Prove, in this exact/semi-exact diagram, that there exists unique homomorphism $\gamma:C\to C'$ satisfying $\gamma\circ g = g'\circ \beta$

Question

Just to remember, an exact sequence is when $Im \ f=Ker \ g$, and semi-exact is when $Im \ f \subset Ker \ g$.

Obviously this exercise requires us to use the exactness of everything in the sequences. Since the lower row is semi exact, then for $c\in Im \ g$, there exists $b'\in B'$ such that $g'(b') = c$. I think that now I must use commutativity of this square to come with some relation. I know that from this part, the homomorphism $\gamma$ must come out some way, and I must show it as being uniquely determined. By the 'determined' we know that it must come naturally, I guess. But I'm having difficulties guessing how.

Answer 1

FYI, I've never seen the terminology "semi-exact" before, and would usually say that the lower row is a "complex." Also, note that your sentence about semi-exactness of the lower row doesn't use semi-exactness at all: you're simply stating what it means for $c\in C'$ to be in the image of $g'$.

That said, let's see first how to define a $\gamma : C\to C'$ making the diagram commute. Let $c\in C$. By exactness of the upper row, $g$ is surjective, so there exists $b\in B$ such that $g(b) = c$. For $\gamma : C\to C'$ to make the diagram commute, we would necessarily need $\gamma(c) = \gamma(g(b)) = g'(\beta(b))$. This gives us a candidate for $\gamma$: for $c\in C$, choose $b\in B$ such that $g(b) = c$, and then define $$ \gamma(c) := g'(\beta(b)). $$

We need to show that this is well-defined. Suppose $b$ and $b'$ both satisfy $g(b) = g(b') = c$. Then we must show that $g'(\beta(b)) = g'(\beta(b'))$. Because $g(b) = g(b')$ and $g$ is a homomorphism of $R$-modules, it follows that $0 = g(b) - g(b') = g(b - b')$, so that $b - b'\in\ker g$, and hence by exactness $b - b'\in\mathrm{im}\, f$. So, there exists $a\in A$ such that $f(a) = b - b'$. Then \begin{align*} g'(\beta(b)) - g'(\beta(b') &= g'(\beta(b) - \beta(b'))\\ &= g'(\beta(b - b'))\\ &= g'(\beta(f(a)))\\ &= g'(f'(\alpha(a)))\\ &= 0. \end{align*} In the first two lines above, we have used the fact that all maps appearing are homomorphisms of $R$-modules, and in the last line we have used semi-exactness of the lower row. This tells us that $g'(\beta(b)) = g'(\beta(b'))$ if $g(b) = g(b')$, so that our $\gamma$ is well-defined.

Lastly, let us show that this $\gamma$ is unique. Suppose $\gamma' : C\to C'$ is an $R$-module homomorphism satisfying $\gamma'\circ g = g'\circ\beta$. Since the upper row is exact, $g$ is surjective. Then we see that the value of $\gamma'$ on any $c\in C$ is given by $\gamma'(c) = \gamma'(g(b)) = g'(\beta(b))$ for any $b\in B$ such that $g(b) = c$. But this is exactly how we defined $\gamma$! Thus, any module homomorphism $\gamma' : C\to C'$ making the diagram commute must agree with the $\gamma$ we defined above.

Answer 2

Define $\gamma : C \to C'$ as follows. Let $c \in C$. Since the top row is exact, the map $g$ is surjective. Choose $x \in g^{-1}(c)$. Define $\gamma(c)$ to be $g'(\beta(x))$.

Obviously, it's unclear that this map is well-defined, so we have to check that any two choices $x,y \in g^{-1}(c)$ map to the same thing in $C'$. Consider the difference $x-y \in B$. Since $g(x-y)=0$, there exists $z \in A$ such that $f(z)=x-y$, by exactness. Now, following the arrows downward, we have $g'(f'(\alpha(z)))=0$, by semi-exactness of the bottom row. But commutativity of the diagram means that $0=g'(f'(\alpha(z)))=g'(\beta(f(z)))=g'(\beta(x-y))=g'(\beta(x))-g'(\beta(y))$. Hence the map $\gamma$ is well-defined and makes the diagram commute by construction.