Basis for singular chains group

Question

The singular chain group $S_p(X)$ is defined as the free abelian group generated by continuous functions $T \in C( \Delta_p , X)$.

What I understand this means is that we define $T' \in S_p(X)$ as $T'(T) = 1$ and $T'(V) = 0, \forall V \in C( \Delta_p , X)$ s.t. $V \neq T$. We therefore take $$\lbrace T' \rbrace_{T \in C(\Delta_p,X)} $$ as a basis. But for example if I take $id : \Delta_1 \rightarrow \mathbb R$ and $(-id) : \Delta_1 \rightarrow \mathbb R$, these should be part of a basis of $S_1(\mathbb R)$. However $id = -(-id)$, so I do not have a unique way of representing $id \in S_1(\mathbb R)$, therefore they cannot be part of a basis.

My question is: how can I write the elements of a basis for $S_p(X)$?

Answer 1

To avoid any suggestive notation, let $\sigma : \Delta_1 \to \Bbb R$ be the identity and $\tau : \Delta_1 \to \Bbb R$ be the map $x \mapsto -x$. $\sigma$ and $\tau$ are distinct elements of $C(\Delta_1, \Bbb R)$ since $\sigma(1) \ne \tau(1)$. By the definition of the free abelian group, any element of $S_1(\Bbb R)$ can be uniquely written as a linear combination of elements of $C(\Delta_1, \Bbb R)$. Therefore $-\sigma$ and $\tau$ are different elements in $S_1(\Bbb R)$ by uniqueness.

It can happen that the equality you seek is true when passing to homology. But $\Bbb R$ is not a very interesting example of this because it's contractible and hence $H_1(\Bbb R)$ is trivial.

Answer 2

Two distinct maps $T,T':\Delta_p \to X$ define different elements of the basis for $S_p(X)$. As you are using $-$ in two senses, you can't write $-(-id)=id$. Similarly in your example of $X= \mathbb R$, $T$ and $2T$ are distinct basis elements, which should be written say $[T], [2T]$ and $[2T] \ne 2 [T]$ in the chain group. So it is an instructive mistake!