Pushout of open map is open I have been struggling with the following problem.
Consider the pushout for topological spaces (or adjunction space) $B \cup_A C$ obtained by gluing together $B$ and $C$ along $A$ by means of continuous maps $f$ and $g$.
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
A & \ra{f} & B\\
\da{g} & & \da{g'}\\
C & \ras{f'} & B \cup_A C\\
\end{array}
$$
show that if $f$ is open and injective then $f'$ (the pushout of $f$) is open.
I have tried to take an open set $U \subseteq C$, and show that $f'(U)$ is open, for this I have to show that $g'^{-1}(f'(U))$ is open in $B$, so I tried it to compare it somehow to $f(g^{-1}(U))$ which is an open in $B$, but have not been able to make any progress after that. 
Also I'm not sure how to use the injectivity of $f$, is it maybe for its inverse? so I can take the equality $f'(g(x)) = g'(f(x))$ and maybe manipulate it somehow to get something like $f'(g(f^-1(x)) = g'(x), x \in B$, but I'm not sure if that is helpful.
Any help would be appreciated.
 A: We have $B \cup_A C = (B \sqcup C)/\sim$, where $\sim$ is the equivalence relation generated by $f(a) \sim g(a)$, $a \in A$. Let $\pi\colon B \sqcup C \to B \cup_A C$ denote the standard projection and we aim to show that $\pi^{-1}f'(U)$ is open, that is $\pi^{-1}f'(U) \cap C$ and $\pi^{-1}f'(U) \cap B$ are open.
Now for $c \in C$ we have $c \in \pi^{-1}f'(U)$ if there is some $u \in U$ with $\pi(u) = \pi(c)$. That is $u \sim c$ and hence there are some elements $a_i \in A$ with $g(a_1) = u$, $f(a_1) = f(a_2)$, $g(a_2) = g(a_3)$, $\ldots$, $f(a_{n-1}) = f(a_n)$, $g(a_n) = c$ by definition of $\sim$. Injectivity of $f$ gives $a_1 = a_2$, $a_3 = a_4$, $\ldots$ and hence all $g(a_i)$ are equal. Therefore we need to have $c = u \in U$. That is $\pi^{-1}f'(U) \cap C = U$, which is open.
Now let $b \in B$, we have $\pi(b) \in f'(U)$ iff if there is some $u \in U$ with $\pi(u) = \pi(b)$. That is $u \sim c$ and hence there are some elements $a_i \in A$ with $g(a_1) = u$, $f(a_1) = f(a_2)$, $g(a_2) = g(a_3)$, $\ldots$, $g(a_{n-1}) = g(a_n)$, $f(a_n) = b$ by definition of $\sim$. Injectivity of $f$ gives $a_1 = a_2$, $a_3 = a_4$, $\ldots$ and hence all $g(a_i)$ are equal. That is $u = g(a_1) = g(a_n)$, $f(a_n) = b$. That is $\exists a \in A \; \mid f(a) = b, g(a) = u$, which is equivalent to $b \in fg^{-1}(U)$. So $\pi^{-1}f'(U) \cap B = fg^{-1}(U)$, which is open, as $g$ is continuous and $f$ is open.
