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You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$.

You are right that there are two possible framings for every 2-handle of $W$, but there is some redundancy and cohomology controls that. The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class (for a proof, see this paper of Hambleton, Kreck, and Teichner).

That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember.

In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable.

When there are no 1-handles, the 2-complex is a bouquet of spheres and hence $H^2(X)$ is a product of one $\mathbb Z/_{2\mathbb Z}$ for each 2-handle.

To answer your question 1: I don't think there is a way in general to distinguish from the two thickenings. To distinguish between them you need to fix a trivialization of the tangent bundle on the 1-skeleton, which is not unique. Think for instance about the simple case where you do a 1-surgery to $S^3 \times S^1$ along the circle $\{p\} \times S^1$. There are two possible choices, there is no way to choose between them, and in fact they both yield the same manifold $S^4$.

You are right that there are two possible ways to perform a 1-surgery, but I don't think that there is a way to choose between them. Think for instance about the simple case where you do a 1-surgery to $S^3 \times S^1$ along the circle $\{p\} \times S^1$. There are two possible choices, but there is no way to choose between them, since there is a self-diffeomorphism of $S^3 \times S^1$ that sends one to the other. They both yield the same manifold $S^4$.

However, you can still resolve this ambiguity in an elegant and simple way by using cohomology and Stiefel - Whitney classes.

You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$. The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class (for a proof, see this paper of Hambleton, Kreck, and Teichner).

That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember.

In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable.

When there are no 1-handles, the 2-complex is a bouquet of spheres and hence $H^2(X)$ is a product of one $\mathbb Z/_{2\mathbb Z}$ for each 2-handle. This is the simple case.

You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$.

You are right that there are two possible framings for every 2-handle of $W$, but there is some redundancy and cohomology controls that. The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class (for a proof, see this paper of Hambleton, Kreck, and Teichner).

That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember.

In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable.

You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$.

You are right that there are two possible framings for every 2-handle of $W$, but there is some redundancy and cohomology controls that. The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class (for a proof, see this paper of Hambleton, Kreck, and Teichner).

That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember.

In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable.

When there are no 1-handles, the 2-complex is a bouquet of spheres and hence $H^2(X)$ is a product of one $\mathbb Z/_{2\mathbb Z}$ for each 2-handle.

To answer your question 1: I don't think there is a way in general to distinguish from the two thickenings. To distinguish between them you need to fix a trivialization of the tangent bundle on the 1-skeleton, which is not unique. Think for instance about the simple case where you do a 1-surgery to $S^3 \times S^1$ along the circle $\{p\} \times S^1$. There are two possible choices, there is no way to choose between them, and in fact they both yield the same manifold $S^4$.

You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$.

You are right that there are two possible framings for every 2-handle of $W$, but there is some redundancy and cohomology controls that. The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class. That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember.

In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable.

You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$.

You are right that there are two possible framings for every 2-handle of $W$, but there is some redundancy and cohomology controls that. The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class (for a proof, see this paper of Hambleton, Kreck, and Teichner). 

That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember.

In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable.