Could someone please check my homology computations and see if I have made a mistake in the following? Also, I'd appreciate it if I could sample your thoughts about the last note!
Problem Consider a 2-simplex $\Delta = [v_0, v_1, v_2]$ with edges $e_1 = [v_0,v_1]$, $e_2 = [v_1,v_2]$ and $e_3 = [v_0,v_2]$. Let $\Delta_1$ be obtained by identifying edges $e_1$ and $e_3$, and identifying $v_1$ and $v_2$. Compute the relative homology groups $H_i(\Delta_1,\{v_0\})$, $H_i(\Delta_1,\{v_0,v_1\})$ and $H_i(\Delta_1,\{v_0,v_1\}\cup e_2)$.
Solution
$\mathbf{H_i(\Delta_1,\{v_0\})}$ $$ \partial_2 \Delta_1 = e_1+e_2-e_1 = e_2\\ \partial_1 e_1 = v_1, \partial_1 e_2 = 0\\ \partial_0 v_1 = 0 $$ Thus, $$ H_2(\Delta_1,\{v_0\}) = 0\\ H_1(\Delta_1,\{v_0\}) = 0\\ H_0(\Delta_1,\{v_0\}) = 0 $$
$\mathbf{H_i(\Delta_1,\{v_0,v_1\})}$ $$ \partial_2 \Delta_1 = e_1+e_2-e_1 = e_2\\ \partial_1 e_1 = 0, \partial_1 e_2 = 0 $$ Thus, $$ H_2(\Delta_1,\{v_0,v_1\}) = 0\\ H_1(\Delta_1,\{v_0,v_1\}) = <e_1>\\ H_0(\Delta_1,\{v_0,v_1\}) = 0 $$
$\mathbf{H_i(\Delta_1,\{v_0,v_1\}\cup e_2)}$ $$ \partial_2 \Delta_1 = e_1-e_1 = 0\\ \partial_1 e_1 = 0 $$ Thus, $$ H_2(\Delta_1,\{v_0,v_1\}\cup e_2) = <\Delta_1>\\ H_1(\Delta_1,\{v_0,v_1\}\cup e_2) = <e_1>\\ H_0(\Delta_1,\{v_0,v_1\}\cup e_2) = 0 $$
Note:
I see that $\{v_0\} \subset \{v_0,v_1\} \subset \{v_0,v_1\}\cup e_2$, and the homology groups are getting bigger and bigger. Because the boundary homomorphisms are mapping into smaller and smaller images, and therefore the kernel of the homomorphisms is bigger and bigger?
Are $H_i(\Delta_1,\{v_0\})$ isomorphic to those for a cone? (I am guessing excision of $v_0$ is applicable here.)
Are $H_i(\Delta_1,\{v_0,v_1\}\cup e_2)$ isomorphic to those for a sphere with two points identified (after excising the identified point)? Fundamental group of $S^2$ with north and south pole identified
Thank you.
