Are topology space $T$ and $L$ homeomorphic? Suppose $T$ and $L$ are both close set in Euclidean space $\mathbb{R}^2$.Is there a homeomorphism between topology space $T$ and $L$(See picture below)?
There are two views fighting in my head.
$Point$ $A$: $T$ and $L$ are homeomorphic. The homeomorphism is given by moving the vertical line of $T$ parallelly to the endpoint of horizontal line of $T$.
$Point$ $B$: $T$ and $L$ are not homeomorphic. We prove it by contradiction.If $T$ and $L$ are homeomorphic, we assume that the homeomorphism maps point $x_0 \in T$(The intersection of vertical and horizontal line) to $y_0 \in L$. Then, $T-\{x_0\}$ is homeomorphic to $L-\{y_0\}$. However, $T-\{x_0\}$ has $3$ connected components so the $0th$ homology group $H_0(T-\{x_0\})=Z \oplus Z \oplus Z$, and similarly $S-\{y_0\}$ has zeroth homology group $Z \oplus Z$ since no matter where is $y_0$, $S-\{y_0\}$ has only $2$ connected components.
Point $A$ vs Point $B$, which is right, or they are both wrong answer?

 A: They're not homeomorphic. $L$ is homeomorphic to a straight line segment, so let's examine whether $T$ is homeomorphic to $[0,1]$. Suppose $f: T\to [0,1]$ is a homeomorphism. Let $a,b,c$ denote the left, right, and bottom ends of $T$, respectively. 
I claim that either $f(a)=0$ or $f(a)=1$. To show this, let $S$ be the line segment in $T$ from $a$ to $x_0$, including $a$, but not including $x_0$. Then there exists a homeomorphism $g: S\to [0,1)$, with $g(a)=0$. Suppose $f(a)=z\in (0,1)$. $h=g\circ f^{-1}: [0,1]\to [0,1)$ is injective and continuous, as it is the composition of two injective continuous functions. Further, $h(z)=0$, while $z\in (0,1)$. Pick $\alpha, \beta \in (0,1)$ with $0<\alpha<z<\beta<1$. Suppose $h(\alpha)\leq h(\beta)$. Then $z\leq \beta$ and $h(z)\leq h(\alpha)\leq h(\beta)$, so by the IVT, there exists $w\in [z,\beta]$ such that $h(w)=h(\alpha)$. But this contradicts the injectivity of $h$. If $f(\beta)\leq f(\alpha)$, this argument works similarly. 
Thus, either $f(a)=0$ or $f(a)=1$. The same holds for $b,c$ by similar reasoning. But then we have three distinct points mapping to two points, which means that $f$ is not injective, a contradiction. Thus, $T$ and $[0,1]$ are not homeomorphic. 
Intuitively: Your function from point $A$ is not a homeomorphism, since it sends both $x_0$ and $b$ to the corner of $L$, so it's not injective. 
A: 
Lemma: Suppose that $X$ and $Y$ are homeomorphic connected spaces. Let $f$ be a homeomorphism from $X$ to $Y$. Then for all $p \in X$ the space $X\setminus \{p\}$ is homeomorphic to $Y \setminus \{f(p)\}$.

Proof: the required homeomorphism is just the restriction of $f$ (which has the restriction of $f^{-1}$ as its inverse, and restrictions of continuous maps are continuous.
Now let $f: T \to L$ be a homeomorphism. Then let $x_0$ be the cross point of $T$.
Note that for any $q \in L$, $L \setminus \{q\}$ has at most two components (1 if $q$ is one of the end-points, $2$ otherwise.
So $T\setminus \{x_0\}$ has $3$ connected components while $L \setminus \{f(x_0)\}$ has at most 2. So these spaces are not homeomorphic. This contradicts the lemma, so $f$ cannot exist.
