7 Clarity!

Hi, Again: I am trying to understand an argument I wrote down a long time ago, to show that a given element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.

EDIT: My previous layout may have been unclear:

The SETUP is that of a genus-3 orientable surface $S_3$ trivially-embedded ( in the sense that it bounds a 3-handlebody) in the 4-sphere $S^4$, and we have a composition $C_4C_3C_4^{-1}$ of Dehn twists defined in the embedded $S_3$ about respective curves $c_3,c_4$ in $S_3$ ; where these two (basic) curves are defined in the paragraph below. We then want to show that the composition of twists on the embedded $S_3$ extends to the whole ambient $S^4$, i.e., we want to show that there is a self-diffeomorphism Phi:$S^4$-->$S^4$ , which extends the composition of the twists $C_4C_3C_4^{-1}$ defined on $S_3$, i.e., $Phi|_{S_3}$ (the restriction of Phi to $S_3$), is the map $C_4C_3C_4^{-1}$,

So consider the orientable, genus-3 surface $S_3$, trivially-embedded in the 4-sphere $S^4$ and two basis curves $c_3$ and $c_4$, which meet at exactly one point (i.e., their algebraic intersection)intersection number is 1), e.g., a meridian and a parallel in the same sub-torus of $S_3$.

Then the map h defined on (the embedded copy of) $S_3$ which we want to be embedded have extended into $S^4$ is defined by :

h:=$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.

The argument showing that h extends had to see with a sort of surgery on $S^4$ , which where we get decompose $S^4$ by joining (identifying) identifying/gluing) a $B^3 \times S^1$ (a 3-ball and a 1-sphere)and an $S^2\times D^2$ (2-sphere and a 2-disk), along their common boundary $S^2\times S^1$ (no problem so far).

NOw######, this is how the original argument went:

We somehow consider a solid torus $D^2\times S^1$ trivially-embedded in the 3-ball $B^3$ --which -where "trivially" means that its complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration)Fibration, so that a small disk $d^2$in an evenly-covered neighborhood lifts to a $d^2\times S^1$; same for the complement of the $d^2$), and we consider a tubular neighborhood U of the solid torus $D^2\times S^1$, given by:

U=$D^2\times S^1 \times S^1$ of the solid torus.

The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a topological $D^2\times S^1\times S^1$ , and the boundary $\partial E^4$ of this $E^4$ is itself a 3-torus, i.e. $\partial E^4 = S^1\times S^1\times S^1$.

