prove that tensor product is pushout I want to show that if $A$ and $B$ are $C$-algebras (all rings are commutative) and $\varphi:A \rightarrow A \otimes_C B$ and $\phi:B \rightarrow A \otimes_C B$ are natural maps, then $A \otimes_C B$ is the pushout. I got stuck in the following place:
For any object $W$ together with maps $\varphi':A \rightarrow W$ and $\phi':B \rightarrow W$ there exist unique map $u: A \otimes_C B \rightarrow W$ which satisfies universal property. So I need $\varphi'=u \circ \varphi$ and $\phi'=u \circ \phi$. 
For any $a \in A$ and $b \in B$, we have $$\varphi'(a)=u \circ \varphi(a)=u(a \otimes 1)$$$$\phi'(b)=u \circ \phi(b)=u(1 \otimes b)$$
From the equalities we see where $u: A\otimes_CB \rightarrow W$ maps points $a \otimes 1$ and $1 \otimes b$.
But, since $A\otimes_CB$ contains other points, where we need to map point $a \otimes b$, i.e. what is $u(a \otimes b)$?
Any hints are welcome! Thank you!
 A: You already know how to get the commutative diagram 
$$ A \longrightarrow B \ \ \ \ \  \\
\downarrow \ \ \ \ \ \ \ \  \downarrow \ \ \ \ \\
C \rightarrow B \otimes_A C, $$
so suppose we have a commutative diagram like this one with $W$ in the place of $B \otimes_A C$. Let $f_B: B \to W$ and $f_C: C \to W$ be the corresponding $k$-algebra homomorphisms. Instead of trying to construct the map directly like you're trying, let's use the universal property of the tensor product:

Suppose $G$ is an abelian group, and $\phi: B \times C \to G$ is a set-function such that $\phi(b_1+b_2,c) = \phi(b_1,c) + \phi(b_2,c)$, $\phi(b,c_1+c_2) = \phi(b,c_1) + \phi(b,c_2)$, and $\phi(ab,c) = \phi(b,ac)$ for all $b,b_1,b_2 \in B$, $c,c_1,c_2 \in C$, and $a \in A$. Then there exists a unique morphism of abelian groups $\psi: B \otimes_A C \to G$ such that $\phi = \psi \circ \otimes$.

$W$ is a $k$-algebra, so in particular an abelian group under addition. Using $f_B$ and $f_C$, we can construct $\phi$ by setting $\phi(b,c) = f_B(b)f_C(c)$. Then linearity in both arguments follows from the additivity of ring homomorphisms, and the last property follows from commutativity of the diagram.
The universal property now gives you the map $B \otimes_A C \to W$ for free! 
