Equal Euler characteristic without homeomorphism

The torus: by [L1] the quotient of the square by the pairings of the word $ab{a}^{-1}{b}^{-1}$ is homeomorphic to $T^2=(\mathbb{R}\mathrm{/}\mathbb{Z}{)}^2$ and is a connected surface schema, so $T^2$ is a nonempty compact connected boundaryless surface, it is orientable, and its Euler characteristic is $\chi (T^2)=0$.

The Klein bottle: by [L2] the quotient of the square by the pairings of the word $aabb$ is the Klein bottle $K$ and is a connected surface schema, so $K$ is a nonempty compact connected boundaryless surface, it is nonorientable, and $\chi (K)=0$.

Orientability is a homeomorphism invariant. Let $h:M\to N$ be a homeomorphism of boundaryless $n$-manifolds and let $\nu$ be an integral orientation of $N$. For $x\in M$ the map $h$ is a homeomorphism of pairs $(M,M\setminus \{x\})\to (N,N\setminus \{h(x)\})$, so by [L4] it induces an isomorphism $h_{\ast x}:H_n(M,M\setminus \{x\};\mathbb{Z})\to H_n(N,N\setminus \{h(x)\};\mathbb{Z})$, and ${\mu }_x:=h_{\ast x}^{-1}({\nu }_{h(x)})$ generates the infinite cyclic fiber at $x$ because ${\nu }_{h(x)}$ generates its fiber and an isomorphism carries generators to generators. For continuity let $x\in M$: by [L3] there are a closed coordinate ball $K$ around $h(x)$ in $N$ and a generator $c$ of $G_K=H_n(N,N\setminus K;\mathbb{Z})$ with $\nu =\pm s_{K,c}$ on $\mathrm{int}K$; applying [L4] to the homeomorphism of pairs $(M,M\setminus h^{-1}(K))\to (N,N\setminus K)$ and then to $(M,M\setminus \{y\})\to (N,N\setminus \{h(y)\})$ for $y\in \mathrm{int}{h}^{-1}(K)$ gives $r_{K,h(y)}(c)=h_{\ast y}(r_{h^{-1}(K),y}(h_{\ast K}^{-1}(c)))$, so on the ball interior $h^{-1}(K)$ the section $\mu$ equals $\pm s_{h^{-1}(K),h_{\ast K}^{-1}(c)}$, which is a basic generator section and hence continuous. Thus $\mu$ is an integral orientation of $M$, and a homeomorphism transports orientability from its target to its source.

Conclusion. If there were a homeomorphism $h:T^2\to K$, then applying step 1.3 to $h^{-1}:K\to T^2$ and an integral orientation of $T^2$, which exists by [L1], would produce an integral orientation of $K$, contradicting that $K$ is nonorientable by [L2]. Hence $T^2$ and $K$ are not homeomorphic, while both are nonempty compact connected boundaryless surfaces with $\chi =0$ by [L1] and [L2]. Therefore equal Euler characteristic does not determine the homeomorphism class of a compact connected surface. Only the two explicit schema computations, the given orientation data and the functoriality of relative homology are used, so no choice principle is used.