We then use a result by Montesinos, whereby maps defined on the bounding $\partial E^4$, which induce certain maps in homology, extend to the whole complement $E^4$, and then we construct , I think, a map from $S^4$ to itself, whose restriction to $S_3$ \partial E^4$induce a map of the Montesinos type. But I cannot make sense of why/how this argument shows that the given map extends. Any Ideas? Thanks in Advance. 6 Rewrote for clarity Hi, Again: I am trying to understand an argument I wrote down a long time ago, to show that a given element of$M_3$(mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of$S_3$in$S^4$extends (into the ambient$S^4$); by trivially embedded we mean that this$S_3$bounds a topological$H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached. EDIT: My previous layout may have been unclear: The SETUP is that of a genus-3 orientable surface$S_3$trivially-embedded ( in the sense that it bounds a 3-handlebody) in the 4-sphere$S^4$, and we have a composition$C_4C_3C_4^{-1}$of Dehn twists defined in the embedded$S_3$about respective curves$c_3,c_4$in$S_3$; where these two (basic) curves are defined in the paragraph below. We then want to show that the composition of twists on the embedded$S_3$extends to the whole ambient$S^4$, i.e., we want to show that there is a self-diffeomorphism Phi:$S^4$-->$S^4$, which extends the composition of the twists$C_4C_3C_4^{-1}$defined on$S_3$, i.e.,$Phi|_{S_3}$(the restriction of Phi to$S_3$), is the map$C_4C_3C_4^{-1}$, So consider the orientable, genus-3 surface$S_3$, trivially-embedded in the 4-sphere$S^4$and two basis curves$c_3$and$c_4$, which meet at exactly one point (algebraic intersection), e.g., a meridian and a parallel in the same sub-torus of$S_3$. Then the map h defined on (the embedded copy of)$S_3$to be embedded is : h:=$C_4C_3C_4^{-1}$, where$C_i$is the Dehn twist about$c_i$. The argument showing that h extends had to see with a sort of surgery on$S^4$, which we get by joining (identifying)$B^3 \times S^1$(a 3-ball and a 1-sphere)and an$S^2\times D^2$(2-sphere and a 2-disk), along their common boundary$S^2\times S^1$(no problem so far). NOw######, this is how the original argument went: We then somehow consider a solid torus$D^2\times S^1$trivially-embedded in the 3-ball$B^3$--which means that its complement in$S^3$is also a solid torus (we can show this last using, e.g., the Hopf Fibration), and we consider the a tubular neighborhood U of the solid torus$D^2\times S^1$, given by: U=$D^2\times S^1 \times S^1$of the solid torus. The complement$S^4 -U$:=$E^4$in$S^4$of this tubular neighborhood U is a topological$D^2\times S^1\times S^1$, and the boundary$\partial E^4 $of this$E^4$is itself a 3-torus, i.e.$\partial E^4 = S^1\times S^1\times S^1$. We then use a result by Montesinos, whereby maps defined on the bounding$\partial E^4$, which induce certain maps in homology, extend to the whole complement$E^4$, and then we construct , I think, a map whose restriction to$S_3$induce a map of the Montesinos type. But I cannot make sense of why/how this argument shows that the given map extends. Any Ideas? Thanks in Advance. 5 Rewrote for clarity; added 171 characters in body Hi, Again: I am trying to understand an argument I wrote down a long time ago, to show that a given element of$M_3$(mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of$S_3$in$S^4$extends (into the ambient$S^4$); by trivially embedded we mean that this$S_3$bounds a topological$H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached. EDIT: My previous layout may have been unclear: The SETUP is that of an a genus-3 orientable surface$S_3$trivially-embedded ( in the sense that it bounds a 3-handlebody) in the 4-sphere$S^4$, and we have a composition$C_4C_3C_4^{-1}$of Dehn twists defined in the embedded$S_3$about respective curves$c_3,c_4$in$S_3$; where these two (basic) curves are defined in the paragraph below. We then want to show that the composition of twists on the embedded$S_3$extends to the whole ambient$S^4$, i.e., we want to show that there is an automorphism a self-diffeomorphism Phi:$S^4$-->$S^4$, which extends the composition of the twists ,$C_4C_3C_4^{-1}$defined on$S_3$, i.e.,$Phi|_{S_3}$(the restriction of Phi to$S_3$), is$C_4C_3C_4^{-1}$,is the map$C_4C_3C_4^{-1}$, So consider the orientable, genus-3 surface$S_3$, trivially-embedded in the 4-sphere$S^4$and two basis curves$c_3$and$c_4$, which meet at exactly one point (algebraic intersection), e.g., a meridian and a parallel in the same sub-torus of$S_3$. Then the map h defined on (the embedded copy of)$S_3$to be embedded is : h:=$C_4C_3C_4^{-1}$, where$C_i$is the Dehn twist about$c_i$. The argument showing that h extends had to see with a sort of surgery on$S^4$, which we get by joining (identifying)$B^3 \times S^1$(a 3-ball and a 1-sphere)and an$S^2\times D^2$(2-sphere and a 2-disk), along their common boundary$S^2\times S^1$(no problem so far). NOw######: We then somehow consider a solid torus$D^2\times S^1$trivially-embedded in$B^3$--which means that its complement in$S^3$is also a solid torus (we can show this last using, e.g., the Hopf Fibration), and we consider the tubular neighborhood U of the solid torus$D^2\times S^1$, given by: U=$D^2\times S^1 \times S^1$of the solid torus. The complement$S^4 -U$:=$E^4$in$S^4$of this tubular neighborhood U is a topological$D^2\times S^1\times S^1$, and the boundary$\partial E^4 $of this$E^4$is itself a 3-torus, i.e.$\partial E^4 = S^1\times S^1\times S^1$. We then use a result by Montesinos, whereby maps defined on the bounding$\partial E^4$, which induce certain maps in homology, extend to the whole complement$E^4$, and then we construct , I think, a map whose restriction to$S_3\$ induce a map of the Montesinos type.

But I cannot make sense of why/how this argument shows that the given map extends.

Any Ideas?