {P_2} & {P_1} & {P_0} & M & 0 \\
	\\
	\cdots & {Q_2} & {Q_1} & {Q_0} & N & 0
	\arrow["f_{-1} \da f", from=1-5, to=3-5]
	\arrow["{\exists f_0}", dashed, from=1-4, to=3-4]
	\arrow["{\exists f_1}", dashed, from=1-3, to=3-3]
	\arrow["{\exists f_2}", dashed, from=1-2, to=3-2]
	\arrow[from=1-1, to=1-2]
	\arrow["{d_2^P}", from=1-2, to=1-3]
	\arrow["{\varepsilon = d_0^P}", from=1-4, to=1-5]
	\arrow[from=1-5, to=1-6]
	\arrow[from=3-1, to=3-2]
	\arrow["{d_2^Q}"', from=3-2, to=3-3]
	\arrow["{d_1^Q}"', from=3-3, to=3-4]
	\arrow["{\eta = d_0^Q}"', from=3-4, to=3-5]
	\arrow[from=3-5, to=3-6]
	\arrow["{d_1^P}", from=1-3, to=1-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)


:::

:::{.remark}
The proof will only use that $P \mapsvia{\epsilon} M$ is a chain complex of projective objects, i.e. $d^2 = 0$, and that \( \epsilon \circ d_1^p = 0 \).
To make the notation more consistent, we'll write $Z_{-1}(P) \da M$ and $Z_{-1}(Q) \da N$.
Toward an induction, suppose that the $f_i$ have been constructed for $i\leq n$, so $f_{i-1} \circ d = d \circ f_i$.
:::

:::{.proof title="Existence"}
A fact about chain maps is that they induce maps on the kernels of the outgoing maps, so there is a map $f_n': Z_n(P) \to Z_n(Q)$.
We get a diagram where the top row is not necessarily exact:

\begin{tikzcd}
	{P_{n+1}} & {} & {Z_n(P)} \\
	\\
	{Q_{n+1}} && {Z_n(Q)} && 0
	\arrow["d", from=1-1, to=1-3]
	\arrow["d", from=3-1, to=3-3]
	\arrow["d", from=3-3, to=3-5]
	\arrow["{f_{n'}}", from=1-3, to=3-3]
	\arrow["{\exists f_{n+1}}"{description}, dashed, from=1-1, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwwXSxbMiwwLCJaX24oUCkiXSxbMCwwLCJQX3tuKzF9Il0sWzAsMiwiUV97bisxfSJdLFsyLDIsIlpfbihRKSJdLFs0LDIsIjAiXSxbMiwxLCJkIl0sWzMsNCwiZCJdLFs0LDUsImQiXSxbMSw0LCJmX3tuJ30iXSxbMiwzLCJcXGV4aXN0cyBmX3tuKzF9JyIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

Using the definition of projective, since $P_{n+1}$ is projective, the map $f_{n+1}: P_{n+1} \to Q_{n+1}$ exists where $d \circ f_{n+1} = f_n' \circ d = f_n \circ d$, since $f_n = f_n'$ on $\im d \subseteq Z_n(P)$.
This yields commutativity of the above square.

:::

:::{.proof title="Uniqueness"}
Suppose $g: P\to Q$ is another lift of $f'$, the consider $h\da f-g$.
This is a chain map $P\to Q$ lifting of $f' - f' = 0$.
We'll construct a chain contraction $\ts{ s_n:; P_n \to Q_{n+1} }$ by induction on $n$:

We have the following diagram:

\begin{tikzcd}
	&& {P_0} && M \\
	\\
	{Q_1} && {Q_0} && N
	\arrow["{f-f'=0}", from=1-5, to=3-5]
	\arrow["\varepsilon", from=1-3, to=1-5]
	\arrow["{h_0 \da f_0 - f_0'}"', from=1-3, to=3-3]
	\arrow["\eta"', from=3-3, to=3-5]
	\arrow["d"', from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbNCwwLCJNIl0sWzIsMCwiUF8wIl0sWzIsMiwiUV8wIl0sWzQsMiwiTiJdLFswLDIsIlFfMSJdLFswLDMsImYtZic9MCJdLFsxLDAsIlxcdmFyZXBzaWxvbiJdLFsxLDIsImhfMCBcXGRhIGZfMCAtIGZfMCciLDJdLFsyLDMsIlxcZXRhIiwyXSxbNCwyLCJkIiwyXV0=)

