induced homomorphism in homology I've read a lot about induced homomorphisms in homology, but I need to see it on some examples. Let's say we have an inclusion $i: S^0 \to D^1$ and it induces the homomorphism $i_*: H_0(S^0) \to H_0(D^1)$. I know that $H_0(S^0) = \mathbb{Z} \times \mathbb{Z}$ and $H_0(D^1) = \mathbb{Z}$. How should I visualise/write this to see exactly how does this homomorphism work? 
 A: In terms of the logical process of inducing maps, you can think in the following way:
A continuous map $f : X\rightarrow Y$ induces chain maps $f_\# : C_k(X) \rightarrow C_k(Y)$ (for each $k$) which induces maps on homology $f_* : H_k(X) \rightarrow H_k(Y)$.
It's important to note that $f_*$ exists only because $f_\#$ sends homologous cycles in $X$ to homologous cycles in $Y$. $f_*$ is then well-defined in mapping the equivalence class of a cycle on $X$ to the equivalence class of the image cycle on $Y$. All this is to say: it can be easier to look at the map $f_\#$ in order to understand $f_*$.
Now for your example. $S^0$ is a disjoint union of two points, say $P$ and $Q$. This implies, as you rightly note, that $H_0(S^0) \cong \mathbb{Z}\times \mathbb{Z}$. Let me write $H_0(S^0) = \mathbb{Z}\cdot [p] \oplus \mathbb{Z}\cdot[q]$ to emphasize the natural basis $\{[p],[q]\}$ (where $p,q \in C_0(S^0)$ are the $0$-cycles corresponding to the points $P,Q$ respectively). We similarly have $H_0(D^1) = \mathbb{Z}\cdot [x]\cong \mathbb{Z}$ where some point $X\in D^1$ has corresponding $0$-cycle $x \in C_0(D^1)$. Now $f : S^0 \rightarrow D^1$ can be defined as $P\mapsto \mathbf{0}$, $Q\mapsto \mathbf{1}$ (thinking of $D^1$ as $[\mathbf{0},\mathbf{1}]$). This induces $f_\# : C_0(S^0) \rightarrow C_0(D^1)$ sending $p$ to $0$ and $q$ to $1$ (where $0\in C_0(D^1)$ is the $0$-cycle corresponding to the point $\mathbf{0} \in D^1$ and similarly for $1\in C_0(D^1)$). This means that $f_*([p]) = [f_\#(p)] = [0]$ and $f_*([q]) = [f_\#(q)] = [1]$. Now it remains to note that $0$, $1$ and $x$ are all homologous $0$-cycles on $D^1$, which means $[0] = [1] = [x]$. So we have fully described the map $H_0(S^0) \rightarrow H_0(D^1)$ which ultimately amounts to $\mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}$ sending both $(1,0)$ and $(0,1)$ to $1$.
