diff --git a/noncoordinate_polynomial_fibre_v2.txt b/noncoordinate_polynomial_fibre_v2.txt
index 34f2e12..aff1ab7 100644
--- a/noncoordinate_polynomial_fibre_v2.txt
+++ b/noncoordinate_polynomial_fibre_v2.txt
@@ -48,8 +48,8 @@ automorphism of CC^n, and f is its first component.
 
 
 For instance z_1 is a coordinate, and so is z_1+z_2^2, because
-(z_1,z_2)mapsto(z_1+z_2^2,z_2) has the polynomial inverse
-(w_1,w_2)mapsto(w_1-w_2^2,w_2).
+(z_1,z_2)|->(z_1+z_2^2,z_2) has the polynomial inverse
+(w_1,w_2)|->(w_1-w_2^2,w_2).
 
 
 If fin R_n is a coordinate, then R_n/(f)congCC^{[n-1]}.
@@ -82,7 +82,7 @@ n=4 (Section~sec:fibre) and, by adjoining variables, for all nge4
 # The polynomial
 
 The starting point is the cusp. Since (u^3)^2+(-u^2)^3=0, the map
-umapsto(u^3,-u^2) parametrizes the curve x^2+y^3=0. We perturb both entries
+u|->(u^3,-u^2) parametrizes the curve x^2+y^3=0. We perturb both entries
 by independent multiples of a new variable h. In R=CC[h,u,v,w] put
 
   x=u^3+hv, y=-u^2+hw,
@@ -125,7 +125,7 @@ produces a third coordinate on that fibre, together with its polynomial inverse.
 The polynomial F of (eq:F) satisfies
 [
   R/(F)congCC^{[3]}
-  qquadandqquad
+   and 
   grad F(2,0,-tfrac12,tfrac12)=0.
 ]
 Consequently F is not a coordinate of R.
@@ -166,7 +166,7 @@ B need not be a domain.
 Let B be a commutative ring and let h,x,y,sin B satisfy x^2+y^3=hs and
 (h,y)=B. Then
 [
-  C=B[U,V,W]/(U^3+hV-x, -U^2+hW-y, S(h,U,V,W)-s)cong B[T].
+  C=B[U,V,W]/(U^3+hV-x, -U^2+hW-y, S(h,U,V,W)-s)~= B[T].
 ]
 
 
@@ -201,8 +201,8 @@ while hW-a=h(W-W_0) and y^2W-b=y^2(W-W_0). So in A[W] the ideal
 It remains to identify A with B[T]. The B-algebra maps Ato B[T] and
 B[T]to A given by
 [
-  Umapsto hT-beta x, Gmapsto yT+alpha x,
-  Tmapstoalpha U+beta G
+  U|-> hT-beta x, G|-> yT+alpha x,
+  T|->alpha U+beta G
 ]
 are inverse. Indeed, in A the relation hG-yU=x and alpha h+beta y=1 give
 [
@@ -260,7 +260,7 @@ s=S(h,u,v,w). Under this substitution the relation x^2+y^3-sh becomes
 identically zero by (eq:cusp), and the relation h-1-p(x,y,s) becomes the
 equation F=0 of (eq:F). Therefore
 
-  R/(F)cong Ccong B[T]congCC[X,Y,T].
+  R/(F)~= Ccong B[T]congCC[X,Y,T].
 
 This proves the first assertion of Theorem~thm:main: X, Y and T are
 polynomial coordinates on the zero fibre. Appendix~app:coordinates records
@@ -284,7 +284,7 @@ partial derivatives of F vanish at P.
 
 To see why this excludes a coordinate, suppose F were the first component of a
 polynomial automorphism Phi=(F,g_2,g_3,g_4) of CC^4 with polynomial
-inverse Psi. The chain rule gives JPhi(P)cdot JPsi(Phi(P))=I, so
+inverse Psi. The chain rule gives JPhi(P)· JPsi(Phi(P))=I, so
 JPhi(P) is invertible. But its first row is grad F(P)=0, so the matrix is
 not invertible. This contradiction proves that F is not a coordinate.