hcju/stacksqa huggingface dataset:
Proof of (1). Note that if $x, y \in X$ have the same image under $r_p$ then $p$ is on the line $\overline{xy}$. Consider the finite type scheme $$ T = \{(y, p) \mid y \in X \setminus \{x\},\ p \in \mathbf{P}(V),\ p \in \overline{xy}\} $$ and the morphisms $T \to X$ and $T \to \mathbf{P}(V)$ given by $(y, p) \mapsto y$ and $(y, p) \mapsto p$. Since each fibre of $T \to X$ is a line, we see that the dimension of $T$ is $\dim(X) + 1 < \dim(\mathbf{P}(V))$. Hence $T \to \mathbf{P}(V)$ is not surjective. On the other hand, consider the finite type scheme $$ T' = \{p \mid p \in \mathbf{P}(V) \setminus \{x\}, \ \overline{xp}\text{ tangent to }X\text{ at }x\} $$ Then the dimension of $T'$ is $\dim(X) < \dim(\mathbf{P}(V))$. Thus the morphism $T' \to \mathbf{P}(V)$ is not surjective either. Let $U \subset \mathbf{P}(V) \setminus X$ be nonempty open and disjoint from these images; such a $U$ exists because the images of $T$ and $T'$ in $\mathbf{P}(V)$ are constructible by Morphisms, Lemma \ref{morphisms-lemma-chevalley}. Then for $p \in U$ closed the projection $r_p|_X : X \to \mathbf{P}(W_p)$ is injective on the tangent space at $x$ and $r_p^{-1}(\{r_p(x)\}) = \{x\}$. This means that $r_p$ is unramified at $x$ (Varieties, Lemma \ref{varieties-lemma-injective-tangent-spaces-unramified}), finite by Lemma \ref{lemma-projection-generically-finite}, and $r_p^{-1}(\{r_p(x)\}) = \{x\}$ thus \'Etale Morphisms, Lemma \ref{etale-lemma-finite-unramified-one-point} applies and there is an open neighbourhood $R$ of $r_p(x)$ in $\mathbf{P}(W_p)$ such that $(r_p|_X)^{-1}(R) \to R$ is a closed immersion which proves (1). \medskip\noindent Proof of (2). In this case we still conclude that the morphism $T' \to \mathbf{P}(V)$ is not surjective. Arguing as above we conclude that for $U \subset \mathbf{P}(V)$ avoiding $X$ and the image of $T'$, the projection $r_p|_X : X \to \mathbf{P}(W_p)$ is \'etale at $x$ and finite.
Set $Y' = Y \setminus Y \cap Z$. Let $y \in Y'$, $z \in Z$ be closed points with $r_p(y) = r_p(z)$. Then $p$ is on the line $\overline{yz}$ passing through $y$ and $z$. Consider the finite type scheme $$ T = \{(y, z, p) \mid y \in Y', z \in Z, p \in \overline{yz}\} $$ and the morphism $T \to \mathbf{P}(V)$ given by $(y, z, p) \mapsto p$. Observe that $T$ is irreducible and that $\dim(T) = \dim(Y) + \dim(Z) + 1$. Hence the general fibre of $T \to \mathbf{P}(V)$ has dimension at most $\dim(Y) + \dim(Z) + 1 - n$, more precisely, there exists a nonempty open $U \subset \mathbf{P}(V) \setminus (Y \cup Z)$ over which the fibre has dimension at most $\dim(Y) + \dim(Z) + 1 - n$ (Varieties, Lemma \ref{varieties-lemma-dimension-fibres-locally-algebraic}). Let $p \in U$ be a closed point and let $F \subset T$ be the fibre of $T \to \mathbf{P}(V)$ over $p$. Then $$ (Y \cap r_p^{-1}(r_p(Z))) \setminus (Y \cap Z) $$ is the image of $F \to Y$, $(y, z, p) \mapsto y$. Again by Varieties, Lemma \ref{varieties-lemma-dimension-fibres-locally-algebraic} the closure of the image of $F \to Y$ has dimension at most $\dim(Y) + \dim(Z) + 1 - n$.
