On Closed maps between metric spaces For any two metric spaces $(X, d_{X})$ and $(Y, d_Y)$, a map $f:X \to Y$ is said to be a closed map if whenever $F$ is closed in $X$ , then $f(F)$ is closed in $Y$ . For any subset $B$ of a metric space, $B$ is given the induced metric.The metric on $X×Y$ is given by
$d\{(x,y), (x',y')\}=\max \{d_X(x,x'), d_Y(y,y')\}  $. Which of the following are true ?
$(a)$ For any subset $A\subseteq X$ , the inclusion map $i : A \to X$ is closed map.
$(b)$ The projection map $p_1 : X×Y \to X$ given by $p_1(x,y)=x$ is closed.
$(c)$ Suppose that $f:X\to Y, g:Y\to Z$ are continous maps . If $g\circ f: X\to Z$ is closed map then $g|_{f(X)} : f(X) \to Z$ is closed map . Here $g|_{f(X)}$ means the restriction of $g$ on $f(X)$
$(d)$ If $f:X \to Y$ takes closed balls into closed sets , then $f$ is closed .
My Thinking :-
$(a)$ Let $A$ be an open subset of $X$.
Then $A$ with the induced metric is closed in itself but
$i(A)=A$ is open in $X$.
So it is false.
$(b)$ I think this is true but  can't prove.
Let $B\subseteq X×Y$ be a closed subset . Then $\overline{B}=B$
Let $y\in \{ p_1(B)\} '$( The derived set) .
Then there exist a sequence $\{p_1(x_n)\} \in  p_1(B) $ such that $\displaystyle \lim_{n\to \infty} p_1(x_n)=y$
Since $p_1$ is continous with respect to the product metric and so by sequential criterion of continuity , it follows that
$\displaystyle \lim_{n\to \infty} p_1(x_n)=p_1(\displaystyle \lim_{n\to \infty}x_n)$ (But the problem here is that $\{x_n\}$ may not be convergent !)
I can't proceed after this.
I have no idea for $(c)$ and $(d)$.
Please give hints.  Thank you.
 A: Some hints:
For b) and d), think of the projection map from $\mathbb{R}^{2}$ to $\mathbb{R}$.
For c), try to use the fact that if $f: X \to Y$ is a continuous function and $A \subset Y$ is closed, then $f^{-1}(A)$ is closed in $X$.
A: a) Your counterexample is a good idea, but is too general. Being closed and open are not mutually disjoint properties, even for nonempty proper subsets. For instance, in $(0,1)\cup (2,3)$ with the subspace metric, $(0,1)$ is open and closed. In many examples, this will not work. For instance, take the specific inclusion $(0,1)\subseteq \mathbb R$. This is not a closed map as $(0,1)$ is closed on itself but not in $\mathbb R$
b) This is actually false. Consider $X=Y=\mathbb R$. Let $A = \{(x,y)\in \mathbb R^2 : xy=1\}$. This is the graph of $y=1/x$, so it is closed. However, $p_1[A] = \mathbb R-\{0\}$, which is not closed in $\mathbb R$.
c) Let $A\subseteq f[X]$ be closed. Then $A= f[f^{-1}[A]]$. As $f$ is continuous, $f^{-1}[A]$ is closed in $X$. Hence, $g[A]= g[f[f^{-1}[A]]]=(g \circ f)[f^{-1}[A]]$ is closed, as $g \circ f$ is closed by assumption.
d) The same counterexample as in part a) works. A closed ball in $(0,1)$ is just a closed interval $[a,b]\subseteq (0,1)$, which is certainly still closed in $\mathbb R$. However, as discussed in a) this inclusion map is not closed.
