If $i$ is an inclusion why is the induced $i_*$ an epimorphism Given the following exact homology sequence of a pair. This is in Example 2 (page 134) from Munkres. This is where I am always stuck computing homology using exact sequence. I cannot grab the last paragraph. I know that $i$ is an inclusion, but why then it implies that $i_*$ maps both generator to the same generator of $H_1(K)$. And why it then implies $i_*$ is epim and the kernel is $\mathbb{Z}$? Can anyone elaborate this. Here $K$ is a complex whose polytope is annulus, and $K_0$ is a subcomplex whose polytope is the union of inner and outer circles of that annulus. Any help is appreciated :)

 A: In applying the Mayer-Vietoris sequence, not only do you need to know how to evaluate homology groups of spaces, but you also need to evaluate induced homology homomorphisms of continuous maps.
In this example, you have correctly evaluated the homology groups $H_1(K) \approx \mathbb{Z}$ and $H_1(K_0) \approx \mathbb{Z} \oplus \mathbb{Z}$. 
The additional information you need is how to evaluate the inclusion induced homology homomorphism $i_* : H_1(K_0) \to H_1(K)$. This is a linear map, with domain $H_1(K_0) \approx \mathbb{Z} \oplus \mathbb{Z}$ and range $H_1(K) \approx \mathbb{Z}$, and you need a formula for it. 
In brief, $K_0$ has two components $C_1,C_2$, each of those two components separately includes into $K$ as a homotopy equivalence, and the induced homology homomorphisms $(i_1)_*, (i_2)_*$ from the homology groups of the components to the homology of $K$ are therefore isomorphisms. 
From this you can derive a formula for $i_* : \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$. It is a linear function. Its restriction to the first coordinate is the isomorphism $(i_1)_*$ and so $i_*(m,0)=m$. Its restriction to the second coordinate is the isomorphism $(i_2)_*$ and so $i_*(0,n)=n$. The general formula is therefore $i_*(m,n)=m+n$. 
Can you take it from here?