Let $B \subset \mathbf{P}$ be the set of ``bad'' points, i.e., those points $[g]$ that violate either (1) or (2). Note that $[\text{id}_V] \not \in B$ by assumption. Moreover, $B$ is closed. Hence it suffices to prove that $\dim(B) \leq \dim(\mathbf{P}) - 2$ (Lemma \ref{lemma-find-lines}). \medskip\noindent First, consider the open $G = \text{PGL}(V) \subset \mathbf{P}$ consisting of points $[g]$ such that $g : V \to V$ is invertible. Since $G$ acts doubly transitively on $\mathbf{P}(V)$ (Lemma \ref{lemma-doubly-transitive}) we see that $$ T = \{(x, y, [g]) \mid x \in X, y \in Y, [g] \in G, r_g(x) = y\} $$ is a locally trivial fibration over $X \times Y$ with fibre equal to the stabilizer of a point in $G$. Hence $T$ is a variety. Observe that the fibre of $T \to G$ over $[g]$ is $r_g(X) \cap Y$. The morphism $T \to G$ is surjective, because any translate of $X$ intersects $Y$ (note that by the assumption that $X$ and $Y$ intersect properly and that $X \cap Y \not = \emptyset$ we see that $\dim(X) + \dim(Y) \geq \dim(\mathbf{P}(V))$ and then Varieties, Lemma \ref{varieties-lemma-intersection-in-projective-space} implies all translates of $X$ intersect $Y$). Since the dimensions of fibres of a dominant morphism of varieties do not jump in codimension $1$ (Varieties, Lemma \ref{varieties-lemma-dimension-fibres-locally-algebraic}) we conclude that $B \cap G$ has codimension $\geq 2$. \medskip\noindent Next we look at the complement $Z = \mathbf{P} \setminus G$. This is an irreducible variety because the determinant is an irreducible polynomial (Lemma \ref{lemma-determinant}). Thus it suffices to prove that $B$ does not contain the generic point of $Z$. For a general point $[g] \in Z$ the cokernel $V \to \Coker(g)$ has dimension $1$, hence $U(g)$ is the complement of a point. Since $X \not = \mathbf{P}(V)$ we see that for a general $[g] \in Z$ we have $X \subset U(g)$. Moreover, the morphism $r_g|_X : X \to r_g(X)$ is finite, hence $\dim(r_g(X)) = \dim(X)$. On the other hand, for such a $g$ the image of $r_g$ is the closed subspace $H = \mathbf{P}(\Im(g)) \subset \mathbf{P}(V)$ which has codimension $1$. For general point of $Z$ we see that $H \cap Y$ has dimension $1$ less than $Y$ (compare with Varieties, Lemma \ref{varieties-lemma-exact-sequence-induction}). Thus we see that we have to show that $r_g(X)$ and $H \cap Y$ intersect properly in $H$. For a fixed choice of $H$, we can by postcomposing $g$ by an automorphism, move $r_g(X)$ by an arbitrary automorphism of $H = \mathbf{P}(\Im(g))$. Thus we can argue as above to conclude that the intersection of $H \cap Y$ with $r_g(X)$ is proper for general $g$ with given $H = \mathbf{P}(\Im(g))$. Some details omitted.
