Let us denote by $T$ the torus and $K$ the bottle of Klein. Let $f:T\to K$ be a continuous function. Prove that $f_*: H_2(T; \mathbb{Z}/2)\to H_2(K; \mathbb{Z}/2)$ is the trivial mapping.
I know that $H_2(T; \mathbb{Z}/2)\cong\mathbb{Z}/2$ and $H_2(K; \mathbb{Z}/2)\cong\mathbb{Z}/2$, but I don't know how to conclude that $f_*$ is trivial, could someone please help me or give me a hint? Thank you.
This can be done by looking at a certain commutative square:
$$\require{AMScd} \begin{CD} H_2(T)@>{f_*}>> H_2(K)\\ @VVV @VVV \\ H_2(T;\mathbb{Z}_2) @>{f_*}>> H_2(K;\mathbb{Z}_2) \end{CD}$$
This is evidently commutative since taking a chain and reducing its coefficients mod 2 commutes with taking a chain and pushing it to a chain in the Klein bottle.
The first vertical map is surjective by the universal coefficient theorem since tensoring with $\mathbb{Z}_2$ is the same as reducing coefficients mod 2. The second vertical map is 0 since $H_2(K)=0$. This means that the lower horizontal map must be 0, as claimed.