Recently, I have encountered a question:
Prove that the map $f_Y:Y\rightarrow Y\coprod_XZ$ defined by $f(y)=[y]$ is continuous.
However, when I tried to prove it using the open sets definition of continuity, it seems really complicated as I need to write down the set definition and I failed to do so. Is there any way to prove this more effectively?
(Off-topic: What are the best way to study topology? It seems drawing graphs is pretty useful but sometimes it is hard to draw a picture for concepts like continuity. Do we just hard-memorise the definitions?)
As Randall comments, $f_Y = \pi \circ j_Y$, where $j_Y : Y \to Y \coprod Z$ is the canonical injection (which is trivially continuous) and $\pi : Y \coprod Z \to Y \coprod_X Z$ is the quotient map (which is continuous by definition).
It seems that you know the concept of a pushout, and this construction assigns to a pair of continuous maps $\phi_Y: X \to Y$ and $\phi_Z: X \to Z$ a commutative diagram of spaces and continuous maps $\require{AMScd}$ \begin{CD} X @>{\phi_Y}>> Y \\ @V{\phi_Z}VV @V{f_Y}VV \\ Z @>{f_Z}>> Y \coprod_X Z\end{CD}
with a certain universal property. Thus $f_Y$ is continuous by the concept of the pushout.
A standard explicit construction is $$Y \coprod_X Z = (Y \coprod Z)/\phi_Y(x) \sim \phi_Z(x),\\ f_Y(y) =[y], f_Z(z) =[z] .$$ Perhaps your question concerns this construction and asks why $f_ Y, f_Z$ are continuous?