Write $\mathbf{P}^N = \mathbf{P}(V_N)$ so $\dim(V_N) = N + 1$ and set $X_N = X$. We are going to choose a sequence of projections from points \begin{align*} & r_N :  \mathbf{P}(V_N) \setminus \{p_N\} \to \mathbf{P}(V_{N - 1}), \\ & r_{N - 1} : \mathbf{P}(V_{N - 1}) \setminus \{p_{N - 1}\} \to \mathbf{P}(V_{N - 2}), \\ & \ldots,\\ & r_{n + 1} : \mathbf{P}(V_{n + 1}) \setminus \{p_{n + 1}\} \to \mathbf{P}(V_n) \end{align*} as in Section \ref{section-projection}. At each step we will choose $p_N, p_{N - 1}, \ldots, p_{n + 1}$ in a suitable Zariski open set. Pick a closed point $x \in Z \subset X$. For every $i$ pick closed points $x_{it} \in T_i \cap Z$, at least one in each irreducible component of $T_i \cap Z$. Taking the composition we obtain a morphism $$ \pi = (r_{n + 1} \circ \ldots \circ r_N)|_X : X \longrightarrow \mathbf{P}(V_n) $$ which has the following properties \begin{enumerate} \item $\pi$ is finite, \item $\pi$ is \'etale at $x$ and all $x_{it}$, \item $\pi|_Z : Z \to \pi(Z)$ is an isomorphism over an open neighbourhood of $\pi(x_{it})$, \item $T_i \cap \pi^{-1}(\pi(Z)) = (T_i \cap Z) \cup E_i$ with $E_i \subset T_i$ closed and $\dim(E_i) \leq d + d' + 1 - (n + 1) = d + d' - n$. \end{enumerate} It follows in a straightforward manner from Lemmas \ref{lemma-projection-generically-finite}, \ref{lemma-projection-generically-immersion}, and \ref{lemma-projection-injective} and induction that we can do this; observe that the last projection is from $\mathbf{P}(V_{n + 1})$ and that $\dim(V_{n + 1}) = n + 2$ which explains the inequality in (4). \medskip\noindent Let $C \subset \mathbf{P}(V_N)$ be the scheme theoretic closure of $(r_{n + 1} \circ \ldots \circ r_N)^{-1}(\pi(Z))$. Because $\pi$ is \'etale at the point $x$ of $Z$, we see that the closed subscheme $C \cap X$ contains $Z$ with multiplicity $1$ (local calculation omitted). Hence by Lemma \ref{lemma-transversal-subschemes} we conclude that $$ C \cdot X = [Z] + \sum m_j[Z_j] $$ for some subvarieties $Z_j \subset X$ of dimension $d$. Note that $$ C \cap X = \pi^{-1}(\pi(Z)) $$ set theoretically. Hence $T_i \cap Z_j \subset T_i \cap \pi^{-1}(\pi(Z)) \subset T_i \cap Z \cup E_i$. For any irreducible component of $T_i \cap Z$ contained in $E_i$ we have the desired dimension bound. Finally, let $V$ be an irreducible component of $T_i \cap Z_j$ which is contained in $T_i \cap Z$. To finish the proof it suffices to show that $V$ does not contain any of the points $x_{it}$, because then $\dim(V) < \dim(Z \cap T_i)$. To show this it suffices to show that $x_{it} \not \in Z_j$ for all $i, t, j$. \medskip\noindent Set $Z' = \pi(Z)$ and $Z'' = \pi^{-1}(Z')$, scheme theoretically. By condition (3) we can find an open $U \subset \mathbf{P}(V_n)$ containing $\pi(x_{it})$ such that $\pi^{-1}(U) \cap Z \to U \cap Z'$ is an isomorphism. In particular, $Z \to Z'$ is a local isomorphism at $x_{it}$. On the other hand, $Z'' \to Z'$ is \'etale at $x_{it}$ by condition (2). Hence the closed immersion $Z \to Z''$ is \'etale at $x_{it}$ (Morphisms, Lemma \ref{morphisms-lemma-etale-permanence}). Thus $Z = Z''$ in a Zariski neighbourhood of $x_{it}$ which proves the assertion.
