Adjunction space is a pushout

Solution 1:

Yes, that's right. On the generating set of $\sim$ we have that $(a,0) \sim (f(a), 1)$ (where the disjoint union I denote with $0$ as the second coordinate for elements of $X$ and $1$ for elements of $Y$) and these yield the same result: either we map under $u$ to $\alpha_1(a) = \alpha_1(i(a))$ under the requirement that $u$ must commute with $\alpha_1$ or to $\alpha_2(f(a))$ under the requirement that $u$ must commute with $\alpha_2$, if we take the second guise. But by commutativity of the diagram with $i$,$f$,$\alpha_1$ and $\alpha_2$, these 2 give the same value in $Q$.

So on the generating set we have no conflict in the definition of $u$, so now you must show that $u$ is well-defined on all classes (which can be larger than 2 points, of course). You could use that $x \sim y$ iff there are finitely many steps via generating pairs (or their inverse, of course) from $x$ to $y$.

The unicity is clear, like I said, from having to commute with $\alpha_1$ and $\alpha_2$ $u$ has to be defined like this; it just remains to verify it is actually well-defined.

[edit] Akhil's remark in the comments above merits mention too, as I forgot: we defined $u$ on the disjoint sum originally, showed we had no conflict with the equivalence relation $\sim$, so that we have a well-defined map from the quotient. As the map from the sum is continuous (it is iff both "summands" $\alpha_1$ and $\alpha_2$ are) the standard theorem on quotient spaces (sometimes called the universal theorem for quotient maps) implies that $u$ is indeed continuous from the quotient to $Q$.