Proof verification: $T\in\mathcal{L}(\mathcal{P}(\bf{R}))$ being injective and $\deg Tp\leq\deg p$ implies $T$ is surjective and $\deg Tp=\deg p$ Notations: $\mathcal P(\mathbf R)$ denotes the polynomials with real coefficients, while $\mathcal{L}(\mathcal P(\mathbf R))$ denotes the linear transformations within $\mathcal P(\mathbf R)$ with respect to the field of real numbers. Also, $\mathcal P_m(\mathbf R)$ denotes the polynomials with real coefficient and degree less than or equal to $m$.

Are the proofs to the following propositions Correct ?
Theorem. Given that $T\in\mathcal{L}(\mathcal{P}(\mathbf{R}))$ is injective and that $\deg Tp\leq\deg p$ for all $p\in\mathcal{P}(\mathbf{R})$ such that $p$ is a non-zero polynomial then
$a)$ $T$ is surjective 
Proof. Let $p$ be arbitrary non-zero polynomial. We may assume without loss of generality that $\deg p = m$ where $m\in\{0,1,2,3,...\}$
Let us now examine the function $T|_{\mathcal{P}_m\mathbf{(R)}}$. The proposition $\forall p\in\mathcal{P}(\mathbf{R})(\deg Tp\leq\deg p)$ together with $T\in\mathcal{L}(\mathcal{P}\mathbf{(R)})$ implies that $T|_{\mathcal{P}_m\mathbf{(R)}}\in\mathcal{L}(\mathcal{P}_m\mathbf{(R)})$, moreover the injectivity of $T$ implies the the injectivity  of $T|_{\mathcal{P}_m\mathbf{(R)}}$ and since injectivity and surjectivity are equivalent for linear operators defined on a finite-dimensional vector space it follows that $T|_{\mathcal{P}_m\mathbf{(R)}}$ is surjective, consequently $\exists q\in \mathcal{P(\mathbf{R})}(Tq=p)$.
$\blacksquare$
$b)$ $\deg Tp=\deg p$ for all non zero polynomial $p\in\mathcal{P}(\mathbf{R})$ 
Proof. We prove the above propostion by recourse to Mathematical-Induction.
Basis-Step: Let $p$ be an arbitrary polynomial with $\deg p=0$ ($p$ is non-zero) we know that $\forall p\in\mathcal{P}(\mathbf{R})(\deg Tp\leq\deg p)$ thus in particular $\deg Tp\leq \deg p$.
Now assume for the purpose of contradiction that $\deg Tp<\deg p$ which implies that $\deg Tp = -\infty$ but since $T$ is injective and therefore $\operatorname{null}T=\{0\}$ it must be that $p=0$ a contradiction consequently $\deg Tp=\deg p=0$.
Inductive Step: Assume that $k$ is arbitrary and that $\deg Tp=\deg p$ where $\deg p\in\{0,1,2,3,...,k\}$.
Consider now an arbitrary polynomial $q$ such that $\deg q=k+1$ assume further that $Tq=r$ and that $\deg r<\deg q$ therefore $\deg r\leq k$ since $T$ is surjective it follows that for some $s\in\mathcal{P(\mathbf{R})}$, $Ts=r$ and by inductive hypothesis $\deg s = \deg r \leq k$ but $Ts=r=Tq$ and $s\neq q$ contradicting the fact that $T$ is injective therefore it must be that $\deg q=\deg r$.
$\blacksquare$
 A: For part $b)$, the degree of $k$ needs to be able to be zero, or else your induction would not be able to prove that the proposition is true for $p$ with degree $1$. You only proved that $P(0)$ is true and that $P(1) \implies P(2)$ etc. without proving that $P(0) \implies P(1)$, where $P$ is the proposition allegedly proved using the induction. Apart from this pedantry, you also need more ink explaining how $\deg r < \deg q$ implies $\deg q = k$. (Actually, you can only say that $\deg q \le k$.)
A: There are problems with the final paragraph. As you have already been told, all you can say about $q$ is that $\deg q\leqslant k$. Besides, how do you reach a contradiction here? It is more simple to say that, since $\deg q\leqslant k$, it follows from what you have already proved that $q=Ts$ for some polynomial $s$ whose degree is also smaller than or equal to $k$. But then $s\neq p$ and therefore $T$ would not be injective.
A: proof-verification (for one of the many edited versions of your proof): 

Theorem. Given that $T\in\mathcal{L}(\mathcal{P}(\mathbf{R}))$ is injective and that $\deg Tp\leq\deg p$ for all $p\in\mathcal{P}(\mathbf{R})$ such that $p$ is a non-zero polynomial then
$a)$ $T$ is surjective 
Proof. Let $p$ (One could choose another notation so that it would not be confused with the free variable p used later.) be arbitrary non-zero polynomial. We may assume without loss of generality that $\deg p = m$ where $m\in\{0,1,2,3,...\}.$ Let us now examine the function $T|_{\mathcal{P}_m\mathbf{(R)}}$. The proposition assumption $$
\forall p\in\mathcal{P}(\mathbf{R})(\deg Tp\leq\deg p)
$$ 
  together with $T\in\mathcal{L}(\mathcal{P}\mathbf{(R)})$ implies that $T|_{\mathcal{P}_m\mathbf{(R)}}\in\mathcal{L}(\mathcal{P}_m\mathbf{(R)})$; moreover the injectivity of $T$ implies the the injectivity  of $T|_{\mathcal{P}_m\mathbf{(R)}}$ and since injectivity and surjectivity are equivalent for linear operators defined on a finite-dimensional vector space it follows that $T|_{\mathcal{P}_m\mathbf{(R)}}$ is surjective, consequently 
  $$
\exists q\in \mathcal{P(\mathbf{R})}(Tq=p).
$$ 
$b)$ $\deg Tp=\deg p$ for all non zero polynomial $p\in\mathcal{P}(\mathbf{R})$.
Proof. We prove the above propostion by recourse to mathematical induction.
Basis-Step: Let $p$ (Again, another notation would be better.) be an arbitrary polynomial with $\deg p=0$ ($p$ is non-zero) we know by assumptions that $\forall p\in\mathcal{P}(\mathbf{R})(\deg Tp\leq\deg p)$ thus in particular $\deg Tp\leq \deg p$. (One can guess though, notations are confusing.)
Now assume for the purpose of contradiction that $\deg Tp<\deg p$ which implies that $\deg Tp = -\infty$ (This is OK if you have defined that the zero polynomisl has degree $-\infty$.) but since $T$ is injective and therefore $\operatorname{null}T=\{0\}$ it must be that $p=0$ a contradiction consequently $\deg Tp=\deg p=0$. (Isn't 0 the only polynomial with degree $-\infty$ ? Why would you need the injectivity of T to make such conclusion?)
Inductive Step: Assume that $k$ is arbitrary and that $\deg Tp=\deg p$ where $\deg p\in\{0,1,2,3,...,k\}.$
Consider now an arbitrary polynomial $q$ such that $\deg q=k+1$ assume further that $Tq=r$ and that $\deg r<\deg q$ therefore $\deg r\leq k$ since $T$ is surjective it follows that for some $s\in\mathcal{P(\mathbf{R})}$, $Ts=r$ and by inductive hypothesis $\deg s = \deg r \leq k$ (and thus $s\neq q$)but $Ts=r=Tq$ and $s\neq q$, contradicting the fact that $T$ is injective therefore it must be that $\deg q=\deg r$.