Setting $P_{-1}\da 0$ and $s_{-1}: P_{-1} \to Q_0$ to be the zero map, we have $\eta \circ h_0 = \eps (f' - f') = 0$.
Using projectivity of $P_0$, there exists an $s_0$ as shown below which satisfies $h_0 = d \circ s_0 = ds_0 + s_{-1} d$ where $s_{-1} d= 0$:

\begin{tikzcd}
	&& {P_0} && {P_{-1} = 0} \\
	\\
	{Q_1} && {d(Q_1)} && 0
	\arrow["{d_0 = 0}", from=1-3, to=1-5]
	\arrow["{s_{-1} = 0}", from=1-5, to=3-3]
	\arrow["{h_0}"', from=1-3, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow[two heads, from=3-1, to=3-3]
	\arrow["{\exists s_1}"', dashed, from=1-3, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwwLCJQXzAiXSxbNCwwLCJQX3stMX0gPSAwIl0sWzIsMiwiZChRXzEpIl0sWzQsMiwiMCJdLFswLDIsIlFfMSJdLFswLDEsImRfMCA9IDAiXSxbMSwyLCJzX3stMX0gPSAwIl0sWzAsMiwia18wIiwyXSxbMiwzXSxbNCwyLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMCw0LCJcXGV4aXN0cyBzXzEiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=)

Proceeding inductively, assume we have maps $s_i: P_i \to Q_{i+1}$ such that $h_{n-1} = d s_{n-1} + s_{n-2} d$, or equivalently $ds_{n-1} = h_{n-1} - s_{n-2} d$.
We want to construct $s_n$ in the following diagram:

\begin{tikzcd}
	&& {P_n} && {P_{n-1}} && {P_{n-2}} \\
	\\
	{Q_{n+1}} && {Q_n} && {Q_{n-1}}
	\arrow["d", from=1-3, to=1-5]
	\arrow["d", from=1-5, to=1-7]
	\arrow["{h_{n-1}}", from=1-5, to=3-5]
	\arrow["{s_{n-2}}"{description}, from=1-7, to=3-5]
	\arrow["{s_{n-1}}"{description}, from=1-5, to=3-3]
	\arrow["d"{description}, from=3-3, to=3-5]
	\arrow["d"{description}, from=3-1, to=3-3]
	\arrow["{h_{n}}"{description}, from=1-3, to=3-3]
	\arrow["{\exists s_n}"{description}, dashed, from=1-3, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJQX24iXSxbNCwwLCJQX3tuLTF9Il0sWzYsMCwiUF97bi0yfSJdLFsyLDIsIlFfbiJdLFs0LDIsIlFfe24tMX0iXSxbMCwyLCJRX3tuKzF9Il0sWzAsMSwiZCJdLFsxLDIsImQiXSxbMSw0LCJoX3tuLTF9Il0sWzIsNCwic197bi0yfSIsMV0sWzEsMywic197bi0xfSIsMV0sWzMsNCwiZCIsMV0sWzUsMywiZCIsMV0sWzAsMywiaF97bn0iLDFdLFswLDUsIlxcZXhpc3RzIHNfbiIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

So consider $h_n - s_{n-1} d: P_n \to Q_n$, which we want to equal $d(s_n)$.
We want exactness, so we need better control of the image!
We have $d(h_n - s_{n-1} d) = d h_n - (h_{n-1} - s_{n-2} d)d$.
But this is equal to $d h_n - h_{n-1}d = 0$ since $h$ is a chain map.
Thus we get $h_n - s_{n-1}d: P_n \to Z_n(Q)$, and thus using projectivity one last time, we obtain the following:

\begin{tikzcd}
	&& {P_n} \\
	\\
	{Q_{n+1}} && {Z_n(Q)} && 0
	\arrow["{\exists s_n}", dashed, from=1-3, to=3-1]
	\arrow["d", from=3-1, to=3-3]
	\arrow["{h_n - s_{n-1}d}", from=1-3, to=3-3]
	\arrow["d", from=3-3, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJRX3tuKzF9Il0sWzIsMiwiWl9uKFEpIl0sWzIsMCwiUF9uIl0sWzQsMiwiMCJdLFsyLDAsIlxcZXhpc3RzIHNfbiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDEsImQiXSxbMiwxLCJoX24gLSBzX3tuLTF9ZCJdLFsxLDMsImQiXV0=)

Since $P_n$ is projective, there exists an $s_n: P_n \to Q_{n+1}$ such that $ds_n = h_n - s_{n-1} d$.
:::