Homological Algebra Lemmas

None of the proofs below are original. My goal here is to practice reading terse homological algebra proofs from Stacks Project and experiment with presenting them in a much more verbose way.

Lemma SP.12.5.11

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TODO

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Lemma SP.12.5.12

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We have $e : \ker f \to \ker k$ (with $j \circ e = g \circ i$) as the coastriction of $g \circ i$ to $\ker k.$ This is valid since

$$k \circ g \circ i = h \circ f \circ i = h \circ 0 = 0.$$

Since we have $0 : \ker k \to y$ and $j : \ker k \to x$ the pullback UP of $w$ gives us $t = \langle j, 0 \rangle.$

Since $f \circ t = 0,$ we have a coastriction $u : \ker k \to \ker f$ such that $i \circ u = t.$

To show isomorphism, we want $u \circ e = \id_{\ker f}$ and $e \circ u = \id_{\ker k}.$

For the first case, we can send probes into $x$ and $y$ and use the joint monomorphism property of a pullback.

Start with $g \circ i \circ u \circ e,$ the $x$-probe. Since $i \circ u = t,$ this is just $g \circ t \circ e.$

Now use $j = g \circ t.$

Now $j \circ e = g \circ i.$

We have what we need from the first probe, $g \circ i \circ u \circ e = g \circ i.$

Now for the second probe, we want $f \circ i \circ u \circ e.$ This is simple.

Since $f \circ i = 0,$ it doesn’t matter whether we include an $u \circ e$ or not, the result remains $0.$

So, since $f \circ i \circ u \circ e = f \circ i$ and $g \circ i \circ u \circ e = g \circ i,$ we can conclude that $i \circ u \circ e = i$ via the joint monomorphism property of $w.$ Since $i$ itself s a monomorphism, this further implies $u \circ e = \id_{\ker f}.$

To prove $e \circ u = \id_{\ker k}$ we will consider $j \circ e \circ u.$ But since $j \circ e = g \circ i$ this is the same as $g \circ i \circ u.$

But now we see that $i \circ u = t.$

Now we use $j = g \circ t.$

So $j \circ e \circ u = j$ and since $j$ is mono, we have $e \circ u = \id_{\ker k}.$

Thus we have an isomorphism $\ker f \cong \ker k.$