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$\newcommand{\RR}{\mathbb{R}}$The other answers are completely general, but there is simpler way if we use the (given) hypothesis that all the action is taking place in $\RR^3$. So, suppose that $T$ is a finite triangulation contained in $\RR^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is:

• the space $|T|$ is a manifold and
• the boundary $\partial\\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere).

EDIT x2 - Here is a discussion of the "bit of work". Suppose that $C = |T|$ is a manifold and $S = \partial C$ is a two-sphere. Then by Alexander's theorem $S$ bounds a ball $B \subset \RR^3$. We need to show that $B$ is equal to $C$. By the Jordan–Brouwer separation theorem there are two possibilities. Either $S$ separates $B$ from $C$ or it does not.

In the separating case form $M = B \cup C$. Thus $M$ is a compact three-manifold without boundary, embedded in $\RR^3$. This contradicts invariance of domain. See Corollary 2B.4 of Hatcher's Algebraic topology.

Suppose instead that $B$ and $C$ are on the same side of $S$. It follows that $C \subset B$. We must prove the opposite inclusion. Suppose that $p$ is a point of $B$. Let $r$ be any point of $S$ that is as close as possible to $p$. Let $I = [p,r]$ be the line segment from $p$ to $r$. So $I \subset B$. Order the points of $I$, from $p$, to $r$. Note that $r \in S$ so $r \in C$. Let $J = I \cap C$. Let $q = \inf J$. Since $C$ is a closed subset of $\RR^3$ the set $J$ is closed and thus $q$ lies in $C$. Since $C$ is a manifold there is a neighborhood $V \subset C$ so that $q \in V$. Show that $V \cap I$ is a neighborhood of $q$ in $I$. Thus $q = p$ and we are done.

I don't see how to do the second half with "invariance of domain" directly. I'll also remark that the "bit of work" has now been greatly expanded, and perhaps unnecessarily so. One is supposed to do this sort of thing once and then not worry about it ever again.

$\newcommand{\RR}{\mathbb{R}}$The other answers are completely general, but there is simpler way if we use the (given) hypothesis that all the action is taking place in $\RR^3$. So, suppose that $T$ is a finite triangulation contained in $\RR^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is the following.

• The space $|T|$ is a manifold and
• the boundary $\partial\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere).

EDIT x2 - Here is a discussion of the "bit of work". Suppose that $C = |T|$ is a manifold and $S = \partial C$ is a two-sphere. Then by Alexander's theorem $S$ bounds a ball $B \subset \RR^3$. We need to show that $B$ is equal to $C$. By the Jordan–Brouwer separation theorem there are two possibilities. Either $S$ separates $B$ from $C$ or it does not.

In the separating case form $M = B \cup C$. Thus $M$ is a compact three-manifold without boundary, embedded in $\RR^3$. This contradicts invariance of domain. See Corollary 2B.4 of Hatcher's Algebraic topology.

Suppose instead that $B$ and $C$ are on the same side of $S$. It follows that $C \subset B$. We must prove the opposite inclusion. Suppose that $p$ is a point of $B$. Let $r$ be any point of $S$ that is as close as possible to $p$. Let $I = [p,r]$ be the line segment from $p$ to $r$. So $I \subset B$. Order the points of $I$, from $p$, to $r$. Note that $r \in S$ so $r \in C$. Let $J = I \cap C$. Let $q = \inf J$. Since $C$ is a closed subset of $\RR^3$ the set $J$ is closed and thus $q$ lies in $C$. Since $C$ is a manifold there is a neighborhood $V \subset C$ so that $q \in V$. Show that $V \cap I$ is a neighborhood of $q$ in $I$. Thus $q = p$ and we are done.

I don't see how to do the second half with "invariance of domain" directly. I'll also remark that the "bit of work" has now been greatly expanded, and perhaps unnecessarily so. One is supposed to do this sort of thing once and then not worry about it ever again.

The other answers work, but are more than is necessary. Suppose that $T$ is a finite triangulation in $R^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is:

• the space $|T|$ is a manifold and
• the boundary $\partial\\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere).

Edit - here is a very brief discussion of the extra work needed. Suppose that $|T|$ is a manifold and $S = \partial\\,|T|$ is a two-sphere. Then by Alexander's theorem $S$ bounds a ball $B$. You need to show that $|T|$ is equal to $B$. This can be done using the fact that $S$ has a "collar" inside of $|T|$.

$\newcommand{\RR}{\mathbb{R}}$The other answers are completely general, but there is simpler way if we use the (given) hypothesis that all the action is taking place in $\RR^3$. So, suppose that $T$ is a finite triangulation contained in $\RR^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is:

• the space $|T|$ is a manifold and
• the boundary $\partial\\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere).

EDIT x2 - Here is a discussion of the "bit of work". Suppose that $C = |T|$ is a manifold and $S = \partial C$ is a two-sphere. Then by Alexander's theorem $S$ bounds a ball $B \subset \RR^3$. We need to show that $B$ is equal to $C$. By the Jordan–Brouwer separation theorem there are two possibilities. Either $S$ separates $B$ from $C$ or it does not.

In the separating case form $M = B \cup C$. Thus $M$ is a compact three-manifold without boundary, embedded in $\RR^3$. This contradicts invariance of domain. See Corollary 2B.4 of Hatcher's Algebraic topology.

Suppose instead that $B$ and $C$ are on the same side of $S$. It follows that $C \subset B$. We must prove the opposite inclusion. Suppose that $p$ is a point of $B$. Let $r$ be any point of $S$ that is as close as possible to $p$. Let $I = [p,r]$ be the line segment from $p$ to $r$. So $I \subset B$. Order the points of $I$, from $p$, to $r$. Note that $r \in S$ so $r \in C$. Let $J = I \cap C$. Let $q = \inf J$. Since $C$ is a closed subset of $\RR^3$ the set $J$ is closed and thus $q$ lies in $C$. Since $C$ is a manifold there is a neighborhood $V \subset C$ so that $q \in V$. Show that $V \cap I$ is a neighborhood of $q$ in $I$. Thus $q = p$ and we are done.

I don't see how to do the second half with "invariance of domain" directly. I'll also remark that the "bit of work" has now been greatly expanded, and perhaps unnecessarily so. One is supposed to do this sort of thing once and then not worry about it ever again.

The other answers work, but are more than is necessary. Suppose that $T$ is a finite triangulation in $R^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is:

• the space $|T|$ is a manifold and
• the boundary $\partial\\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should bs a sphere or a disk) and the boundary (it should be a sphere).

The other answers work, but are more than is necessary. Suppose that $T$ is a finite triangulation in $R^3$. Let $|T|$ be the underlying space for $T$. A necessary and sufficient condition for $|T|$ to be a three-ball is:

• the space $|T|$ is a manifold and
• the boundary $\partial\\,|T|$ is a two-sphere.

These are clearly necessary. That they suffice is a theorem of Alexander, plus a bit of work. Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it. It is more correct to think in terms of recognizing surfaces. Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere).

Edit - here is a very brief discussion of the extra work needed. Suppose that $|T|$ is a manifold and $S = \partial\\,|T|$ is a two-sphere. Then by Alexander's theorem $S$ bounds a ball $B$. You need to show that $|T|$ is equal to $B$. This can be done using the fact that $S$ has a "collar" inside of $|T|$.