If $C \cap X = \emptyset$, then we take the constant family $C' = C \times \mathbf{P}^1$. Thus we may and do assume $C \cap X \not = \emptyset$. \medskip\noindent Write $\mathbf{P}^N = \mathbf{P}(V)$ so $\dim(V) = N + 1$. Let $E = \text{End}(V)$. Let $E^\vee = \Hom(E, \mathbf{C})$. Set $\mathbf{P} = \mathbf{P}(E^\vee)$ as in Lemma \ref{lemma-make-family}. Choose a general line $\ell \subset \mathbf{P}$ passing through $\text{id}_V$. Set $C' \subset \ell \times \mathbf{P}(V)$ equal to the closed subscheme having fibre $r_g(C)$ over $[g] \in \ell$. More precisely, $C'$ is the image of $$ \ell \times C \subset \mathbf{P} \times \mathbf{P}(V) $$ under the morphism (\ref{equation-r-psi}). By Lemma \ref{lemma-make-family} this makes sense, i.e., $\ell \times C \subset U(\psi)$. The morphism $\ell \times C \to C'$ is finite and $C'_{[g]} = r_g(C)$ set theoretically for all $[g] \in \ell$. Parts (1) and (2) are clear with $0 = [\text{id}_V] \in \ell$. Part (3) follows from the fact that $r_g(C)$ and $X$ intersect properly for all $[g] \in \ell$. Part (4) follows from the fact that a general point $\infty = [g] \in \ell$ is a general point of $\mathbf{P}$ and for such as point $r_g(C) \cap T$ is proper for any closed subvariety $T$ of $\mathbf{P}(V)$. Details omitted.
Write $\beta = \sum n_i[T_i]$ for some subvarieties $T_i \subset X$ of dimension $s$. By linearity we may assume that $\alpha = [Z]$ for some irreducible closed subvariety $Z \subset X$ of dimension $r$. We will prove the lemma by induction on the maximum $e$ of the integers $$ \dim(Z \cap T_i) $$ The base case is $e = r + s - \dim(X)$. In this case $Z$ intersects $\beta$ properly and the lemma is trivial. \medskip\noindent Induction step. Assume that $e > r + s - \dim(X)$. Choose an embedding $X \subset \mathbf{P}^N$ and apply Lemma \ref{lemma-moving} to find a closed subvariety $C \subset \mathbf{P}^N$ such that $C \cdot X = [Z] + \sum m_j[Z_j]$ and such that the induction hypothesis applies to each $Z_j$. Next, apply Lemma \ref{lemma-move} to $C$, $X$, $T_i$ to find $C' \subset \mathbf{P}^N \times \mathbf{P}^1$. Let $\gamma = C' \cdot X \times \mathbf{P}^1$ viewed as a cycle on $X \times \mathbf{P}^1$. By Lemma \ref{lemma-transfer} we have $$ [Z] + \sum m_j[Z_j] = \text{pr}_{X, *}(\gamma \cdot X \times 0) $$ On the other hand the cycle $\gamma_\infty = \text{pr}_{X, *}(\gamma \cdot X \times \infty)$ is supported on $C'_\infty \cap X$ hence intersects $\beta$ transversally. Thus we see that $[Z] \sim_{rat} - \sum m_j[Z_j] + \gamma_\infty$ by Lemma \ref{lemma-rational-equivalence-and-intersection}. Since by induction each $[Z_j]$ is rationally equivalent to a cycle which properly intersects $\beta$ this finishes the proof.
