Background. Suppose that $M$ is an oriented, connected, closed manifold of dimension $d$ with fundamental class $\mu \in H_d(M;\Bbb Z)$. Let $\Delta : M \to M \times M$ be the diagonal map. Then the pushforward $\Delta_\ast\mu \in H_d(M\times M)$ is defined.
If we apply the intersection pairing to $\Delta_\ast\mu$ with itself, we obtain the self-intersection class
$$\Delta_\ast \mu \cdot \Delta_\ast \mu \in H_0(M\times M) \cong \Bbb Z .$$
If I'm not mistaken, then $\Delta_\ast \mu \cdot \Delta_\ast \mu$ is the Euler characteristic
of $M$ multiplied by a suitable generator.
I know of two proofs, but both of them use a certain amount of geometry. The first proof
interprets the intersection of the cycles representing $\Delta_\ast \mu$ (i.e., the image of the diagonal map) as the transversal self-intersection of the
diagonal with itself. The latter is supported in a tubular neighborhood of the diagonal.
Then one makes use of the tubular neighborhood theorem to interpret the that self-intersection as the transversal intersection of the zero section of tangent bundle with itself. Lastly, one uses the Poincaré–Hopf index theorem to interpret the latter as the Euler characteristic.
The other proof I know of is along the lines of Milnor and Stasheff Chapter 11. However, the latter would also seem to require the tubular neighborhood theorem to interpret the pullback $\Delta^\ast(u'')$ of the diagonal cohomology class $u'' \in H^d(M\times M)$
as the Euler class of the tangent bundle of $M$.
In summary, as far as I know, there is no proof that $\Delta_\ast \mu \cdot \Delta_\ast \mu$ is the Euler characteristic of $M$ which doesn't make use of the tubular neighborhood theorem. Note too that the intersection $\Delta_\ast \mu \cdot \Delta_\ast \mu$ does not require transversality to define—one may take the Poincaré dual of $\Delta_\ast \mu$, then apply the cup square and then apply Poincaré duality again.
My Question. Suppose now that $M$ is merely an oriented connected Poincaré duality space
of dimension $d$, with fundamental class $\mu$.
Does $\Delta_\ast \mu \cdot \Delta_\ast \mu$ coincide with the Euler characteristic of $M$?
Note that the analogue of the tubular neighborhood theorem is not generally known to hold in the Poincaré case (i.e., there may not be a Poincaré embedding of the diagonal map).
Best Answer
It follows from Poincaré duality in $M\times M$. I'll suppress gradings and be vague about signs here.
Let $x_i$ be a rational homology basis for $M$. Let $a_i$ be the cohomology basis that is dual to this basis in the linear algebra sense: $\langle a_i,x_j\rangle=\delta_{ij}$. Let $y_i$ be the homology basis that corresponds to $a_i$ by Poincaré duality. In other words, the intersection product of $x_i$ and $y_j$ is $\pm \delta_{ij}$.
Then $\Delta_\ast\mu$ is the sum of $\pm x_i\times y_i$, and it's also the sum of $\pm y_j\times x_j$, so the intersection product of $\Delta_\ast\mu$ with itself is the sum, over $i$ and $j$, of $\pm\delta_{ij}\delta_{ji}$, i.e. the sum of $\pm\delta_{ii}$. This is the Euler characteristic, if you get the signs right.