TITLE: Which of the given statements are true?
QUESTION [5 upvotes]: Which of the following statements are true?
a. Consider the subspace $S^1 = \{(x, y)\in \mathbb{R}^2:x^2 + y^2 = 1\}$ of $\mathbb{R}^2$. Then,
there exists a continuous function $f : S^1\to \mathbb{R}$ which is onto.
b. There exists a continuous function $f : S^1 \to \mathbb{R}$ which is one-one.
c. Let $X = \{A = (a_{ij})\in M_2(\mathbb{R}):tr(A) = 0,~ |a_{ij}|\le2~\forall~1\le i,j\le 2\}$
Let $Y = \{\det(A) :A\in X\}$. Then, there exist $\alpha < 0$ and $\beta> 0$ such
that $Y = [\alpha,\beta].$

My attempt:
a) False ($S^1$ is compact but $\mathbb{R}$ is not), 
b) False (For otherwise $S^1$ being compact $f(S^1)$ is homeomorphic to $S^1$ since any bijection from a compact set to a $T_2$ space is has continuous inverse. Also $f(S_1)$ being connected and compact and non-empty must be of the form $[a, b]$ where $a < b.$ Now removing any point from $S_1$ leave $S_1$ connected. The same doesn’t hold for $[a, b]$ except the points $a$ and $b$), 
c) True ($X$ is compact,since $X$ is bounded [easy to see]  and $X=tr^{-1} (0)∩[-2,2]^4$  is closed since closeness is productive property and trace is continuous determinant is continuous so $\det X$  is closed and bounded.Also $X$ is path connected,determinant is continuous so $\det X$  is connected & hence an interval also there' s both 
positive and negetive image)
Am I right?

REPLY [3 votes]: Yes, you're spot on. Good work.
The only thing you might want to add in order to complete or nail the "existence" argument in part $(c)$, is a specification of the endpoints of interval $Y$: 
We have $\alpha = -8 < 0, \;\beta > 4$, so as Did points out in the comment, $Y = [-8, 4]$.