All of these follow readily from the results above. \medskip\noindent For (1) it suffices to show that $\text{pr}_{X,*}( \Gamma_f \cdot \alpha \cdot \beta) = \text{pr}_{X,*}(\Gamma_f \cdot \alpha) \cdot \text{pr}_{X,*}(\Gamma_f \cdot \beta)$ for cycles $\alpha$, $\beta$ on $X \times Y$. If $\alpha$ is a cycle on $X \times Y$ which intersects $\Gamma_f$ properly, then it is easy to see that $$ \Gamma_f \cdot \alpha = \Gamma_f \cdot \text{pr}_X^*(\text{pr}_{X,*}(\Gamma_f \cdot \alpha)) $$ as cycles because $\Gamma_f$ is a graph. Thus we get the first equality in \begin{align*} \text{pr}_{X,*}(\Gamma_f \cdot \alpha \cdot \beta) & = \text{pr}_{X,*}( \Gamma_f \cdot \text{pr}_X^*(\text{pr}_{X,*}(\Gamma_f \cdot \alpha)) \cdot \beta) \\ & = \text{pr}_{X,*}(\text{pr}_X^*(\text{pr}_{X,*}(\Gamma_f \cdot \alpha)) \cdot (\Gamma_f \cdot \beta)) \\ & = \text{pr}_{X,*}(\Gamma_f \cdot \alpha) \cdot \text{pr}_{X,*}(\Gamma_f \cdot \beta) \end{align*} the last step by the projection formula in the flat case (Lemma \ref{lemma-projection-formula-flat}). \medskip\noindent If $g : Y \to Z$ then property (2) follows formally from the observation that $$ \Gamma = \text{pr}_{X \times Y}^*\Gamma_f \cdot \text{pr}_{Y \times Z}^*\Gamma_g $$ in $Z_*(X \times Y \times Z)$ where $\Gamma = \{(x, f(x), g(f(x))\}$ and maps isomorphically to $\Gamma_{g \circ f}$ in $X \times Z$. The equality follows from the scheme theoretic equality and Lemma \ref{lemma-transversal}. \medskip\noindent For (3) we use the projection formula for flat maps twice \begin{align*} f_*(\alpha \cdot pr_{X, *}(\Gamma_f \cdot pr_Y^*(\beta))) & = f_*(pr_{X, *}(pr_X^*\alpha \cdot \Gamma_f \cdot pr_Y^*(\beta))) \\ & = pr_{Y, *}(pr_X^*\alpha \cdot \Gamma_f \cdot pr_Y^*(\beta))) \\ & = pt_{Y, *}(pr_X^*\alpha \cdot \Gamma_f) \cdot \beta \\ & = f_*(\alpha) \cdot \beta \end{align*} where in the last equality we use the remark on graphs made above. This proves (3). \medskip\noindent Property (4) rests on identifying the intersection product $\Gamma_f \cdot pr_Y^*\alpha$ in the case $f$ is flat. Namely, in this case if $V \subset Y$ is a closed subvariety, then every generic point $\xi$ of the scheme $f^{-1}(V) \cong \Gamma_f \cap pr_Y^{-1}(V)$ lies over the generic point of $V$. Hence the local ring of $pr_Y^{-1}(V) = X \times V$ at $\xi$ is Cohen-Macaulay. Since $\Gamma_f \subset X \times Y$ is a regular immersion (as a morphism of smooth projective varieties) we find that $$ \Gamma_f \cdot pr_Y^*[V] = [\Gamma_f \cap pr_Y^{-1}(V)]_d $$ with $d$ the dimension of $\Gamma_f \cap pr_Y^{-1}(V)$, see Lemma \ref{lemma-multiplicity-lci-CM}. Since $\Gamma_f \cap pr_Y^{-1}(V)$ maps isomorphically to $f^{-1}(V)$ we conclude.
By Lemma \ref{lemma-reduction-diagonal} we have $$ e(X, V \cdot W, Z) = e(X \times X, \Delta \cdot V \times W, \Delta(Z)) $$ Since $\Delta : X \to X \times X$ is a regular immersion (see Lemma \ref{lemma-diagonal-regular-immersion}), we see that $e(X \times X, \Delta \cdot V \times W, \Delta(Z))$ is a positive integer by Lemma \ref{lemma-multiplicity-with-lci}.
