Prove that $v(C)$ is closed in $X \coprod _f Y$ iff $(f(C \cap A) \cup (C \cap Y))$ closed in $Y$.

Question

I'm self studying Rotman's Algebraic topology and I've come across this problem.

I'm trying to show the implication in either direction and I can't figure out what to do next.

$\Rightarrow$

I can see that $v(C)$ being closed implies $v^{-1}v(C)$ is closed in $X \coprod Y$ and equals the set in the hint, but from here I can't figure anything further.

$\Leftarrow$

$(C \cap Y) \cup (f(C \cap A)$ closed in $Y$ implies $f^{-1}(f(C \cap A)) \cup f^{-1}(C \cap Y)$ closed in $A \Rightarrow$ closed in $X$. But from here I'm stuck.

Anyone have any ideas?

Answer 1

By the definition of the sum and quotient topologies ,$v[C]$ is closed iff $v^{-1}[v[C]]$ is closed in $X \coprod Y$ iff $v^{-1}[v[C]] \cap X$ is closed in $X$ and $v^{-1}[v[C]] \cap Y$ is closed in $Y$. Now use the hint?