Algebraically I do understand the usual proof of the proposition, like the on in the accepted answer: If $g \circ f$ is injective, so is $g$

But I still am missing something. When I draw the composition of $g$ and $f$, I can draw $g \circ f$ as injective while $f$ is merely surjective. What am I doing wrong here? According to the proposition this should not be possible.

In the figure, $f$ is blue $g$ is red and thus $g \circ f$ is violet. Notice how $g \circ f$ is injective (while $g$ alone is not), but $f$ is not injective.

enter image description here

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    $\begingroup$ how is $g\circ f$ injective on your picture ? the top $2$ left-most points have the same image under $g\circ f$. $\endgroup$ – Gabriel Romon Oct 31 '17 at 8:58
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    $\begingroup$ $g\circ f$ has same domain as $f$, so I don't think you really mean that $g\circ f$ is violet. The combination of a blue arrow and a violet arrow is $g\circ f$. $\endgroup$ – Gabriel Romon Oct 31 '17 at 9:03
  • $\begingroup$ But $g \circ f$ is the violet arrows. How is that not injective? $\endgroup$ – user3578468 Oct 31 '17 at 9:03
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    $\begingroup$ $g \circ f$ goes from the blue arrow to the violet arrow. $\endgroup$ – Siong Thye Goh Oct 31 '17 at 9:05
  • $\begingroup$ Oh, ok, now I see. That was my error. Thanks. $\endgroup$ – user3578468 Oct 31 '17 at 9:08

In your picture, $f$ is not surjective.

The beginning of the red arrow has no preimage, hence $f$ is not surjective but your result require $f$ to be a surjective function.

Edit: Also, as mentioned in the comment, $g\circ f$ is not injective as well.

btw, to make sure your function has a chance to be injective, make sure your codomain is at least as large as your domain. Notice that domain of $g\circ f$ is equal to the domain of $f$.

Strictly speaking, the second part is just the function $g$ and the whole picture is $g \circ f$.

  • $\begingroup$ Hm, right, I got that wrong, but then simply imagine the red arrow wasn't there. The issue remains, does it not? $f$ is still not injective then. (I did not downvote your answer btw.) $\endgroup$ – user3578468 Oct 31 '17 at 9:02
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    $\begingroup$ OP believes the violet arrows are $g\circ f$, which leads him to believe that $g\circ f$ is injective. $\endgroup$ – Gabriel Romon Oct 31 '17 at 9:05

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