Q.If $d$ and $e$ are positive integer and $T:R^{d} → R^{e}$ be a linear transformation then
(a) $T$ is open map if and only if $T$ is surjective
(b) $T$ is closed map if and only if $T$ is either zero or injective
Here is an argument .
For (a)
Suppose $T$ is an open map. We want to show that $T$ is surjective. To do this, suppose that $T$ is not surjective. Then, there exists some vector $y$ in $R^{e}$ such that there is no $x$ in $R^{d}$ such that $T(x) = y$. Consider the set $U = {y}^c$, the complement of ${y}$ in $R^{e}$. Since $T$ is not surjective, $U$ is non-empty. Now, note that $T(U)$ is an open set in $R^{e}$, since $T$ is an open map. This follows from the fact that $T(U)$ is the image of an open set under a continuous linear transformation. But, since $y$ is not in $U$, we have that $y$ is not in $T(U)$. Thus, $T(U)$ is a non-empty open subset of $R^{e}$ that does not contain $y$. This contradicts the fact that $y$ is in the closure of $T(U)$ (since every non-empty open set contains a point in its closure). Therefore, we must have that T is surjective.
Suppose $T$ is surjective. We want to show that $T$ is an open map. To do this, let $U$ be an open set in $R^{d}$. We want to show that $T(U)$ is an open set in $R^{e}$. Let $y$ be a point in $T(U)$, so that there exists some $x$ in $U$ such that $T(x) = y$. Since $U$ is open, there exists some $r > 0$ such that the open ball centered at $x$ with radius $r$, $B(x, r)$, is contained in $U$. Now, consider the image of $B(x, r)$ under $T$. We have that:
$T(B(x, r)) = {T(x + h) : ||h|| < r} = {T(x) + T(h) : ||h|| < r} = {y + T(h) : ||h|| < r}$
Note that this last set is the translation of the open ball centered at $y$ with radius $r$, $B(y, r)$, by the vector $T(h)$, which varies over all vectors of norm less than $r$. Since $T$ is surjective, we have that for any $z$ in $R^{e}$, there exists some $x$ in $R^{d}$ such that $T(x) = z$. Therefore, for any point $w$ in $B(y, r)$, there exists some $h$ in $R^{d}$ such that $T(h) = w - y$. Thus, we can write:
$w = y + T(h) = T(x) + T(h) = T(x + h)$
Since $x + h$ is in $B(x, r)$, we have that $w$ is in $T(B(x, r))$. This shows that $B(y, r)$ is a subset of $T(B(x, r))$. But, since $T$ is a linear transformation, it is continuous. Therefore, $T(B(x, r))$ is an open set in $R^{e}$. Thus, we have shown that for any point $y$ in $T(U)$, there exists an open ball around $y$ that is contained in $T(U)$. This shows that $T(U)$ is an open set in $R^{e}$, and thus, $T$ is an open map.
For (b)
Suppose $T$ is a closed map. We want to show that $T$ is either zero or injective.
$Case 1$ : $T$ is the zero map. In this case, $T$ is clearly a closed map, and $T$ is also injective.
$Case 2$: $T$ is not the zero map.
In this case, there exists some nonzero vector $x$ in $R^{d}$ such that $T(x)$ is nonzero. Since $T$ is not the zero map, there exists some vector $y$ in $R^{d}$ such that $T(y)$ is nonzero. Since $T$ is closed, we have that whenever a sequence ${x_n}$ in $R^{d}$ converges to some limit $x$ in $R^{d}$, the sequence ${T(x_n)}$ converges to some limit $y$ in $R^{e}$. Now, consider the sequence ${x + (1/n)y}$ in $R^{d}$. This sequence converges to $x$, but ${T(x + (1/n)y)} = {T(x) + (1/n)T(y)}$ does not converge to $T(x)$ as $n$ goes to $∞$, since $T(y)$ is nonzero. Therefore, we have a sequence in $R^{d}$ that converges to $x$, but whose image under $T$ does not converge to $T(x)$. This implies that $T$ is not a closed map. Therefore, we must have that $T$ is injective.
