Show that $[l_1 \cdot l_2 \cdot l_3 ] = [l_1 + l_2 + l_3] \in H_1(X)$ The first Homology group of X Let $l_1$ , $l_2$ and $l_3$ be three paths in X with $l_1 (0) =
l_3 (1)$, $l_1 (1) = l_2 (0)$ and $l_2 (1) = l_3 (0)$. Define the loop $l = l_1 \cdot l_2 \cdot l_3 $ (based at $l_1 (0)$).
Show that $l$ and $l_1 + l_2 + l_3$ are 1-cycles. (This can just be shown by applying the boundary map, $\partial$, on the paths are to get $0$. Hence they are in $\ker(\partial)$, which defines them as 1-cycles).
Furthermore, show that $[l] = [l_1 + l_2 + l_3] \in H_1(X)$
My confusion is regarding treating this as an algebraic problem, or proving it through the definition of homotopy equivalence, i.e using the definition of the class $[l]$ as everything homotopy equivalent to the path $l$
 A: What you are asked to show, namely that $[l] = [l_1 + l_2 + l_3]$ in $H_1(X)$, means precisely that $$l_1+l_2+l_3-l=\partial(\text{something})$$ where $\text{something}\in C_2(X)$, the free abelian group on the set of singular $2$-simplices. That is, you will need to define certain maps $\Delta^2\to X$ and a linear combination of them ($\in C_2(X)$) whose boundary is $l_1+l_2+l_3-l$. Below is a photograph of a drawing of a solution.

The drawing represents two continuous maps $x,y:\Delta^2\to X$. They are constructed as follows: on the boundary $\partial\Delta^2$ they are defined using the paths $l_1,l_2,l_3$, and in the interior, they are defined to be constant along the lighter lines I drew inside the triangles, equal to whatever value they take on the border $\partial \Delta^2$. The vertices of $\Delta^2$ are numbered $0,1,2$ to indicate how the border is calculated.
A: Consider a cylinder $[0,1]\times S^1$ based an a circle $S^1$, where the "left" boundary $\{0\}\times S^1$ is partitioned into a single $0$-cell and a single $1$-cell, while the "right" boundary component $\{1\}\times S^1$ is partitioned into three $1$-cells by three $0$-cells.  Clearly this can be extended to a triangulation of the cylinder.  Now map this to the manifold $X$ by a function $f: [0,1]\times S^1 \to X$, where $f(s,t)$ is independent of $s\in [0,1]$ and only depends on $t\in S^1$.  This induces a 2-chain showing that the two cycles you defined are homologous.  Hence they define the same homology class in $H_1(X)$.
