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Let $X_w$ be the presentation complex of a one-relator group $\langle x_1,\dotsc,x_n\mid w\rangle$, with $w$ cyclically reduced, i.e., $X_w=R\cup_w D$, with $R$ the rose with $n$ petals labeled $x_1,\dotsc,x_n$, $D$ the unit disk, and $w\colon\partial D=S^1\to R$ is an immersion representing the word $w$.

Let $f\colon E\to X_w$ be a reduced disk diagram in $X_w$; $E$ is a compact (connected) simply-connected planar 2-complex whose (oriented) edges are labeled with the $x_i$ so that the (cyclic) word $w$ (or $w^{-1}$) is read off when traversing the boundary of each two-cell of $E$ in such a way that for any pair of two-cells meeting in an edge labeled $x$, those occurrences of $x$ are in different places in $w$. For each such $E$, let $\hat E$ be the planar two-complex obtained by forgetting the labels and removing all valence two vertices (except perhaps one in the case where $E$ has only one two-cell and is already essentially the disk).

A theorem of Weinbaum's says that no proper subword of $w$ is in the normal closure of $w$, which has the consequence that the closure of every two-cell in $E$ (and therefore $\hat E$) is the closed two-cell. (Equivalently the dual graph has no monogons.)

Now consider a compact (connected) simply-connected planar two-complex $E'$ with no valence-two vertices (and no internal valence one-vertices, to avoid trivial counterexamples which would require the word $w$ to be non-reduced), such that the closure of every two-cell is the closed two-cell. Is there a word $w$ and a reduced disk diagram $f\colon E\to X_w$ such that $E'$ and $\hat E$ are isomorphic as two-complexes? Is Weinbaum's theorem the only restriction on the shape of a reduced disk diagram in a one-relator group?

Obviously the answer is no but I cannot prove anything.

Let $X_w$ be the presentation complex of a one-relator group $\langle x_1,\dotsc,x_n\mid w\rangle$, with $w$ cyclically reduced, i.e., $X_w=R\cup_w D$, with $R$ the rose with $n$ petals labeled $x_1,\dotsc,x_n$, $D$ the unit disk, and $w\colon\partial D=S^1\to R$ is an immersion representing the word $w$.

Let $f\colon E\to X_w$ be a reduced disk diagram in $X_w$; $E$ is a compact (connected) simply-connected planar 2-complex whose (oriented) edges are labeled with the $x_i$ so that the (cyclic) word $w$ (or $w^{-1}$) is read off when traversing the boundary of each two-cell of $E$ in such a way that for any pair of two-cells meeting in an edge labeled $x$, those occurrences of $x$ are in different places in $w$, and don't differ by a rotation when $w$ is a proper power. For each such $E$, let $\hat E$ be the planar two-complex obtained by forgetting the labels and removing all valence two vertices (except perhaps one in the case where $E$ has only one two-cell and is already essentially the disk).

A theorem of Weinbaum's says that no proper subword of $w$ is in the normal closure of $w$, which has the consequence that the closure of every two-cell in $E$ (and therefore $\hat E$) is the closed two-cell. (Equivalently the dual graph has no monogons.)

Now consider a compact (connected) simply-connected planar two-complex $E'$ with no valence-two vertices (and no internal valence one-vertices, to avoid trivial counterexamples which would require the word $w$ to be non-reduced), such that the closure of every two-cell is the closed two-cell. Is there a word $w$ and a reduced disk diagram $f\colon E\to X_w$ such that $E'$ and $\hat E$ are isomorphic as two-complexes? Is Weinbaum's theorem the only restriction on the shape of a reduced disk diagram in a one-relator group?

Obviously the answer is no but I cannot prove anything.

Let $X_w$ be the presentation complex of a one-relator group $\langle x_1,\dotsc,x_n\mid w\rangle$, with $w$ cyclically reduced, i.e., $X_w=R\cup_w D$, with $R$ the rose with $n$ petals labeled $x_1,\dotsc,x_n$, $D$ the unit disk, and $w\colon\partial D=S^1\to R$ is an immersion representing the word $w$.

Let $f\colon E\to X_w$ be a reduced disk diagram in $X_w$; $E$ is a compact (connected) simply-connected planar 2-complex whose (oriented) edges are labeled with the $x_i$ so that the (cyclic) word $w$ (or $w^{-1}$) is read off when traversing the boundary of each two-cell of $E$ in such a way that for any pair of two-cells meeting in an edge labeled $x$, those occurrences of $x$ are in different places in $w$. For each such $E$, let $\hat E$ be the planar two-complex obtained by forgetting the labels and removing all valence two vertices (except perhaps one in the case where $E$ has only one two-cell and is already essentially the disk).

A theorem of Weinbaum's says that no proper subword of $w$ is in the normal closure of $w$, which has the consequence that the closure of every two-cell in $E$ (and therefore $\hat E$) is the closed two-cell. (Equivalently the dual graph has no monogons.)

Now consider a compact (connected) simply-connected planar two-complex $E'$ with no valence-two vertices (and no internal valence one-vertices, to avoid trivial counterexamples which would require the word $w$ to be non-reduced), such that the closure of every two-cell is the closed two-cell. Is there a word $w$ and a reduced disk diagram $f\colon E\to X_w$ such that $E'$ and $\hat E$ are isomorphic as two-complexes? Is Weinbaum's theorem the only restriction on the shape of a reduced disk diagram in a one-relator group?

Obviously the answer is no but I cannot prove anything.

Let $X_w$ be the presentation complex of a one-relator group $\langle x_1,\dotsc,x_n\mid w\rangle$, with $w$ cyclically reduced, i.e., $X_w=R\cup_w D$, with $R$ the rose with $n$ petals labeled $x_1,\dotsc,x_n$, $D$ the unit disk, and $w\colon\partial D=S^1\to R$ is an immersion representing the word $w$.

Let $f\colon E\to X_w$ be a reduced disk diagram in $X_w$; $E$ is a compact (connected) simply-connected planar 2-complex whose (oriented) edges are labeled with the $x_i$ so that the (cyclic) word $w$ (or $w^{-1}$) is read off when traversing the boundary of each two-cell of $E$ in such a way that for any pair of two-cells meeting in an edge labeled $x$, those occurrences of $x$ are in different places in $w$. For each such $E$, let $\hat E$ be the planar two-complex obtained by forgetting the labels and removing all valence two vertices (except perhaps one in the case where $E$ has only one two-cell and is already essentially the disk).

A theorem of Weinbaum's says that no proper subword of $w$ is in the normal closure of $w$, which has the consequence that the closure of every two-cell in $E$ (and therefore $\hat E$) is the closed two-cell. (Equivalently the dual graph has no monogons.)

Now consider a compact (connected) simply-connected planar two-complex $E'$ with no valence-two vertices (and no internal valence one-vertices, to avoid trivial counterexamples which would require the word $w$ to be non-reduced), such that the closure of every two-cell is the closed two-cell. Is there a word $w$ and a reduced disk diagram $f\colon E\to X_w$ such that $E'$ and $\hat E$ are isomorphic as two-complexes? Is Weinbaum's theorem the only restriction on the shape of a reduced disk diagram in a one-relator group?

Obviously the answer is no but I cannot prove anything.