l3lab/ntp-mathlib huggingface dataset:
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    

Proof state:
case mk
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d :
  (CokernelCofork.ofπ (biprod.desc f (-g))
        (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0)).pt ⟶
    R
hd :
  Cofork.π
        (CokernelCofork.ofπ (biprod.desc f (-g))
          (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0)) ≫
      d =
    u
⊢ e = 0

Next tactic:
dsimp at d
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; 

Proof state:
case mk
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd :
  Cofork.π
        (CokernelCofork.ofπ (biprod.desc f (-g))
          (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0)) ≫
      d =
    u
⊢ e = 0

Next tactic:
dsimp at hd
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    

Proof state:
case mk
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
⊢ e = 0

Next tactic:
have : f ≫ d = 0
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    

Proof state:
case this
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
⊢ f ≫ d = 0
case mk
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this✝ :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
this : f ≫ d = 0
⊢ e = 0

Next tactic:
calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by rw [Category.assoc, hd]
      _ = 0 := biprod.inl_desc _ _
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by 

Proof state:
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
⊢ f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d

Next tactic:
rw [biprod.inl_desc]
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by 

Proof state:
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
⊢ (biprod.inl ≫ biprod.desc f (-g)) ≫ d = biprod.inl ≫ u

Next tactic:
rw [Category.assoc, hd]
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by rw [Category.assoc, hd]
      _ = 0 := biprod.inl_desc _ _
    -- But f is an epimorphism, so d = 0...
    

Proof state:
case mk
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this✝ :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
this : f ≫ d = 0
⊢ e = 0

Next tactic:
have : d = 0 := (cancel_epi f).1 (by simpa)
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by rw [Category.assoc, hd]
      _ = 0 := biprod.inl_desc _ _
    -- But f is an epimorphism, so d = 0...
    have : d = 0 := (cancel_epi f).1 (by 

Proof state:
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this✝ :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
this : f ≫ d = 0
⊢ f ≫ d = f ≫ 0

Next tactic:
simpa
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by rw [Category.assoc, hd]
      _ = 0 := biprod.inl_desc _ _
    -- But f is an epimorphism, so d = 0...
    have : d = 0 := (cancel_epi f).1 (by simpa)
    -- ...or, in other words, e = 0.
    

Proof state:
case mk
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this✝¹ :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
this✝ : f ≫ d = 0
this : d = 0
⊢ e = 0

Next tactic:
calc
      e = biprod.inr ≫ biprod.desc (0 : X ⟶ R) e := by rw [biprod.inr_desc]
      _ = biprod.inr ≫ biprod.desc f (-g) ≫ d := by rw [← hd]
      _ = biprod.inr ≫ biprod.desc f (-g) ≫ 0 := by rw [this]
      _ = (biprod.inr ≫ biprod.desc f (-g)) ≫ 0 := by rw [← Category.assoc]
      _ = 0 := HasZeroMorphisms.comp_zero _ _
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by rw [Category.assoc, hd]
      _ = 0 := biprod.inl_desc _ _
    -- But f is an epimorphism, so d = 0...
    have : d = 0 := (cancel_epi f).1 (by simpa)
    -- ...or, in other words, e = 0.
    calc
      e = biprod.inr ≫ biprod.desc (0 : X ⟶ R) e := by 

Proof state:
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this✝¹ :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
this✝ : f ≫ d = 0
this : d = 0
⊢ e = biprod.inr ≫ biprod.desc 0 e

Next tactic:
rw [biprod.inr_desc]
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by rw [Category.assoc, hd]
      _ = 0 := biprod.inl_desc _ _
    -- But f is an epimorphism, so d = 0...
    have : d = 0 := (cancel_epi f).1 (by simpa)
    -- ...or, in other words, e = 0.
    calc
      e = biprod.inr ≫ biprod.desc (0 : X ⟶ R) e := by rw [biprod.inr_desc]
      _ = biprod.inr ≫ biprod.desc f (-g) ≫ d := by 

Proof state:
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this✝¹ :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
this✝ : f ≫ d = 0
this : d = 0
⊢ biprod.inr ≫ biprod.desc 0 e = biprod.inr ≫ biprod.desc f (-g) ≫ d

Next tactic:
rw [← hd]
/-- In an abelian category, the pullback of an epimorphism is an epimorphism.
    Proof from [aluffi2016, IX.2.3], cf. [borceux-vol2, 1.7.6] -/
instance epi_pullback_of_epi_f [Epi f] : Epi (pullback.snd : pullback f g ⟶ Y) :=
  -- It will suffice to consider some morphism e : Y ⟶ R such that
    -- pullback.snd ≫ e = 0 and show that e = 0.
    epi_of_cancel_zero _ fun {R} e h => by
    -- Consider the morphism u := (0, e) : X ⊞ Y⟶ R.
    let u := biprod.desc (0 : X ⟶ R) e
    -- The composite pullback f g ⟶ X ⊞ Y ⟶ R is zero by assumption.
    have hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0 := by simpa
    -- pullback_to_biproduct f g is a kernel of (f, -g), so (f, -g) is a
    -- cokernel of pullback_to_biproduct f g
    have :=
      epiIsCokernelOfKernel _
        (PullbackToBiproductIsKernel.isLimitPullbackToBiproduct f g)
    -- We use this fact to obtain a factorization of u through (f, -g) via some d : Z ⟶ R.
    obtain ⟨d, hd⟩ := CokernelCofork.IsColimit.desc' this u hu
    dsimp at d; dsimp at hd
    -- But then f ≫ d = 0:
    have : f ≫ d = 0;
    calc
      f ≫ d = (biprod.inl ≫ biprod.desc f (-g)) ≫ d := by rw [biprod.inl_desc]
      _ = biprod.inl ≫ u := by rw [Category.assoc, hd]
      _ = 0 := biprod.inl_desc _ _
    -- But f is an epimorphism, so d = 0...
    have : d = 0 := (cancel_epi f).1 (by simpa)
    -- ...or, in other words, e = 0.
    calc
      e = biprod.inr ≫ biprod.desc (0 : X ⟶ R) e := by rw [biprod.inr_desc]
      _ = biprod.inr ≫ biprod.desc f (-g) ≫ d := by rw [← hd]
      _ = biprod.inr ≫ biprod.desc f (-g) ≫ 0 := by 

Proof state:
C : Type u
inst✝³ : Category.{v, u} C
inst✝² : Abelian C
inst✝¹ : HasPullbacks C
W X Y Z : C
f : X ⟶ Z
g : Y ⟶ Z
inst✝ : Epi f
R : C
e : Y ⟶ R
h : pullback.snd ≫ e = 0
u : X ⊞ Y ⟶ R := biprod.desc 0 e
hu : PullbackToBiproductIsKernel.pullbackToBiproduct f g ≫ u = 0
this✝¹ :
  IsColimit
    (CokernelCofork.ofπ (biprod.desc f (-g))
      (_ : Fork.ι (PullbackToBiproductIsKernel.pullbackToBiproductFork f g) ≫ biprod.desc f (-g) = 0))
d : Z ⟶ R
hd : biprod.desc f (-g) ≫ d = biprod.desc 0 e
this✝ : f ≫ d = 0
this : d = 0
⊢ biprod.inr ≫ biprod.desc f (-g) ≫ d = biprod.inr ≫ biprod.desc f (-g) ≫ 0

Next tactic:
rw [this]