Solution
= Solution
An $N$-dimensional <stratified pseudomanifold> has a closed filtration $\varnothing=X_{-1}\subseteq X_0\subseteq\cdots\subseteq X_N=X$ with each nonempty $X_s\setminus X_{s-1}$ an $s$-dimensional <manifold>, dense top stratum, and $X_{N-1}=X_{N-2}$. Each point of an $s$-dimensional stratum has a stratum-preserving conical neighbourhood $\mathbb R^s\times cL$, where $L$ is a compact $(N-s-1)$-dimensional <stratified pseudomanifold>. The definition is inductive, starting with zero-dimensional spaces. There are no singular codimension-one strata; connectedness of links is not required. One may work in the usual triangulable category used for <Witt bordism>.
A rational <Witt space> is such a <stratified pseudomanifold> for which every stratum of odd codimension $2r+1$ has link $L$ satisfying $IH_r^{\bar m}(L;\mathbb Q)=0$. No extra condition is imposed at even codimension. A <Witt space with collared boundary> has Witt interior and Witt boundary, and a stratum-preserving collar $\partial X\times[0,\varepsilon)$ with the product stratification. An <orientation> means an <orientation> of the regular stratum, inducing the boundary <orientation> on its collar.
A three-pronged graph, with its central vertex and open edges as strata, is a <stratified space> but not a one-dimensional <pseudomanifold>: the central vertex is a singular codimension-one stratum, and deleting it leaves three local branches instead of the two branches of a one-manifold. For a non-Witt <pseudomanifold>, take the <suspension> of $T^2$. Its two vertices have codimension three and link $T^2$, whose middle group is $IH_1(T^2;\mathbb Q)=H_1(T^2;\mathbb Q)=\mathbb Q^2$. It violates the <Witt space> condition.
The rational <Witt bordism classification> in dimensions divisible by four states that, for every $k\ge1$, the middle <intersection form> induces an isomorphism
$$
\boxed{\Omega_{4k}^{\mathrm{Witt}}\xrightarrow{\ \sim\ }W(\mathbb Q),\qquad [X]\longmapsto[IH^{2k}(X;\mathbb Q),\lambda_X].}
$$
Here $W(\mathbb Q)$ is the <Witt group of a field>, so a nonsingular symmetric form represents zero precisely when it is metabolic, that is, when it has a half-dimensional totally isotropic subspace. The <Complex projective plane> gives a nonzero example: its middle group is generated by the degree-two class $h$, with $\langle h^2,[\mathbb {CP}^2]\rangle=1$. Its form is $\langle1\rangle$, of <signature> one; a metabolic form has <signature> zero. Thus $[\mathbb {CP}^2]\ne0$ in $\Omega_4^{\mathrm{Witt}}$.
For a closed oriented <Witt space> $X$ of dimension $2k+1$, take the closed cone
$$
Y=(X\times[0,1])/(X\times\{0\}),\qquad\partial Y=X\times\{1\}.
$$
Away from its vertex it is a stratified product with an interval, so all old <Witt space> conditions persist. At the new vertex the codimension is $2k+2$, which is even; hence its odd-dimensional link $X$ creates no new Witt condition. The collar $X\times(1-\varepsilon,1]$ is explicit, and the product <orientation> may be chosen to induce the given <orientation> on $X$. This is a compact oriented <Witt space with collared boundary>, proving
$$
\boxed{\Omega_{2k+1}^{\mathrm{Witt}}=0.}
$$
Now let $\dim X=2$. Its singular strata are zero-dimensional. Their links are compact one-dimensional <pseudomanifolds>, which have no singular strata and are therefore \b[nonempty finite disjoint unions of circles]. A single circle is allowed, since a chosen stratum need not be intrinsically singular. The Witt condition is automatic in dimension two, because singular strata have even codimension.
Construct the <normalization of a two-dimensional pseudomanifold> concretely. Remove disjoint small conical neighbourhoods of the finitely many zero-stratum points. The remainder is a compact oriented <surface with boundary>, with one boundary circle for each local branch. Cap every such circle with a separate disk. The result $\widetilde X$ is a closed compact oriented topological <surface>. Map each new disk homeomorphically onto its branch of the original cone, sending all the disk centres belonging to one old singular point to that point. Along the disk boundaries this agrees with the identification with the remainder, so it defines a continuous surjection $\pi:\widetilde X\to X$. It is a <homeomorphism> off these finitely many centres, in particular over the prescribed open stratum. Every fibre is a point except over a zero-stratum point, where it is the finite set of its branch centres. Thus every fibre is zero-dimensional, as required.