Suppose $F$ is additive. A direct sum $z$ of $x$ and $y$ is characterized by having morphisms $i : x \to z$, $j : y \to z$, $p : z \to x$ and $q : z \to y$ such that $p \circ i = \text{id}_x$, $q \circ j = \text{id}_y$, $p \circ j = 0$, $q \circ i = 0$ and $i \circ p + j \circ q = \text{id}_z$, according to Remark \ref{remark-direct-sum}. Clearly $F(x), F(y), F(z)$ and the morphisms $F(i), F(j), F(p), F(q)$ satisfy exactly the same relations (by additivity) and we see that $F(z)$ is a direct sum of $F(x)$ and $F(y)$. Hence, $F$ transforms direct sums to direct sums. \medskip\noindent To see that $F$ transforms zero to zero, use the characterization (3) of the zero object in Lemma \ref{lemma-preadditive-zero}.
There is a canonical morphism $\Coim(f) \to y$ because $\Ker(f) \to x \to y$ is zero. The composition $\Coim(f) \to y \to \Coker(f)$ is zero, because it is the unique morphism which gives rise to the morphism $x \to y \to \Coker(f)$ which is zero (the uniqueness follows from Lemma \ref{lemma-kernel-mono} (3)). Hence $\Coim(f) \to y$ factors uniquely through $\Im(f) \to y$, which gives us the desired map.
Assume (1) and let $p : z \to z$ be as in (3). Let $x = \Ker(p)$ and $y = \Ker(1 - p)$. There are maps $x \to z$ and $y \to z$. Since $(1 - p)p = 0$ we see that $p : z \to z$ factors through $y$, hence we obtain a morphism $z \to y$. Similarly we obtain a morphism $z \to x$. We omit the verification that these four morphisms induce an isomorphism $x = y \oplus z$ as in Remark \ref{remark-direct-sum}. Thus (1) $\Rightarrow$ (3). The implication (2) $\Rightarrow$ (3) is dual. Finally, condition (3) implies (1) and (2) by Lemma \ref{lemma-additive-cat-biproduct-kernel}.
Let $X$ be an object of $\mathcal{D}$ and let $e : X \to X$ be an idempotent. The functor $$ W \longmapsto \Ker( \Mor_\mathcal{D}(W, X) \xrightarrow{e} \Mor_\mathcal{D}(W, X) ) $$ if representable if and only if $e$ has a kernel. Note that for any abelian group $A$ and idempotent endomorphism $e : A \to A$ we have $$ \Ker(e : A \to A) = \Ker(\Phi : \prod\nolimits_{n \in \mathbf{N}} A \to \prod\nolimits_{n \in \mathbf{N}} A ) $$ where $$ \Phi(a_1, a_2, a_3, \ldots) = (ea_1 + (1 - e)a_2, ea_2 + (1 - e)a_3, \ldots) $$ Moreover, $\Phi$ has the right inverse $$ \Psi(a_1, a_2, a_3, \ldots) = (a_1, (1 - e)a_1 + ea_2, (1 - e)a_2 + ea_3, \ldots). $$ Hence (1) holds. The proof of (2) is dual (using the dual definition of a Karoubian category, namely condition (2) of Lemma \ref{lemma-karoubian}).
Suppose the diagram is cartesian and $k$ is an epimorphism.  Let $u = (g, f) : w \to x \oplus y$ and let $v = (k, -h) : x \oplus y \to z$.  As $k$ is an epimorphism, $v$ is an epimorphism, too. Therefore  and by Lemma \ref{lemma-characterize-cartesian}, the sequence  $0\to w\overset{u}\to x\oplus y\overset{v}\to z\to 0$ is exact. Thus, the  diagram is cocartesian by Lemma \ref{lemma-characterize-cartesian}. Finally,  $f$ is an epimorphism by Lemma \ref{lemma-cartesian-kernel} and  Lemma \ref{lemma-characterize-injective}. This proves (1), and (2)  follows by duality.
