$T$ is a torus which is CW complex with one $0$- cell $v$, two $1$-cell $a,b$ and one $2$-cell $e_2$.
Here in the picture of torus, if we identify the circle $a$, intuitively it will look line $S^2/\{p,q\}$ for any two distinct points in $S^2$. Then by using a well know result regarding HEP and homotopy invariance (which says if $(X,A)$ has HEP and $A$ is contractible then $X/A\simeq X$) we can deduce that $S^2/\{p,q\}\simeq S^2\vee S^1$. (Here $\simeq$ denotes homotopy equivalent)
But I am unable to write down a proof of $T/S^1_a\simeq S^2/\{p,q\}$. Can anyone help me in this regard? Thanks for help in advance.
It is possible to show that $T/\mathbb{S}^1_a$ is homeomorphic to $\mathbb{S}^2/\{p,q\}$. Define $C=\mathbb{S}^1 \times [0,1]$, if $\mathbb{S^1}_j=\mathbb{S}^1\times\{j\}$ for $j=0,1$, and $\sim$ denotes the relation that identifies $\mathbb{S}^1_0$ and $\mathbb{S}^1_1$ with points we have that: $$\mathbb{S}^2 \cong C/{\sim},$$
with $p=\mathbb{S}^1_0/{\sim}$ and $q=\mathbb{S}^1_1/{\sim}$. One way to see this is using the homogeneity lemma and the fact that the annulus is homemorphic to the cylinder. Furthermore, if $\pi:C \to C/{\sim}$ denotes the canonical projection we have that: $$ \mathbb{S}^2/\{p,q\}=(C/{\sim})/\pi(\mathbb{S}^1_0) \sqcup \pi(\mathbb{S}^1_1) $$
Now let's write down the homeomorphism $\phi:\mathbb{S}^2/\{p,q\} \to T/\mathbb{S}^1_a$. Define $\tilde{\pi}: (C/{\sim}) \to \mathbb{S}^2/\{p,q\}$ as the canonical projection and notice that we have a map: $$C\rightarrow^{\pi}\ C/{\sim} \rightarrow^{\tilde{\pi}} \mathbb{S}^2/\{p,q\}.$$ It's easy to see that this map descends to:
$$\phi: C/{\mathbb{S}^1_0 \sqcup \mathbb{S}^1_1} \to \mathbb{S}^2/\{p,q\}.$$
On the other hand:
$$T/\mathbb{S}^1_a = C/{\mathbb{S}^1_0 \sqcup \mathbb{S}^1_1}.$$
So indeed $\phi$ is well defined on the spaces that we need. Finally to see that $\phi $ is a homeomorphism we use the same argument as above to construct a map $\psi:T/\mathbb{S}^1_a \to \mathbb{S}^2/\{p,q\}$ that will be a inverse to $\phi$.