Each component of $\widetilde X$ bounds an oriented <handlebody>: a <surface> of genus $g$ is the boundary of a three-ball with $g$ one-handles attached. Let $H$ be the disjoint union of these handlebodies, oriented so that $\partial H=\widetilde X$. Form
$$
Z=H\cup_{\pi}(X\times[0,1]),
$$
gluing $x\in\partial H$ to $(\pi(x),0)$. The quotient is compact and Hausdorff, and its only remaining boundary is $X\times\{1\}$, with an explicit product collar. Near regular points of the gluing locus, two half-ball neighbourhoods glue to a three-ball; the induced <orientations> match after choosing the collar <orientation> in the usual boundary-gluing convention.
It remains to check the singular links, rather than merely assert that this quotient is a <Witt space>. If an old point $p$ has $r$ circle branches, the new one-dimensional stratum $\{p\}\times(0,1)$ has link the disjoint union of those $r$ circles. This has codimension two and imposes no Witt condition. At $(p,0)$, each branch supplies a half-three-ball from $H$ and another from the collar. The halves glue to a full three-ball; the different branches share only the positive collar axis. Intersecting with a small link therefore gives $r$ copies of $S^2$ identified at one common point. This is a two-dimensional <pseudomanifold> with zero middle <intersection homology>: an allowable one-cycle avoids the common point, so decomposes into cycles in the $r$ punctured spheres, each homeomorphic to $\mathbb R^2$; each such cycle bounds there. Consequently $IH_1(\bigvee_{a=1}^r S^2;\mathbb Q)=0$. This is exactly the Witt condition at the new codimension-three stratum. The boundary has the required product stratification. Hence $Z$ is a three-dimensional <Witt space with collared boundary> with $\partial Z\cong X$, proving
$$
\boxed{\Omega_2^{\mathrm{Witt}}=0.}
$$
For the final product assertion, “product” means a nontrivial product with at least two positive-dimensional factors; a single four-dimensional factor would allow the preceding $\mathbb {CP}^2$ counterexample. State the standard rational intersection-homology product fact used here: products of <Witt spaces> are Witt, their <intersection cohomology> satisfies the <Künneth theorem>, and the complementary-degree intersection pairing on elementary tensors is the product of the factor pairings, with the sign $(-1)^{\deg\beta\deg\alpha'}$ for $\alpha\otimes\beta$ and $\alpha'\otimes\beta'$. Terms whose degrees in a factor do not add to that factor's dimension pair to zero. Regrouping factors reduces total dimension four to cases $1+3$ or $2+2$.
In the $1+3$ case, write $X=A\times B$. The middle <intersection cohomology> is
$$
IH^2(X)=(IH^0(A)\otimes IH^2(B))\oplus(IH^1(A)\otimes IH^1(B)).
$$
Each summand is totally isotropic, because the degrees in $A$ are respectively $0+0$ and $1+1$, neither equal to one. They pair perfectly with one another by factor <Poincare duality>. Thus either summand has half the total dimension and is a Lagrangian for the middle symmetric form.
For dimensions $2+2$, the middle group is the direct sum
$$
(IH^0(A)\otimes IH^2(B))\oplus(IH^1(A)\otimes IH^1(B))\oplus(IH^2(A)\otimes IH^0(B)).
$$
The outside summands are isotropic and pair perfectly with one another. Each factor's degree-one pairing is a nonsingular skew-symmetric form, so choose a <Lagrangian subspace> $L_A\subset IH^1(A)$. Inside the central summand, $L_A\otimes IH^1(B)$ is isotropic and has half its dimension. It is orthogonal to both outside summands by their degrees. Therefore
$$
(IH^0(A)\otimes IH^2(B))\oplus(L_A\otimes IH^1(B))
$$
is an isotropic subspace of half the full middle dimension. In both cases the form is metabolic. The <Witt bordism classification> now proves the desired \b[$[X]=0$ in $\Omega_4^{\mathrm{Witt}}$]. This also proves the assertion for more factors, using the regrouping just described.