Prove that constant 1-simplex is a 1-boundary

I have been given the following question: Given a topological space $X$, for any $x_0 \in X$ we have defined $\varphi_{x_0}:\sigma_1 \rightarrow X: (t_0,t_1) \rightarrow x_0$. Prove that $\varphi_{x_0} + B_1(X) = 0 + B_1(X)$.

I already have found that \begin{align*} \varphi_{x_0} + B_1(X) = 0 + B_1(X) &\iff \exists d \in B_1(X): \varphi_{x_0} = 0+d\\ &\iff \varphi_{x_0} \in B_1(X) \\ &\iff \exists e \in S_2(X): \varphi_{x_0} = \partial(e) \end{align*}

But now I'm stuck, because I don't know how to explicitly or generally find such an $e$. Can anyone lead me in the right direction?

Solution 1:

You want to show that the constant path $c_x$ is a boundary.
In particular, $c_x$ is the boundary of the $2$-singular simplex $\sigma$ which has the same image of $c_x$: $$\partial_2(\sigma)=\sigma\circ(\hat E_0,E_1,E_2)-\sigma\circ (E_0,\hat E_1,E_2)+\sigma \circ(E_0,E_1,\hat E_2)=c_x-c_x+c_x=c_x.$$ Remember that the $q$-th boundary operator is defined as the map $$\partial_q:S_q(X)\to S_{q-1}(X)\text{ such that}\\\partial_q(\sigma):=\sum_{j=0}^q(-)^j\sigma\circ F_q^j\text{, where }F_q^j:\Delta_{q-1}\to\Delta_q$$ is the face operator that maps the $(q-1)$-standard simplex $[E_0,\dots,E_{q-1}]$ to the $j$-th face of the
$q$-standard simplex.