Hello I want to prove that the abelianization of $G=\langle t_1,...,t_n | t_1^2...t_n^2 \rangle$ is $H=\mathbb{Z}^{n-1} \oplus \mathbb{Z}/2\mathbb{Z}$.
In my textbook they display an inverse of the homomorphism $\phi: H \rightarrow G_{\text{ab}}$ defined by $\phi (a_1,...,a_n)=(a_1 + a_n,...,a_{n-1}+a_n, a_n)$ and conclude.
I proceeded differently and I would like to know if my solution is correct.
proof attempt: Given a word in $m\in G$ we define $\omega_i (m)$ as the sum of the exponents of $t_i$ present in $m$.
For example $w_3(t_3^{-2} t_1t_5t_3^8)=-2+8=6$.
Let $\sigma:G\rightarrow H$ be defined as $\sigma(m)=(w_1(m),...,w_{n-1}(m),[w_{n}(m)]_{\mathbb{Z}/2\mathbb{Z}})$. We see immediatly that $\sigma$ is an homomorphism since when you glue words the $w_i$ will be summed.
Let $q:G\rightarrow G_{\text{ab}}$ be the quotient map. The universal property of abelianization gives a unique map $\psi:G_{\text{ab}}\rightarrow H$ between abelian groups such that $\sigma = \psi \circ q$.
We show that $\psi$ is an isomorphism.
For $x=(a_1,...,a_{n-1},[a_n]_{\mathbb{Z}/2\mathbb{Z}})\in H$ we have $\psi([t_1^{a_1}...t_{n-1}^{a_{n-1}}t_n^{a_n}])=x$ so $\psi$ is surjective.
Now suppose we have $[m]\in \ker \psi$, by moving the first $t_i$ around for $1\leq i \leq n-1$ we have that $[m]=[t_1^0...t_{n-1}^0t_n^{2k}]=[t_n^{2k}]$ for some $k\in\mathbb{Z}$.
Now the relation in the presentation of $G$ gives $[t_n^{2k}]=[t_{n-1}^{-2k}...t_1^{-2k}]$
Hence
\begin{gather} \psi ([m])=\psi([t_n^{2k}])=\psi([t_{n-1}^{-2k}...t_1^{-2k}])=0 \end{gather}
So $\sigma (t_{n-1}^{-2k}...t_1^{-2k})=0$ thus we get that necessarily $k=0$ by definition of the $\omega_i$.
Consequently $[m]=[e]$ so $\psi$ is also injective.
Therefore $G_{\text{ab}} \cong \mathbb{Z}^{n-1} \oplus \mathbb{Z}/2\mathbb{Z}$.
The solution is flawed because the map $\sigma$ isn't a homomorphism.
Suppose otherwise. Then the element $\sigma(t_1^2\cdots t_{n-1}^2)$ is non-trivial, as it is contained in the $\mathbb{Z}^{n-1}$ part. However, as both $\sigma(t_1^2\cdots t_n^2)$ and $\sigma(t_n^2)$ are trivial, we see that $\sigma(t_1^2\cdots t_n^2)\sigma(t_n^2)^{-1}=\sigma(t_1^2\cdots t_{n-1}^2)$ is trivial, which is a contradiction.
Some intuition here: the map $\sigma$ singles out $t_n$ to have order $2$ in the abelianisation. However, $t_n$ is not special and we can define a similar map to make any fixed $t_i$ have order $2$: Define $\sigma_i$ so that $\sigma_i(t_i)$ maps to the element of order $2$ in $\mathbb{Z}^{n-1}\times\mathbb{Z}/2\mathbb{Z}$ and for $i\neq j$, $\sigma_i(t_j)$ maps to an infinite-order element. These maps $\sigma_i$ cannot all be homomorphisms, as then every generator would have order $2$ in the abelianisation, and yet one is a homomorphism if and only if they all are because no $t_i$ is special (more formally, for all $i, j$ there is an automorphism of your group taking $t_i$ to $t_j$).
The answers to this question explain how to calculate the abelianisation of group given by a one-relator presentation, as is the case here. Using the method in my answer, you get the presentation $\langle t_1, \ldots, t_n\mid t_1^2\cdots t_{n-1}^2(t_nt_1^{-1}\cdots t_{n-1}^{-1})^2\rangle$, which clearly has the abelianisation you want ($\langle t_1, \ldots, t_n\mid t_n^2\rangle$ as the other $t_i$ cancel).