Let $i:\Ker(g)\to y$ be the canonical injection. Let  $p:x\to\Coim(f)$ be the canonical projection. Let  $j:\Im(f)\to\Ker(g)$ be the canonical injection. \medskip\noindent Suppose (1) holds. Let $h:w\to y$ with $g\circ h=0$. There exists  $c:w\to\Ker(g)$ with $i\circ c=h$.  Let $v=x\times_{\Ker(g)}w$ with canonical projections  $k:v\to w$ and $l:v\to x$, so that $c\circ k=j\circ p\circ l$.  Then, $h\circ k=i\circ c\circ k=i\circ j\circ p\circ l=f\circ l$.  As $j\circ p$ is an epimorphism by hypothesis, $k$ is an  epimorphism by Lemma \ref{lemma-cartesian-cocartesian}. This implies (2). \medskip\noindent Suppose (2) holds. Then, $g\circ i=0$. So, there are an object  $w$, an epimorphism $k:w\to\Ker(g)$ and a morphism  $l:w\to x$ with $f\circ l=i\circ k$. It follows  $i\circ j\circ p\circ l=f\circ l=i\circ k$. Since $i$ is a  monomorphism we see that $j\circ p\circ l=k$ is an epimorphism.  So, $j$ is an epimorphisms and thus an isomorphism. This implies (1).
Suppose the first row is exact and $k$ is a monomorphism. Let  $a:\Ker(\alpha)\to\Ker(\beta)$ and  $b:\Ker(\beta)\to\Ker(\gamma)$ be the induced morphisms.  Let $h:\Ker(\alpha)\to x$, $i:\Ker(\beta)\to y$ and  $j:\Ker(\gamma)\to z$ be the canonical injections. As $j$ is  a monomorphism we have $b\circ a=0$. Let $c:s\to\Ker(\beta)$  with $b\circ c=0$. Then, $g\circ i\circ c=j\circ b\circ c=0$. By  Lemma \ref{lemma-check-exactness-fibre-product} there are an object $t$, an  epimorphism $d:t\to s$ and a morphism $e:t\to x$ with  $i\circ c\circ d=f\circ e$. Then,  $k\circ \alpha\circ e=\beta\circ f\circ e=\beta\circ i\circ c\circ d=0$.  As $k$ is a monomorphism we get $\alpha\circ e=0$. So, there exists  $m:t\to\Ker(\alpha)$ with $h\circ m=e$. It follows  $i\circ a\circ m=f\circ h\circ m=f\circ e=i\circ c\circ d$.  As $i$ is a monomorphism we get $a\circ m=c\circ d$. Thus,  Lemma \ref{lemma-check-exactness-fibre-product} implies (1), and then  (2) follows by duality.
Assume $\alpha, \gamma$ are surjective and $\delta$ is injective. We may replace $w'$ by $\Im(w' \to x')$, i.e., we may assume that $w' \to x'$ is injective. We may replace $z$ by $\Im(y \to z)$, i.e., we may assume that $y \to z$ is surjective. Then we may apply Lemma \ref{lemma-snake} to $$ \xymatrix{ & \Ker(y \to z) \ar[r] \ar[d] & y \ar[r] \ar[d] & z \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \Ker(y' \to z') \ar[r] & y' \ar[r] & z' } $$ to conclude that $\Ker(y \to z) \to \Ker(y' \to z')$ is surjective. Finally, we apply Lemma \ref{lemma-snake} to $$ \xymatrix{ & w \ar[r] \ar[d] & x \ar[r] \ar[d] & \Ker(y \to z) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & w' \ar[r] & x' \ar[r] & \Ker(y' \to z') } $$ to conclude that $x \to x'$ is surjective. This proves (1). The proof of (2) is dual to this.
