Why $gcd(\alpha,26)\neq 1$ means that $ax\equiv y \pmod{26}$ is not injective? Let's consider the equivalence:
 $ax\equiv y \pmod{26}$
If $gcd(\alpha,26)\equiv 1$ I can multiply each member by $a^{-1}$ and i can obtain $x$ in function of $y$.
If $gcd(\alpha,26)\neq 1$ i can't find the modular inverse, however what assures us that the function is not injective in this situation ?
 A: If $\ 1< c\mid a,m\,$ then $\bmod m\!:\  a(\color{#c00}{m/c}) = (a/c)m\equiv 0\,$ so $\,ax\equiv 0\,$ for both $\,0\not\equiv \color{#c00}{m/c}$
$\ \ $  e.g. $\,\ 2\mid 6,26\ $ thus $\bmod 26\!:\ 6(\color{#c00}{13})\equiv 0\equiv 6(0)\,$ so $\,f(\color{#c00}{13})\equiv f(0)\,$ for $\,f(x) = 6x$
Alternatively see $\,(3)\Rightarrow(4)\Rightarrow(1)\,$ below (the contrapositive of your claim)
Theorem $\ $ The following are equivalent for integers $\rm\:a, m.$
$(1)\rm\ \ \ gcd(a,m) = 1$
$(2)\rm\ \ \ a\:$ is invertible $\rm\ \ \ \ \: (mod\ m)$
$(3)\rm\ \ \ x\,\mapsto\, ax\:$ is $\:1$-$1\:$ $\rm\,(mod\ m)$
$(4)\rm\ \ \ x\,\mapsto\, ax\:$ is onto $\rm\,(mod\ m)$ 
Proof $\  (1\Rightarrow 2)\  $ By Bezout $\rm\, gcd(a,m)\! =\! 1\Rightarrow ja\!+\!km =\! 1\,$ for $\rm\,j,k\in\Bbb Z\,$ $\rm\Rightarrow ja\equiv 1\!\pmod{\! m}$
$(2\Rightarrow 3)\ \ \ \rm ax \equiv ay\,\Rightarrow\,x\equiv y\,$ by scaling by $\rm\,a^{-1}$
$(3\Rightarrow 4)\ \  $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4\Rightarrow 1)\ \ \ \rm x\to ax\,$ onto $\,\Rightarrow\rm \exists\,j\!:\, aj\equiv 1\,$ $\rm\Rightarrow\exists\,j,k\!:\ aj\!+\!mk = 1$ $\,\Rightarrow\,\rm\gcd(a,m)\!=\!1$
See here for a conceptual proof of said Bezout identity for the gcd.
