So for those who are looking this question comes from Weibel 1.3.3. but really the question is really about how to think about exact sequences in general abelian categories.
Consider the following diagram:
where both rows are exact and the arrow $A' \to A$ is epic and the arrows $B' \to B$ and $D' \to D$ are monic. We wish to show the arrow $C' \to C$ is monic as well.
If we assume we are dealing with a category like $R-mod$ it is not overly difficult to see why this is true tracing an element from $C'$ through the diagram to show injectivity. I've used this to model the thinking for proving the four lemme in a general abelian category which is not necessarily small. Where I run into trouble is thinking about what exactness means in a categorical sense. Here is what I've thought of so far:
Consider an appropriate arrow $f:X \to C'$ such that $cf = 0$ where $c:C' \to C$. Then, by the commutativity of the diagram we have $d \gamma' f = 0$ where $\gamma':C' \to D'$ and $d:D' \to D$. Since $d$ is monic we know $\gamma' f = 0$. Thus, $f$ factors through the kernel of $\gamma'$.
Problem
Now, here is where I get unsure. Since the top row is exact we know? that the image of $\beta':B' \to C'$ is isomorphic to the kernel of $\gamma'$. What I am guessing is happening here is there an isomorphism between the object serving as the domain of the kernel for $\gamma'$ and the codomain of the image of $\beta'$. What I'd like to be able to do is back up through $\beta'$ in some way so as to take advantage of the monic nature of $b:B' \to B$ and the epic nature of $a:A' to A$ in a similar as one does when they have elements to push around. So how do I think about exactness in this more general context? And how might the next step of this argument go. Please don't give the answer to the four lemma in completion as I'd like to figure out as much as possible out on my own and I'm really struggling with the ideas in this step only I believe.
Relevant Posts:
Exact sequences and proving the five lemma
Best Answer
$\DeclareMathOperator{\im}{im}$ One rather abusive way of solving this problem is as follows. Consider $f : W \to C'$ s.t. $c \circ f = 0$. Take the least full subcategory containing $W$ and the above diagram which is closed under finite products, kernels, and cokernels. Assuming the abelian category is locally small, the full subcategory will be a small abelian category. Then, one may apply the Mitchell Embedding Theorem.
There is even a complicated way of getting around the largeness of the category based on the completeness of first-order logic and the fact that every consistent theory has a small model. If it were logically consistent that $c$ is not monic, then because the entire situation can be expressed in first-order logic, there would be a small Abelian category in which we would have $c$ not monic. But in the small case, we can apply the Mitchell Embedding Theorem. Thus, it must be logically inconsistent that $c$ is not monic - that is, there must exist some first-order logic proof that $c$ is monic. This guarantees that if you look around enough, you'll eventually find a proof that $c$ is monic in the general case.
Edit: original poster wants to avoid the embedding theorem, so I've added an explanation of how to do that.
First, let's walk through the proof in the case of $R-mod$.
Consider some $f : W \to C'$ s.t. $c \circ f = 0$. We wish to show that $f = 0$.
We have $d \circ \gamma' \circ f = \gamma \circ c \circ f = 0$. Thus, $\gamma' \circ f = 0$. Therefore, $f$ factors through $\ker(\gamma')$.
Now, consider some $x \in W$. Since $f(x) \in \ker(\gamma') = \im(\beta')$, we may take $y \in B'$ s.t. $\beta'(y) = f(x)$.
We have $\beta(b(y)) = c(\beta'(y)) = c(f(x)) = 0$. Then $b(y) \in \ker(\beta)$.
Since $b(y) \in \ker(\beta) = \im(\alpha)$, take $z \in A$ s.t. $\alpha(z) = b(y)$.
Since $a$ is epi, it is surjective. Thus, we may take $w \in A'$ s.t. $a(w) = z$.
$b(\alpha'(w)) = \alpha(a(w)) = \alpha(z) = b(y)$. Therefore, $\alpha'(w) = y$.
Since $y = \alpha'(w)$, we have $y \in \im(\alpha') = \ker(\beta')$. Therefore, $f(x) = \beta'(y) = 0$.
Then $f = 0$. Thus, $c$ is monic.
Now, let's think about how to generalise this proof to hold in an arbitrary Abelian category.
Clearly, steps 1-2 generalise flawlessly. The real challenge is to generalise steps the other steps. Consider the following revised proof, starting with step 3. The trick is that every time we introduce a new variable (w, y, and z), we replace this by forming a pullback.
We abusively write $f : W \to \ker(\gamma') = \im(\beta')$, $\beta' : B' \to \im(\beta')$. Then we may form the pullback $P = \{(x, y) : f(x) = \beta'(y)\}$ with morphisms $p_1 : P \to W$, $p_2 : P \to B'$. Since $\beta' : B' \to \im(\beta')$ is surjective (epi), so too is its pullback $p_1$. Thus, it suffices to show that $f \circ p_1 = 0$; that is, to show that $\beta' \circ p_2 = 0$.
We have $\beta \circ b \circ p_2 = c \circ \beta' \circ p_2 = c \circ f \circ p_1 = 0$. That is, $b \circ p_2$ factors through $\ker(\beta)$.
We may again abusively write $b \circ p_2 : P \to \ker(\beta) = \im(\alpha)$ and $\alpha : A \to \im(\alpha)$. Then we may form the pullback $Q = \{(p, z) : \alpha(z) = b(p_2(p))\} = \{((x, y), z) : \alpha(z) = b(y)$ and $\beta'(y) = f(x)\}$ with morphisms $q_1 : Q \to P$ and $q_2 : Q \to A$. Since $\alpha : A \to \im(\alpha)$ is epi, so too is its pullback $q_1$.
We repeat this trick one more time to produce the pullback $R = \{(q, w) : q_2(p) = a(w)\}$ and morphisms $r_1 : R \to Q$, $r_2 : R \to A$. Since $a$ is epi, so too is its pullback $r_1$.
We have $b \circ \alpha' \circ r_2 = \alpha \circ a \circ r_2 = \alpha \circ q_2 \circ r_1 = b \circ p_2 \circ q_1 \circ r_1$. Since $b$ is monic, we have $\alpha' \circ r_2 = p_2 \circ q_1 \circ r_1$.
Therefore, we have $\beta' \circ p_2 \circ q_1 \circ r_1 = \beta' \circ \alpha' \circ r_2 = 0$. Since $q_1$ and $r_1$ are epi, we have $\beta' \circ p_2 = 0$.
We discussed in step 3 that this implies $f = 0$. Then $c$ is monic.
Hopefully, my revised steps 3-8 make it clear how to deal with the problem in the general setting of Abelian categories. The only nontrivial fact we really need is that pullbacks of epis are epi in Abelian categories.