TITLE: Thoughts that guide a proof in abstract algebra
QUESTION [6 upvotes]: I would like to know which strategies you use when proving abstract theorems. I have a particular example in mind. I worked some hours on a proof of the following:
In the given diagram modules with corresponding module homomorphisms are shown. The homomorphisms f and g are in fact isomorphisms. Furthermore the rows are exact sequences. Task: Show that also h is an isomorphism.
\begin{array}{ccccccc}
0 & \to & A & \to & C & \to & B & \to & 0 \\
& & \downarrow f & & \downarrow h & & \downarrow g \\
0 & \to & A' & \to & C' & \to & B' & \to & 0
\end{array}
I would like to not present my thoughts right from the start, so I don't spoil any nice ideas. To me it is very interesting which strategies you use to prove the theorem. Do you start working in a mechanical way? Do you try to convince yourself of the truth of the theorem using some heuristic reasoning? Do you forget about the goal at first and derive some properties that might help? Do you work with some mental images?

REPLY [6 votes]: I would definitely start by trying to convince myself the truth of the theorem using some heuristic reasoning.  Among other things, I would mentally discard the short exact sequences in favor of submodules and cosets.  In this case, we have a homomorphism $h\colon C\to C'$ between two modules with the following properties:

$h$ maps some submodule $A$ of $C$ isomorphically to a submodule $A'$ of $C'$.
Moreover, $h$ induces a bijection between the cosets of $A$ and the cosets of $A'$.

From this reasoning, it seems clear that $h$ will be a bijection, since the map from each coset of $A$ to the corresponding coset of $A'$ will resemble the isomorphism from $A$ to $A'$.
Next I would write out the diagram-chasing proof.  However, I would be thinking in terms of submodules and cosets, and change the language to short exact sequences as I write.  Here's the argument that would go through my head:
Injective:  Suppose $h(c_1) = h(c_2)$ for some $c_1$ and $c_2$.  Since $h$ acts bijectively on cosets, $c_1$ and $c_2$ must be in the same coset of $A$.  Then $c_1 - c_2$ is an element of $A$ that maps to $0$ under $h$, so $c_1 = c_2$.
Surjective: Let $c' \in C'$.  How can we find an element of $C$ that maps to it under $h$?  Well, we know that $h$ acts bijectively on cosets, so we can find a coset $c+A$ that maps to $c'+A'$ under $H$.  If we choose a representative $c$ of this coset, it won't necessarily be the case that $h(c) = c'$.  However, it is true that $h(c) = c' + a'$ for some $a' \in A'$.  Then $h(c-a) = c'$, where $a$ is the element of $A$ that maps to $a'$.
Here's what it would look like written out:
\begin{array}{ccccccc}
0 & \to & A & \xrightarrow{i} & C & \xrightarrow{\pi} & B & \to & 0 \\
& & \downarrow f & & \downarrow h & & \downarrow g \\
0 & \to & A' & \xrightarrow{i'} & C' & \xrightarrow{\pi'} & B' & \to & 0
\end{array}
Injective:  Let $c_1,c_2\in C$, and suppose that $h(c_1) = h(c_2)$.  Then
$$
g(\pi(c_1)) \;=\; \pi'(h(c_1)) \;=\; \pi'(h(c_2)) = g(\pi(c_2)).
$$
Since $g$ is injective, it follows that $\pi(c_1) = \pi(c_2)$, so $\pi(c_1 - c_2) = 0$.  It follows that $c_1 - c_2 = i(a)$ for some $a\in A$.  Then $i'(f(a)) = h(i(a)) = h(c_1-c_2) = 0$.  Since $i'$ and $f$ are injective, it follow that $a = 0$, so $c_1 = c_2$.
Surjective:  Let $c'\in C'$.  Since $g \circ \pi$ is surjective, there exists a $c\in C$ so that $g(\pi(b)) = \pi'(c')$.  Then $\pi'(h(b)) = \pi' (c')$, so $h(b) - c' \in \ker(\pi')$.  It follows that $h(c) - c' = i'(a')$ for some $a' \in A'$.  Since $f$ is surjective, there exists an $a\in A$ so that $f(a) = a'$.  Then
$$
h(c-i(a)) \;=\; h(c) - h(i(a)) \;=\; h(c) - i'(f(a)) \;=\; h(c) - i'(a') \;=\; h(c) - (h(c) - c') \;=\; c'.
$$