Additive functors commute with direct sums by Lemma \ref{lemma-additive-additive} hence (1) implies (2) and (3). On the other hand (2) and (3) are equivalent because the composition $F(A) \oplus F(B) \to F(A \oplus B) \to F(A) \oplus F(B)$ is the identity map. Assume (2) and (3) hold. Let $f, g : A \to B$ be maps. Then $f + g$ is equal to the composition $$ A \to A \oplus A \xrightarrow{\text{diag}(f, g)} B \oplus B \to B $$ Apply the functor $F$ and consider the following diagram $$ \xymatrix{ F(A) \ar[r] \ar[rd] & F(A \oplus A) \ar[rr]_{F(\text{diag}(f, g))} & & F(B \oplus B) \ar[r] \ar[d] & F(B) \\ & F(A) \oplus F(A) \ar[u] \ar[rr]^{\text{diag}(F(f), F(g))} & & F(B) \oplus F(B) \ar[ru] } $$ We claim this is commutative. For the middle square we can verify it separately for each of the for induced maps $F(A) \to F(B)$ where it follows from the fact that $F$ is a functor (in other words this square commutes even if $F$ does not satisfy any properties beyond being a functor). For the triangle on the left, we use that $F(A \oplus A) \to F(A) \oplus F(A)$ is an isomorphism to see that it suffice to check after composition with this map and this check is trivial. Dually for the other triangle. Thus going around the bottom is equal to $F(f + g)$ and we conclude.
If $F$ is left exact, i.e., $F$ commutes with finite limits, then $F$ sends products to products, hence $F$ preserved direct sums, hence $F$ is additive by Lemma \ref{lemma-additive-functor}. On the other hand, suppose that for every short exact sequence $0 \to A \to B \to C \to 0$ the sequence $0 \to F(A) \to F(B) \to F(C)$ is exact. Let $A, B$ be two objects. Then we have a short exact sequence $$ 0 \to A \to A \oplus B \to B \to 0 $$ see for example Lemma \ref{lemma-additive-cat-biproduct-kernel}. By assumption, the lower row in the commutative diagram $$ \xymatrix{ 0 \ar[r] & F(A) \ar[d] \ar[r] & F(A) \oplus F(B) \ar[r] \ar[d] & F(B) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & F(A) \ar[r] & F(A \oplus B) \ar[r] & F(B) } $$ is exact. Hence by the snake lemma (Lemma \ref{lemma-snake}) we conclude that $F(A) \oplus F(B) \to F(A \oplus B)$ is an isomorphism. Hence $F$ is additive in this case as well. Thus for the rest of the proof we may assume $F$ is additive. \medskip\noindent Denote $f : B \to C$ a map from $B$ to $C$. Exactness of $0 \to A \to B \to C$ just means that $A = \Ker(f)$. Clearly the kernel of $f$ is the equalizer of the two maps $f$ and $0$ from $B$ to $C$. Hence if $F$ commutes with limits, then $F(\Ker(f)) = \Ker(F(f))$ which exactly means that $0 \to F(A) \to F(B) \to F(C)$ is exact. \medskip\noindent Conversely, suppose that $F$ is additive and transforms any short exact sequence $0 \to A \to B \to C \to 0$ into an exact sequence $0 \to F(A) \to F(B) \to F(C)$. Because it is additive it commutes with direct sums and hence finite products in $\mathcal{A}$. To show it commutes with finite limits it therefore suffices to show that it commutes with equalizers. But equalizers in an abelian category are the same as the kernel of the difference map, hence it suffices to show that $F$ commutes with taking kernels. Let $f : A \to B$ be a morphism. Factor $f$ as $A \to I \to B$ with $f' : A \to I$ surjective and $i : I \to B$ injective. (This is possible by the definition of an abelian category.) Then it is clear that $\Ker(f) = \Ker(f')$. Also $0 \to \Ker(f') \to A \to I \to 0$ and $0 \to I \to B \to B/I \to 0$ are short exact. By the condition imposed on $F$ we see that $0 \to F(\Ker(f')) \to F(A) \to F(I)$ and $0 \to F(I) \to F(B) \to F(B/I)$ are exact. Hence it is also the case that $F(\Ker(f'))$ is the kernel of the map $F(A) \to F(B)$, and we win. \medskip\noindent The proof of (3) is similar to the proof of (2). Statement (4) is a combination of (2) and (3).