Can $H_1(\partial \lvert \mathcal R \rvert ) \to H_1(\lvert \mathcal R \rvert)$ be non injective?

Suppose we are in the following situation:

• we have (the image of a) closed simple curve $C$ in the plane;
• we have a simplicial complex $\mathcal R$ over a finite set of vertices such that the boundary of the realization of $\mathcal R$ is precisely $C$
• we know that $H_1(\mathcal R) \ne 0$.

Can the map $H_1(C) \to H_1(\mathcal R)$ be non injective? Equivalently, can $H_2(\mathcal R, C) \to H_1(C)$ be non zero?

Solution 1:

Yes. For example let $T$ be a triangulation of the torus and let $R$ be obtained from $T$ by removing the interior of one 2-simplex. Then the boundary of the realization of $T$ is given by $C$, the three edges of a triangle, and this can certainly be embedded as a simple closed curve in the plane. A computation shows that the inclusion $C \to R$ induces the zero map on $H_1$. Alternatively, $H_*(R,C)$ is isomorphic to $\tilde{H}_*(T)$, and thus $H_2(R,C)\cong \mathbb{Z}$. On the other hand $H_2(R)=0$, so $H_2(R,C)\to H_1(C)$ is nonzero.