Suppose $T$ is either zero or injective. We want to show that $T$ is a closed map.
$Case 1$: $T$ is the zero map. In this case, $T$ is clearly a closed map.
$Case 2$: $T$ is injective. Let ${x_n}$ be a sequence in $R^{d}$ that converges to some limit $x$ in $R^{d}$, and let ${T(x_n)}$ be the corresponding sequence in $R^{e}$. We want to show that ${T(x_n)}$ converges to some limit $y$ in $R^{e}$. Since $T$ is injective, we know that $T(x_n)$ is a Cauchy sequence in $R^{e}$. Therefore, there exists some limit $y$ in $R^{e}$ such that ${T(x_n)}$ converges to $y$. Now, we need to show that $y = T(x)$. Since $T$ is injective, we have that $T(x_n) = T(x)$ if and only if $x_n = x$. Therefore, if we assume that $T(x_n)$ converges to $y$, then we must have that $x_n$ converges to $x$. Therefore, we have shown that ${T(x_n)}$ converges to $T(x)$ if and only if ${x_n}$ converges to $x$. Since ${x_n}$ converges to $x$ by assumption, we have that ${T(x_n)}$ converges to $T(x)$. Therefore, $T$ is a closed map.
I didn't find this question in stack, here is my argument Please have a look. Thank you
The argument can be substantially simplified.
Concerning (a), If $T$ is open then $T(\{x\,:\,\|x\|<1\})$ contains a ball $\{y\,:\,\|y\|<r\},$ for some $r>0.$ By homogeneity of $T$ we get $\mathbb{R}^e\subset {\rm Im}\,T.$
For the converse direction, by homogeneity and additivity is suffices to show that $T(B_X(0,1))$ contains a ball $B_Y(0,r)$ for some $r>0.$ Let $v_1,v_2,\ldots, v_e$ be a linear basis of $\mathbb{R}^e.$ There exist $u_1,u_2,\ldots, u_e\in \mathbb{R}^d$ such that $T(u_e)=v_e.$ There is $t>0$ such that $t\|u_i\|<1$ for all $1\le i\le e.$ Then $B_X(0,1)$ contains all vectors of the form $a_1u_1+\ldots +a_eu_e,$ where $|a_1|+\ldots +|a_e|<t.$ Hence contains all vectors $$T(B_X(0,1))\supset\{a_1v_1+\ldots +a_ev_e\,:\, \quad |a_1|+\ldots +|a_e|<t\}\quad (*)$$ It is not specified what norm OP meant in $\mathbb{R}^e.$ Since all norms are equivalent, the property of a set being open is independent of the norm. Define the norm on $\mathbb{R}^e$ by $$\|a_1v_1+\ldots +a_ev_e\|=|a_1|+\ldots +|a_e|$$ By $(*)$ we get $T(B_X(0,1))\supset B_Y(0,t).$
Concerning (b), the range of $T$ is a linear subspace of $\mathbb{R}^d.$ If $T$ is injective then $T$ is a linear homeomorphism from $\mathbb{R}^d$ onto ${\rm Im }\,T$, as every invertible linear mapping is a homeomorphism. In particular $T$ is closed from $\mathbb{R}^d$ to ${\rm Im}T,$ and hence to $\mathbb{R}^e$, as ${\rm Im}T$ is a closed subspace.
Assume $T$ is nonzero and not injective. Thus there exists $u\notin \ker T$ and $0\neq v\in \ker T.$ Consider the set $$C=\{ tv+(\arctan t)u\,:\, t\in \mathbb{R}\}$$ The set $C$ is closed in the linear span of $u$ and $v,$ hence $C$ is closed. On the other hand $T(C)=\{\alpha Tu\,:\, -{\pi\over 2}<\alpha <{\pi \over 2}\},$ hence it is not closed.