Can you compute relative homology using simplicial chain complex? So I had my Algebraic Topology exam yesterday and one of the questions asked to compute the homology groups $H_*(M)$, $H_*(\partial M)$ and $H_*(M, \partial M)$ where $M$ is the Möbius strip and $\partial M$ its boundary.
So I computed the first two using simplicial homology and a $\Delta$-complex structure on M. Then I computed the relative homology using what I thought was the definition, I took the quotient groups $C^{\Delta}_n(M) / C^{\Delta}_n(\partial M)$. But I realised today that relative homology is defined by taking the quotient of the singular chain complex. So my question is, are the two equivalent?
EDIT: I guess this method might not always make sense but here there was a subcomplex which formed a $\Delta$-complex structure on $\partial M$ so $C^{\Delta}_n(\partial M)$ formed a subgroup of $C^{\Delta}_n (M)$.
 A: There are relative simplicial homology groups $H^\Delta_*(X,A)$ where $A$ is a subcomplex of a $Δ$-complex. These are the homology groups of the chain complex in the third row of the following commutative diagram, where $Δ_n(X,A)$ is the quotient $Δ_n(X)/Δ_n(A)$, so the columns are exact. This also gives rise to a long exact sequence of simplicial homology groups as in the singular case. 
$\qquad$

$$0\to H^Δ_2(M,∂M)→H^Δ_1(∂M)→H^Δ_1(M)→H^Δ_1(M,∂M)→\tilde H^Δ_0(∂M)=0$$
The key is to compute the map $H^Δ_1(∂M)\to H^Δ_1(M)$. Looking at the $Δ$-complex structure as in the picture on the right, we see that $H^Δ_1(M)$ is generated by $c$, and $H^Δ_1(∂M)$ is generated by $a+a'$, which in $H^\Delta_1(M)$ is homologous to $(b+a')+(a-b)\sim c+c$. That means the map is multiplication by $2$, and this determines  the relative groups $H^Δ(M,∂M)$, using exactness of the sequence.
Edit: I just see that your question was about equivalence of singular and simplicial relative homology groups. This is Theorem 2.27 in Hatcher's Algebraic Topology.
