I'm studying "Singular Homology Groups" in J. Lee's Introduction to Topological Manifolds on my own. When I try to compute some concrete examples, confusion arises:
Let $p$ and $q$ be two paths in a topological space $X$ such that $p(1)=q(0)$. Then "intuitively", $p+q-pq$ should lie in $B_1(X)$, the group of 1-boundaries.
I don't know how to prove this, but if I consider a special case: let $X=S^1$ and $q$ be inverse of $p$ (that is, $q(t)=p(1-t)$), then $$p+q=p(1)-p(0)+q(1)-q(0)=p(1)-q(0)+(-p(0)+q(1))=0+0=0.$$ That is, $p+q$ is in $Z_1(X)$ but not $B_1(X)$.
Where did I go wrong? Also, I wonder if someone could show me how to prove that simple example above (when $p$ and $q$ are arbitrary paths).
Yes, $p+q-p\cdot q$ is always a $1$-boundary. We need to find a singular $2$-simplex $\sigma\colon\Delta^2\to X$ that has boundary $p+q-p\cdot q$. Imagine $\Delta^2$ sitting on a line just like the symbol "$\Delta$" does, with vertex labels 0 on the bottom left, 1 on the top and 2 on the bottom right. If we project this downwards orthogonally to the line we end up with "_", a picture of the unit interval $I$. This describes a map $T:\Delta^2\to I$ where the bottom edge is mapped bijectively to $I$ and the two top edges are mapped bijectively to $[0,\tfrac 1 2]$ and $[\tfrac 1 2, 1]$, respectively. Now consider the singular $2$-simplex given by $$ \sigma\colon \Delta^2 \overset{T}\longrightarrow I \overset{p\cdot q}\longrightarrow X. $$ By construction, the boundary of this $2$-simplex is $q-p\cdot q+p$.